text:
# Two-photon photoassociation spectroscopy of CsYb: Ground-state interaction potential and interspecies scattering lengths

@article{Guttridge2018TwophotonPS,
title={Two-photon photoassociation spectroscopy of CsYb: Ground-state interaction potential and interspecies scattering lengths},
author={Alexander Guttridge and Matthew D. Frye and Baochun Yang and Jeremy M. Hutson and Simon L. Cornish},
journal={Physical Review A},
year={2018}
}
• Published 1 June 2018
• Physics
• Physical Review A
We perform two-photon photoassociation spectroscopy of the heteronuclear CsYb molecule to measure the binding energies of near-threshold vibrational levels of the $X{\phantom{\rule{3.33333pt}{0ex}}}^{2}{\mathrm{\ensuremath{\Sigma}}}_{1/2}^{+}$ molecular ground state. We report results for $^{133}\mathrm{Cs}^{170}\mathrm{Yb}$, $^{133}\mathrm{Cs}^{173}\mathrm{Yb}$, and $^{133}\mathrm{Cs}^{174}\mathrm{Yb}$, in each case determining the energy of several vibrational levels including the least-bound…

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### Ground-state properties of Ca 2 from narrow-line two-color photoassociation

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### Two-color photoassociation spectroscopy of ytterbium atoms and the precise determinations of s -wave scattering lengths

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• 2008
By performing high-resolution two-color photoassociation spectroscopy, we have successfully determined the binding energies of several of the last bound states of the homonuclear dimers of six

### High Resolution Molecular Spectroscopy for Producing Ultracold Absolute Ground-State $^{23}$Na$^{87}$Rb Molecules

• Physics, Chemistry
• 2017
We report a detailed molecular spectroscopy study on the lowest excited electronic states of $^{23}\rm{Na}^{87}\rm{Rb}$ for producing ultracold $^{23}\rm{Na}^{87}\rm{Rb}$ molecules in the electronic,

### Tunable dual-species Bose-Einstein condensates ofK39andRb87

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• 2015
We present the production of dual-species Bose-Einstein condensates of $^{39}\mathrm{K}$ and $^{87}\mathrm{Rb}$. Preparation of both species in the $\left| F=1,m_F=-1 \right\rangle$ state enabled us

### Two-photon photoassociation spectroscopy of an ultracold heteronuclear molecule

• Physics, Chemistry
• 2017
We report on two-photon photoassociation (PA) spectroscopy of ultracold heteronuclear LiRb molecules. This is used to determine the binding energies of the loosely bound levels of the electronic

### Creation of an Ultracold Gas of Ground-State Dipolar ^{23}Na^{87}Rb Molecules.

• Physics, Chemistry
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• 2016
The results pave the way toward investigation of ultracold molecular collisions in a fully controlled manner and possibly to quantum gases of ultrACold bosonic molecules with strong dipolar interactions.

### Effects of Configuration Interaction on Intensities and Phase Shifts

The interference of a discrete autoionized state with a continuum gives rise to characteristically asymmetric peaks in excitation spectra. The earlier qualitative interpretation of this phenomenon is

### Quantum degenerate mixtures of strontium and rubidium atoms

• Physics
• 2013
We report on the realization of quantum degenerate gas mixtures of the alkaline-earth-metal element strontium with the alkali-metal element rubidium. A key ingredient of our scheme is sympathetic

### Two-photon photoassociative spectroscopy of ultracold 88-Sr

• Physics
• 2008
We present results from two-photon photoassociative spectroscopy of the least-bound vibrational level of the X$^1\Sigma_g^+$ state of the $^{88}$Sr$_2$ dimer. Measurement of the binding energy allows

### Two-Color Photoassociation Spectroscopy of Ground State Rb2

• Physics
• 1997
We determine the energies of twelve vibrational levels lying within 20GHz of the lowest dissociation limit of {sup 85}Rb{sub 2} with two-color photoassociation spectroscopy of ultracold {sup 85}Rb

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text:
Home > Micro concepts, Perfect Competition > The price taking firm

The price taking firm

August 13, 2011

In a perfectly competitive market, the firm is a price-taker, it cannot influence the market price through the quantity it produces. In practice this means the firm is so small in proportion to the overall market that it has no market power, so it can sell any quantity it is able to produce at the market price.

The overall market price is determined by the market supply (provided by all the firms in the market) and the market demand (demanded by all the consumers in the market). The firm can sell any quantity it can produce at the market price, so even though the market is likely to face a downward sloping market demand curve, the individual firm faces a horizontal market demand curve at the market price. Because the firm can sell as much as it can produce at the market price, the marginal revenue for each unit sold is equal to the price, and the average revenue is also equal to the price, as every unit costs the same, so when you divide the total revenue by the number of units sold, you get the price.

The price-taking firm will observe two rules:

Marginal output rule – the firm will produce at an output where the price is equal to the marginal cost of production (MR = MC, and here P = MR so P = MC)

Shutdown rule – the firm will shut down if the average revenue is lower than the average cost at all output levels, so as the price equals average revenue, it will shut down if the price is lower than the average total cost at all levels.

We can look at this in terms of graphs:

This is the market supply and market demand. The equilibrium market price is P1.

This is the firm’s supply and demand graph. The firm’s supply curve is the marginal cost of production, and it faces a horizontal demand curve at the market price of P1. So it produces quantity q1, the point where price equals marginal cost. Here the firm is able to make some profit $\pi$ because at point q1 the average revenue (the price) is greater than the average total cost. This is a short run situation, because when other people see that there are profits to be made in the industry, it will attract the entry of new firms to the market.

The arrival of the new firms expands the market supply to S2, which drives down the equilibrium market price to P2.

At P2, the price has been driven to the level where the firm produces at point q2, where the price is equal to the marginal cost and the average total cost. At this point because the average revenue (price) is equal to the average cost, there is zero profit. So now you reach an equilibrium point. No new firms will enter the industry as there are no profits to be made, firms are just breaking even. But no firms will leave the industry as they are not making losses and are not in the shutdown position. Remember that perfect competition assumes that the firms are identical and face identical cost functions.

So firms in a perfectly competitive market can make profits in the short run, but will make zero profit in the long run.


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# 20ème conférence Claude Itzykson - Random Surfaces and Random Geometry

Europe/Paris
Description

The Itzykson Conference is held every year at the Institut de Physique Théorique in Saclay to honour the memory of Claude Itzykson. The 20th edition of this conference, which took place on June 10-12 2015, was devoted to random surfaces and random geometry.

A first theme was the study of random maps, or discretized surfaces, which has recently seen major advances using tools from algebra and geometry (topological recursion, discrete integrability), combinatorics (unified bijections) and probability theory (rigorous construction of the Brownian map, the scaling limit of random maps, and of their local limits).

A second theme was the study of Liouville field theory, a continuous approach to quantum gravity, string theory and random surfaces initiated by Polyakov in 1981, whose mathematical understanding was recently sharpened. Unifying this approach with the discrete one via random maps is a hot topic of research.

A third theme concerned random geometries in higher dimension, in the context of quantum gravity (tensor models, random triangulations, etc) and string theories, which forms another very active area.

Finally, questions concerning random surfaces and random geometries in condensed matter, soft matter, biophysics (interfaces and membranes) and social sciences (complex networks) were also tackled.

Organizing committee: Jérémie Bouttier, François David, Bertrand Eynard and Laure Sauboy (secretary).
Sponsors and benefactors: IPhT (CEA and CNRS)DSM, LabEx LMH, LabEx PALM, ANR Cartaplus, ANR Graal, Fondation Daniel Iagolnitzer, Michael and Victoria Bershadsky

Participants
• Alessandro Codello
• Angelo raffaele Fazio
• Anne Itzykson
• Anshuman Pal
• Asaf Nachmias
• Bernard Julia
• Bertrand Duplantier
• Bertrand Eynard
• Bianca Dittrich
• Calin Lazaroiu
• Cecile Monthus
• Chenlin GU
• Christian H. Schmidt
• Cédric Boutillier
• Dario Benedetti
• David García
• David Turton
• Denis BERNARD
• Des Johnston
• Dominique Poulalhon
• Elena Mirela Babalic
• Eleni Katifori
• Emmanuel GUITTER
• Eric Vernier
• Florent Chevallier
• François David
• Gernot Akemann
• Ghislaine Itzykson
• Gilles Schaeffer
• Giulia CARRA
• Grégoire Misguich
• Guillaume LE TREUT
• Henk Hilhorst
• Henri Orland
• Iakov Kononov
• Iosif Bena
• Ivan Dornic
• Jean-Bernard ZUBER
• Jean-François LE GALL
• Jean-Michel Drouffe
• Jiaowen Yang
• Jingzhou Sun
• Johan Blåbäck
• Jonas Kahn
• Joonas Turunen
• Jérémie Bouttier
• Karen Vásquez
• Kay Wiese
• Konstantin Izyurov
• Laure Dumaz
• Laurent Menard
• Leonid Chekhov
• Linxiao Chen
• Loïc Richier
• Luca Lionni
• Maciej Dołęga
• Marc Barthelemy
• Marie Albenque
• Mark Bowick
• Michael Kozdron
• Michel Bauer
• Michel Lapidus
• Mireille Bousquet-Mélou
• Nicolas Curien
• Noam Zeilberger
• Olivier Babelon
• Olivier Golinelli
• Philippe Di Francesco
• Pierre Vanhove
• Raphael Itzykson
• Remi Cocou Avohou
• Renate Loll
• Richard Kenyon
• Rinat Kedem
• Ruben Minasian
• Rémi Rhodes
• Shigenori Seki
• Steffen Rohde
• Stéphane Dartois
• Séverin Charbonnier
• Thibault Delepouve
• Thordur Jonsson
• Timothy Budd
• Valentin Bonzom
• Vincent Pasquier
• Vincent Rivasseau
• Wenjie Fang
• Zoryana Usatenko
• Édouard Brézin
• Édouard Maurel-Segala
• Wednesday, 10 June
• 09:30 09:45
Registration 15m
• 09:45 10:15
Welcome 30m
Speakers: Anne Itzykson, Michel Bauer (IPhT), Raphaël Itzykson
• 10:15 11:00
Counting doodles 45m
Speaker: Jean-Bernard Zuber (LPTHE, Université Pierre et Marie Curie)
• 11:00 11:30
Coffee break 30m
• 11:30 12:15
The Potts model on planar maps 45m
(joint work with Olivier Bernardi, Brandeis University) Let q be an integer. We address the enumeration of q-colored planar maps (planar graphs embedded in the sphere), counted by the total number of edges and the number of monochromatic edges (those that have the same colour at both ends). In physics terms, we are averaging the partition function of the Potts model over all maps of a given size. We prove that the associated generating function is algebraic when q is of the form 2 + 2 cos(jπ/m), for integers j and m (but distinct from 0 and 4). This includes the two integer values q = 2 and q = 3, for which we give explicit algebraic equations. For a generic value of q, we prove that the generating function satisfies an explicit system of differential equations. Both results hold as well for planar triangulations, with a strikingly similar system of differential equations. The starting point of our approach is a recursive construction of q-coloured maps, in the spirit of what W. Tutte did in the seventies and eighties for properly coloured triangulations. This model has also been addressed by other authors and other methods (Bonnet & Eynard in 1999, and more recently Guionnet, Jones, Shlyakhtenko & Zinn-Justin, and Borot, Bouttier & Guitter), but our results are of a different nature and seem more explicit.
Speaker: Mireille Bousquet-Mélou (CNRS, LaBRI, Université de Bordeaux)
• 12:15 13:00
Exponential bounds on the number of 3d causal triangulations 45m
We discuss the problem of bounding the number of distinct triangulations of the 3-dimensional sphere. We prove that the number of causal triangulations of the 3-sphere is bounded by an exponential function of the number of tetrahedra. We describe how the argument might be extended to 4 dimensions.
Speaker: Thordur Jonsson (University of Iceland)
• 13:00 14:30
Lunch 1h 30m
• 14:30 15:15
Extracting Hidden Hierarchies in Weighted Distribution Networks 45m
Natural and man-made transport webs are frequently dominated by dense sets of nested cycles. The architecture of these networks, as defined by the topology and edge weights, determines how efficiently the networks perform their function. Yet, the set of tools that can characterize such a weighted cycle-rich architecture in a physically relevant, mathematically compact way is sparse. In order to fill this void, we have developed an algorithm that hierarchically decomposes (filters) the graph and characterizes the graph based on the decomposition process. The algorithm starts by identifying a physically meaningful tiling of the network, proceeds to sequentially remove the weakest links as determined by the edge weight, merge neighboring plaquettes and and finally produces a tree characterizing this merging process. The properties of this characteristic tree can provide the physical and topological data required to describe the architecture of the network and to build physical models. We show how this new algorithm can be used for automated phenotypic characterization of 2D and 3D weighted networks the structure of which is dominated by cycles, such as the vasculature of leaves and complex transportation webs.
Speaker: Eleni Katifori (University of Pennsylvania and Max Planck Institute for Dynamics and Self-Organization)
• 15:15 16:00
Pseudoknots and Knots in RNA 45m
We present some algorithms for the prediction of RNA pseudo knots based on their topological classification. In addition, we present an analysis of knots in RNA and show that unlike DNA or proteins where they are abundant, they are probably non-existent in RNA.
Speaker: Henri Orland (IPhT)
• 16:00 16:30
Tea break 30m
• 16:30 17:15
Conformal representations of Random Maps and Surfaces 45m
Whereas the combinatorial and metric structure of random maps are somewhat understood, a mathematical understanding of their conformal structure is still missing. Motivated by the quest to find a "conformal map" from the Riemann sphere to a Liouville quantum-gravity sphere, I will talk about uniformization of discrete maps (based on joint work with Don Marshall), and discuss an analog of Mario Bonk's carpet uniformization in the setting of the Conformal Loop Ensemble CLE (based on joint work with Brent Werness).
Speaker: Steffen Rohde (University of Washington)
• 17:15 18:00
Generalized & Quantum SLE Multifractality 45m
We describe some recent advances in the multifractality of the Schramm-Loewner evolution. This includes a generalized notion of integral means spectrum for unbounded whole-plane SLE, depending on two moments, with a phase transition between a new spectrum and the usual spectrum, originally obtained via quantum gravity. Conversely, we show that any such multifractal spectrum in the plane has a universal analogue in Liouville quantum gravity, with specific applications to SLE. Based on joint works with H. Ho, B. Le & M. Zinsmeister ; and G. Borot & J. Miller.
Speaker: Bertrand Duplantier (IPhT)
• Thursday, 11 June
• 09:30 10:15
Bijective proof of Hurwitz formula 45m
In 1891, Hurwitz gives a formula for the number of some branched coverings of the sphere by itself, corresponding to transitive $$m$$-tuples of transpositions such that their product has a prescribed cycle type. I will present a combinatorial proof of this quite simple formula, reminiscent of combinatorial constructions for planar maps. This proof explains in particular why, as formulas for planar maps involve numbers of plane trees (aka Catalan trees), Hurwitz formula involves the number of Cayley trees, a fact that the many previous proofs of the formula do not explain. Moreover, the construction can be extended to a more general setting, leading to results about double Hurwitz numbers of any genus. This is a joint work with Enrica Duchi and Gilles Schaeffer.
Speaker: Dominique Poulalhon (LIAFA, Université Paris Diderot)
• 10:15 11:00
A bivariate two-point function for planar bicolored maps 45m
I will show how to compute the distance-dependent two-point function of vertex-bicolored planar maps with, in addition to the usual control on the faces degrees, a separate control on the numbers of vertices of both colors. This bivariate two-point function is obtained via a technique of slice decomposition and by use of the Stieltjes-type continued fraction formalism. If time allows, I will also discuss another bivariate two-point function in relation with another type of continued fractions. This is joint work with Éric Fusy.
Speaker: Emmanuel Guitter (IPhT)
• 11:00 11:30
Coffee break 30m
• 11:30 12:15
Graphene Statistical Mechanics 45m
Graphene provides an ideal system to test the statistical mechanics of thermally fluctuating elastic membranes. The high Young’s modulus of graphene means that thermal fluctuations over even small length scales significantly stiffen the renormalized bending rigidity. We study the effect of thermal fluctuations on graphene ribbons of width W and length L, pinned at one end, via coarse-grained Molecular Dynamics simulations and compare with analytic predictions of the scaling of width-averaged root-mean-squared height fluctuations as a function of distance along the ribbon. Scaling collapse as a function of W and L also allows us to extract the scaling exponent eta governing the long-wavelength stiffening of the bending rigidity. A full understanding of the geometry-dependent mechanical properties of graphene, including arrays of cuts, may allow the design of a variety of modular elements with desired mechanical properties starting from pure graphene alone.
Speaker: Mark Bowick (Syracuse University)
• 12:15 13:00
Scaling laws for large deviations in Voronoi tessellations 45m
In 1984 Drouffe and Itzykson asked about the probability that a randomly picked two-dimensional Poisson-Voronoi cell have exactly n edges. I will discuss the answer to this question, obtained in 2005, and the developments that have taken place since, including recent results.
Speaker: Henk Hilhorst (LPT, Université Paris-Sud)
• 13:00 14:30
Lunch 1h 30m
• 14:30 15:15
Quantum space time engineering 45m
We will consider lattice gravity approaches such as loop quantum gravity and spin foams, in which quantum geometry is defined via the assignment of quantum geometric variables to a (fixed) lattice. The main problem is then to construct the refinement limit and in this way to loose the dependence on the choice of lattice. I will discuss a renormalization framework in which such a refinement limit can be constructed and present first results for a range of (toy) theories.
Speaker: Bianca Dittrich (Perimeter Institute)
• 15:15 16:00
Random Tensors 45m
Random tensors generalize random matrices and group field theory.Their Feynman perturbative expansion sums over all manifolds and a restricted class of quasi-manifolds, hence may be used to probe random geometries in higher dimensions. We shall review the 1/N expansion of random tensors. It is indexed by a new parameter (Gurau's degree) which is not a topological invariant of the underlying manifold, hence suggests new questions in enumerative combinatorics. Finally we shall discuss briefly the associated class of non-local tensor quantum field theories which generalize non-commutative field theories.
Speaker: Vincent Rivasseau (LPT, Université Paris-Sud)
• 16:00 16:30
Tea break 30m
• 16:30 17:15
Fixed-energy harmonic functions and acyclic orientations 45m
Speaker: Richard Kenyon (Brown University)
• 17:15 18:00
Hurwitz numbers and matrix models 45m
Hurwitz numbers enumerate combinatorial classes of mapping of genus g Riemann surfaces on the complex projective line with branchings at a fixed number of points (at three points for the case of Belyi pairs and Grothendieck's dessins d'enfant and at n points for hypergeometric Hurwitz numbers). The first variant of a matrix-model description of such mappings was proposed by Itzykson and Di Francesco in 1993. We construct the matrix model describing the general situation of n branching points with ramification data fixed at two of them. All these models are tau functions of the KP hierarchy and upon some constraints on their generating functions their solutions can be attained using the topological recursion technique for special chains of Hermitian matrices. The corresponding systems are conjecturally related to TQFTs. (Based on joint papers with Jan Ambjørn, NBI, Copenhagen)
Speaker: Leonid Chekhov (Steklov Mathematical Institute)
• 20:00 22:00
Conference dinner, Restaurant "Bœuf sur le toit" 2h

Restaurant "Bœuf sur le toit"
34 rue du Colisée
75008 Paris
http://www.boeufsurletoit.com/en/

• Friday, 12 June
• 09:30 10:15
Claude, the early days 45m
Speaker: Edouard Brézin (Ecole normale supérieure)
• 10:15 11:00
Liouville quantum gravity on Riemann surfaces 45m
I will present a generic way to construct rigorously Liouville quantum field theory on Riemann surfaces with emphasis on the case of the Riemann sphere. The construction is based on Polyakov’s functional integral and yield non trivial conformal field theories. Then I will explain its main properties, the relation with the uniformization theorem for 2d Riemann surfaces and relate it via precise conjectures to the scaling limit of random planar maps conformally embedded onto the Riemann sphere. Based on joint works with F. David, Y. Huang, A. Kupiainen, H. Lacoin, V. Vargas.
Speaker: Rémi Rhodes (Université Paris-Est)
• 11:00 11:30
Coffee break 30m
• 11:30 12:15
Causal Dynamical Triangulations in 4D - the plot thickens 45m
Causal Dynamical Triangulations (CDT) is a framework for defining a nonperturbative path integral for quantum gravity, based on random geometries with a built-in space-time anisotropy, related to the presence of a local causal (Lorentzian) structure. In four dimensions, several nontrivial results highlight that CDT is a serious contender for THE theory of quantum gravity, including the presence of second-order phase transitions and the demonstration that a renormalization group analysis can be performed despite the absence of a fixed background. I will report on recent results.
Speaker: Renate Loll (Radboud University, Nijmegen)
• 12:15 13:00
Some practical problems about planar graphs: time evolution and typology. 45m
Planar graphs pervade many aspects of science: they are the subject of numerous studies in graph theory, in combinatorics, in quantum gravity, and in biology and botanics. Planar networks are also extensively used to represent various infrastructure networks. In particular, transportation networks and streets patterns are the subject of many studies that are trying to characterize both topological (degree distribution, clustering, etc.) and geometrical (angles, segment length, face area distribution, etc.) aspects of these networks. I will illustrate in this talk some of the problems that are encountered in these studies such as characterizing the structure of simplest paths, how to describe the time evolution of road networks and the possibility of a typology of street patterns.
Speaker: Marc Barthelemy (IPhT)
• 13:00 14:30
Lunch 1h 30m
• 14:30 15:15
The field theory of avalanches 45m
When elastic systems like contact lines on a rough substrate, domain walls in disordered magnets, or tectonic plates are driven slowly, they remain immobile most of the time, before responding with strong intermittent motion, termed avalanche. I will describe the field theory behind these phenomena, explain why its effective action has a cusp, and how such intricate objects as the temporal shape of an avalanche can be obtained.
Speaker: Kay Wiese (Ecole normale supérieure)
• 15:15 16:00
The circle packing of random hyperbolic triangulations 45m
We study random hyperbolic planar triangulations via their circle packing embedding in order to connect their geometry to that of the hyperbolic plane. This leads to several results: Identification of the Poisson and geometric boundaries, a connection between hyperbolicity and a form of non-amenability, and a new proof of the Benjamini-Schramm recurrence result. Based on works with subsets of Omer Angel, Martin Barlow, Ori Gurel-Gurevich, Tom Hutchcroft and Gourab Ray.
Speaker: Asaf Nachmias (University of British Columbia and Tel Aviv University)
• 16:00 16:30
Tea break 30m
• 16:30 17:15
Peeling of infinite Boltzmann planar maps 45m
For a long time it has been known that distances in random surfaces can be conveniently studied by considering associated peeling processes. Inspired by recent results by Curien and Le Gall in the case of random triangulations, I will give a very simple description of a particular peeling process, and its scaling limit, in the general setting of infinite Boltzmann planar maps (IBPM), where one includes independent Boltzmann weights on the faces of different degrees. I will show how using this description one can (at least heuristically) derive many explicit asymptotic relations between various quantities associated to the IBPM, including graph distance, dual graph distance, first-passage time and hop count.
Speaker: Timothy Budd (Niels Bohr Institute)
• 17:15 18:00
Random planar geometry 45m
We will survey recent results showing that the random metric space called the Brownian map appears as the continuous limit of various classes of large discrete random graphs embedded in the plane. These results indicate that the Brownian map is a universal model of random geometry in two dimensions, which has fractal dimension four although it has the topology of the sphere. If time permits, we will also discuss a recent work with Nicolas Curien, which shows that the Brownian map still appears when one considers local perturbations of the graph distance, for instance when one assigns random lengths to the edges of the graphs.
Speaker: Jean-François Le Gall (Université Paris-Sud et Institut universitaire de France)


text:
Question: Strange problem in example check of package
0
3 months ago by
Kyle Johnsen20
United States/Brigham Young University
Kyle Johnsen20 wrote:

I'm not sure what's going on. The examples in my documentation seem to run okay, then I get a weird error at the end apparently outside of my code. I thought something about the example itself could have been messing things up, but as you can see, the error persists even with a commented-out example. I have previously tried commenting out the only part of my examples that ever use graphics, and that didn't fix the problem either. Does anyone have any idea what the problem could be?

Here's the Travis log with the problem: https://travis-ci.org/kjohnsen/MMAPPR2/builds/437132158

Running examples in ‘MMAPPR2-Ex.R’ failed
The error most likely occurred in:
> ### Name: prePeak
> ### Title: Identify chromosomes containing peaks
> ### Aliases: prePeak
>
> ### ** Examples
>
> # postPrePeakMD <- prePeak(postLoessMD)
>
>
>
> ### * <FOOTER>
> ###
> cleanEx()
> options(digits = 7L)
> base::cat("Time elapsed: ", proc.time() - base::get("ptime", pos = 'CheckExEnv'),"\n")
Time elapsed:  422.32 24.256 474.836 0.008 0.128
> grDevices::dev.off()
Error in grDevices::dev.off() :
cannot shut down device 1 (the null device)
modified 3 months ago • written 3 months ago by Kyle Johnsen20
Answer: Strange problem in example check of package
1
3 months ago by
Kyle Johnsen20
United States/Brigham Young University
Kyle Johnsen20 wrote:

After some investigation, I discovered that my examples that interacted with graphics did cause the problem. I edited those parts out of the examples (I missed one place when I tried this the first time) and it worked. I guess this is a problem with R itself--my function calls graphics.off() to clean up after itself after writing PDFs and such, and I suppose R, ignorant of the fact, thought it still needed to call dev.off().

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# X-shooter spectroscopy of young stellar objects in Lupus. Accretion properties of class II and transitional objects

@article{Alcal2016XshooterSO,
title={X-shooter spectroscopy of young stellar objects in Lupus. Accretion properties of class II and transitional objects},
author={J. Alcal{\'a} and C. Manara and A. Natta and A. Frasca and L. Testi and B. Nisini and B. Stelzer and J. G. Williams and S. Antoniucci and K. Biazzo and E. Covino and M. Esposito and Fedor I. Getman and E. Rigliaco},
journal={Astronomy and Astrophysics},
year={2016},
volume={600}
}
• J. Alcalá, +11 authors E. Rigliaco
• Published 2016
• Physics
• Astronomy and Astrophysics
• The mass accretion rate, Ṁ acc , is a key quantity for the understanding of the physical processes governing the evolution of accretion discs around young low-mass ( M ⋆  ≲ 2.0 M ⊙ ) stars and substellar objects (YSOs). We present here the results of a study of the stellar and accretion properties of the (almost) complete sample of class II and transitional YSOs in the Lupus I, II, III and IV clouds, based on spectroscopic data acquired with the VLT/X-shooter spectrograph. Our study combines… CONTINUE READING
112 Citations

#### References

SHOWING 1-10 OF 95 REFERENCES


text:
• ### Design and expected performance of the MICE demonstration of ionization cooling(1701.06403)

Jan. 27, 2017 physics.acc-ph
Muon beams of low emittance provide the basis for the intense, well-characterised neutrino beams necessary to elucidate the physics of flavour at a neutrino factory and to provide lepton-antilepton collisions at energies of up to several TeV at a muon collider. The international Muon Ionization Cooling Experiment (MICE) aims to demonstrate ionization cooling, the technique by which it is proposed to reduce the phase-space volume occupied by the muon beam at such facilities. In an ionization-cooling channel, the muon beam passes through a material in which it loses energy. The energy lost is then replaced using RF cavities. The combined effect of energy loss and re-acceleration is to reduce the transverse emittance of the beam (transverse cooling). A major revision of the scope of the project was carried out over the summer of 2014. The revised experiment can deliver a demonstration of ionization cooling. The design of the cooling demonstration experiment will be described together with its predicted cooling performance.
• ### Pion contamination in the MICE muon beam(1511.00556)

Feb. 10, 2016 hep-ex, physics.ins-det
The international Muon Ionization Cooling Experiment (MICE) will perform a systematic investigation of ionization cooling with muon beams of momentum between 140 and 240\,MeV/c at the Rutherford Appleton Laboratory ISIS facility. The measurement of ionization cooling in MICE relies on the selection of a pure sample of muons that traverse the experiment. To make this selection, the MICE Muon Beam is designed to deliver a beam of muons with less than $\sim$1\% contamination. To make the final muon selection, MICE employs a particle-identification (PID) system upstream and downstream of the cooling cell. The PID system includes time-of-flight hodoscopes, threshold-Cherenkov counters and calorimetry. The upper limit for the pion contamination measured in this paper is $f_\pi < 1.4\%$ at 90\% C.L., including systematic uncertainties. Therefore, the MICE Muon Beam is able to meet the stringent pion-contamination requirements of the study of ionization cooling.
• ### Electron-Muon Ranger: performance in the MICE Muon Beam(1510.08306)

Nov. 3, 2015 hep-ex, physics.ins-det
The Muon Ionization Cooling Experiment (MICE) will perform a detailed study of ionization cooling to evaluate the feasibility of the technique. To carry out this program, MICE requires an efficient particle-identification (PID) system to identify muons. The Electron-Muon Ranger (EMR) is a fully-active tracking-calorimeter that forms part of the PID system and tags muons that traverse the cooling channel without decaying. The detector is capable of identifying electrons with an efficiency of 98.6%, providing a purity for the MICE beam that exceeds 99.8%. The EMR also proved to be a powerful tool for the reconstruction of muon momenta in the range 100-280 MeV/$c$.
• A novel single-particle technique to measure emittance has been developed and used to characterise seventeen different muon beams for the Muon Ionisation Cooling Experiment (MICE). The muon beams, whose mean momenta vary from 171 to 281 MeV/c, have emittances of approximately 1.5--2.3 \pi mm-rad horizontally and 0.6--1.0 \pi mm-rad vertically, a horizontal dispersion of 90--190 mm and momentum spreads of about 25 MeV/c. There is reasonable agreement between the measured parameters of the beams and the results of simulations. The beams are found to meet the requirements of MICE.
• ### Use of the European Data Grid software in the framework of the BaBar distributed computing model(physics/0306082)

June 10, 2003 physics.comp-ph
We present an evaluation of the European Data Grid software in the framework of the BaBar experiment. Two kinds of applications have been considered: first, a typical data analysis on real data producing physics n-tuples, and second, a distributed Monte-Carlo production on a computational grid. Both applications will be crucial in a near future in order to make an optimal use of the distributed computing resources available throughout the collaboration.


text:
What’s new¶

v0.6.2 (August 2022)¶

ECG analysis - PR 68

1. Added the yasa.hypno_find_periods() function to find sequences of consecutive values in hypnogram that are longer than a certain duration. This is a flexible function that can be used to detect NREM/REM periods.

2. Added the yasa.hrv_stage() function, which calculates heart rate (HR) and heart rate variability (HRV) by stage and periods.

3. Added a new dataset containing 8 hours of ECG data. The dataset is in compressed NumPy format and can be found in notebooks/data_ECG_8hrs_200Hz.npz. The dataset also includes an upsampled hypnogram.

4. Added a new Jupyter notebook to calculate EEG-HRV overnight coupling based on the yasa.hrv_stage() function. https://github.com/raphaelvallat/yasa/blob/master/notebooks/16_EEG-HRV_coupling.ipynb

Spindles & slow-waves detection - PR 71

1. Added the yasa.compare_detection() function to determine the correctness of detected events against ground-truth events. It calculates the true positive, false positives and false negatives, and from those, the precision, recall and F1-scores. The input should be the indices of the onset of the event, in samples. It includes a max_distance argument which specifies the tolerance window (in number of samples) for two events to be considered the same.

2. Added the yasa.SpindlesResults.compare_detection() and yasa.SWResults.compare_detection() method. This is a powerful and flexible function that allows to calculate the performance of the current detection against a) another detection or b) ground-truth annotations. For example, we can compare the output of the spindles detection with different thresholds.

3. Added the yasa.SpindlesResults.compare_channels() and yasa.SWResults.compare_channels() methods to compare the overlap of the detected events between channels. Agreement is calculated using the F1-score (default), precision or recall.

4. Add vmin and vmax parameters to yasa.plot_spectrogram(). PR 75

5. Better handling of flat data in yasa.spindles_detect(). The function previously returned a division by zero error if part of the data was flat. See issue 85

6. When using an MNE.Raw object, conversion of the data from Volts to micro-Volts is now performed within MNE. PR 70

7. Use black code formatting.

Others

1. When using an MNE.Raw object, conversion of the data from Volts to micro-Volts is now performed within MNE. PR 70

2. Added SleepECG to the dependencies. SleepECG is used for the heartbeats detection in yasa.hrv_stage().

3. YASA now requires MNE>0.23

v0.6.1 (March 2022)¶

This release fixes a CRITICAL BUG with the spindles detection. Specifically, the yasa.spindles_detect() could return different results depending on the sampling rate of the data. For example, downsampling the data from 256 Hz to 128 Hz may have significantly reduced the number of detected spindles. As explained in issue 54, this bug was caused by a floating-point error in numpy.convolve() when calculating the soft spindle threshold. Tests seem to indicate that only certain sampling frequencies were impacted, such as 200 Hz, 256 Hz or 400 Hz. Other sampling frequencies such as 100 Hz and 500 Hz were seemingly not affected by this bug. Please double-check any results obtained with yasa.spindles_detect()!

Warning

We recommend all users to upgrade to this new version ASAP and check any results obtained with the yasa.spindles_detect() function!

v0.6.0 (February 2022)¶

This is a MAJOR release with several API-breaking changes, new functions, bugfixes and a new section in the documentation.

Documentation

• Added a Quickstart section to illustrate the main functions of YASA. Make sure to check it out!

Plotting

Slow oscillations—sigma coupling

IMPORTANT — The default behavior of coupling=True in yasa.sw_detect() has been changed:

• YASA now uses a ± 1 second window around the negative peak of the slow-waves (2 sec total) to calculate the coupling, instead of a ± 2 sec window. Overall, this tends to increase the ndPAC values because of the higher temporal specificity. To keep a 4-sec window, use coupling_params['time'] = 2.

• We’ve enabled the statistical thresholding in the ndPAC calculation. Practically, this means that events with a weak/unreliable coupling are assigned an ndPAC value of zero. Statistical thresholding can be disabled with coupling_params['p'] = None.

Warning

Because of these changes, the coupling values are therefore not comparable with previous versions of YASA. Please make sure to re-run your analyses with the new default parameters.

Events detection

Others

v0.5.1 (August 2021)¶

This is a bugfix release. The latest pre-trained classifiers for yasa.SleepStaging were accidentally missing from the previous release. They have now been included in this release.

v0.5.0 (August 2021)¶

This is a major release with an important bugfix for the slow-waves detection as well as API-breaking changes in the automatic sleep staging module. We recommend all users to upgrade to this version with pip install –upgrade yasa.

Slow-waves detection

We have fixed a critical bug in yasa.sw_detect() in which the detection could keep slow-waves with invalid duration (e.g. several tens of seconds). We have now added extra safety checks to make sure that the total duration of the slow-waves does not exceed the maximum duration allowed by the dur_neg and dur_pos parameters (default = 2.5 seconds).

Warning

Please make sure to double-check any results obtained with yasa.sw_detect().

Sleep staging

Recently, we have published a preprint article describing YASA’s sleep staging algorithm and its validation across hundreds of polysomnography recordings. In July 2021, we have received comments from three reviewers, which have led us to implement several changes to the sleep staging algorithm. The most significant change is that the time lengths of the rolling windows have been updated from 5.5 minutes centered / 5 minutes past to 7.5 minutes centered / 2 min past, leading to slight improvements in accuracy. Furthermore, we have also updated the training database and the parameters of the LightGBM classifier. Unfortunately, these changes mean that the new version of the algorithm is no longer compatible with the previous version (0.4.0 or 0.4.1). Therefore, if you’re running a longitudinal study with YASA’s sleep staging, we either recommend to keep the previous version of YASA, or to update to the new version and reprocess all your nights with the new algorithm for consistency.

Sleep statistics

Artefact and Unscored epochs are now excluded from the calculation of the total sleep time (TST) in yasa.sleep_statistics(). Previously, YASA calculated TST as SPT - WASO, thus including Art and Uns. TST is now calculated as the sum of all REM and NREM sleep in SPT.

New FAQ

The online documentation now has a brand new FAQ section! Make sure to check it out at https://raphaelvallat.com/yasa/build/html/faq.html

New function: coincidence matrix

We have added the yasa.SpindlesResults.get_coincidence_matrix() and yasa.SWResults.get_coincidence_matrix() methods to calculate the (scaled) coincidence matrix. The coincidence matrix gives, for each pair of channel, the number of samples that were marked as an event (spindles or slow-waves) in both channels. In other words, it gives an indication of whether events (spindles or slow-waves) are co-occuring for any pair of channel. The scaled version of the coincidence matrix can then be used to define functional networks or quickly find outlier channels.

Minor enhancements

1. Minor speed improvements in yasa.SleepStaging.

2. Updated dependency to pyRiemann>=0.2.7, which solves the version conflict with scikit-learn (see issue 33).

3. flake8 requirements for max line length has been changed from 80 to 100 characters.

v0.4.1 (March 2021)¶

New functions

1. Added yasa.topoplot(), a wrapper around mne.viz.plot_topomap(). See 15_topoplot.ipynb

Enhancements

1. The default frequency range for slow-waves in yasa.sw_detect() is now 0.3-1.5 Hz instead of 0.3-2 Hz. Indeed, most slow-waves have a frequency below 1Hz. This may result in slightly different coupling values when coupling=True so make sure to homogenize your slow-waves detection pipeline across all nights in your dataset.

2. yasa.trimbothstd() now handles missing values in input array.

3. yasa.bandpower_from_psd() and yasa.bandpower_from_psd_ndarray() now print a warning if the PSD contains negative values. See issue 29.

4. Upon loading, YASA will now use the outdated package to check and warn the user if a newer stable version is available.

5. YASA now uses the antropy package to calculate non-linear features in the automatic sleep staging module. Previously, YASA was using EntroPy, which could not be installed using pip.

v0.4.0 (November 2020)¶

This is a major release with several new functions, the biggest of which is the addition of an automatic sleep staging module (yasa.SleepStaging). This means that YASA can now automatically score the sleep stages of your raw EEG data. The classifier was trained and validated on more than 3000 nights from the National Sleep Research Resource (NSRR) website.

Briefly, the algorithm works by calculating a set of features for each 30-sec epochs from a central EEG channel (required), as well as an EOG channel (optional) and an EMG channel (optional). For best performance, users can also specify the age and the sex of the participants. Pre-trained classifiers are already included in YASA. The automatic sleep staging algorithm requires the LightGBM and antropy package.

Other changes

1. yasa.SpindlesResults() and yasa.SWResults() now have a plot_detection method which allows to interactively display the raw data with an overlay of the detected spindles. For now, this only works with Jupyter and it requires the ipywidgets package.

2. Added hue input parameter to yasa.SpindlesResults.plot_average(), yasa.SWResults.plot_average() to allow plotting by stage.

3. The get_sync_events() method now also returns the sleep stage when available.

4. The yasa.sw_detect() now also returns the timestamp of the sigma peak in the SW-through-locked 4-seconds epochs. The timestamp is expressed in seconds from the beginning of the recording and can be found in the SigmaPeak column.

Dependencies

2. Added ipywidgets, LightGBM and entropy to dependencies.

v0.3.0 (May 2020)¶

This is a major release with several API-breaking changes in the spindles, slow-waves and REMs detection.

First, the yasa.spindles_detect_multi() and yasa.sw_detect_multi() have been removed. Instead, the yasa.spindles_detect() and yasa.sw_detect() functions can now handle both single and multi-channel data.

Second, I was getting some feedback that it was difficult to get summary statistics from the detection dataframe. For instance, how can you get the average duration of the detected spindles, per channel and/or per stage? Similarly, how can you get the slow-waves count and density per stage and channel? To address these issues, I’ve now modified the output of the yasa.spindles_detect(), yasa.sw_detect() and yasa.rem_detect() functions, which is now a class (= object) and not a simple Pandas DataFrame. The advantage is that the new output allows you to quickly get the raw data or summary statistics grouped by channel and/or sleep stage using the .summary() method.

>>> sp = yasa.spindles_detect(...)
>>> sp.summary()  # Returns the full detection dataframe
>>> sp.summary(grp_chan=True, grp_stage=True, aggfunc='mean')

Similarly, the yasa.get_bool_vector() and yasa.get_sync_events() functions are now directly implemented into the output, i.e.

>>> sw = yasa.sw_detect(...)
>>> sw.summary()
>>> sw.get_sync_events(center='NegPeak', time_before=0.4, time_after=0.8)

One can also quickly plot an average “template” of all the detected events:

>>> sw.plot_average(center="NegPeak", time_before=0.4, time_after=0.8)

For more details, please refer to the documentation of yasa.SpindlesResults(), yasa.SWResults() and yasa.REMResults().

Important

This is an experimental feature, and it’s likely that these functions will be modified, renamed, or even deprecated in future releases based on feedbacks from users. Please make sure to let me know what you think about the new output of the detection functions!

Other changes

1. The coupling argument has been removed from the yasa.spindles_detect() function. Instead, slow-oscillations / sigma coupling can only be calculated from the slow-waves detection, which is 1) the most standard way, 2) better because PAC assumptions require a strong oscillatory component in the lower frequency range (slow-oscillations). This also avoids unecessary confusion between spindles-derived coupling and slow-waves-derived coupling. For more details, refer to the Jupyter notebooks.

2. Downsampling of data in detection functions has been removed. In other words, YASA will no longer downsample the data to 100 / 128 Hz before applying the events detection. If the detection is too slow, we recommend that you manually downsample your data before applying the detection. See for example mne.filter.resample().

3. yasa.trimbothstd() can now work with multi-dimensional arrays. The trimmed standard deviation will always be calculated on the last axis of the array.

4. Filtering and Hilbert transform are now applied at once on all channels (instead of looping across individual channels) in the yasa.spindles_detect() and yasa.sw_detect() functions. This should lead to some improvements in computation time.


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# PSEB 11th Class Physics Solutions Chapter 14 Oscillations

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 14 Oscillations Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 11 Physics Chapter 14 Oscillations

PSEB 11th Class Physics Guide Oscillations Textbook Questions and Answers

Question 1.
Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its center of mass.
(d) An arrow released from a bow.
Solution:
(b) and (c)
Explanations :
(a) The swimmer’s motion is not periodic. The motion of the swimmer between the banks of a river is back and forth. However, it does not have a definite period. This is because the time taken by the swimmer during his back and forth journey may not be the same.

(b) The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic. This is because the magnet oscillates about its position with a definite period of time.

(c) When a hydrogen molecule rotates about its center of mass, it comes to the same position again and again after an equal interval of time. Such motion is periodic.

(d) An arrow released from a bow moves only in the forward direction. It does not come backward. Hence, this motion is not a periodic.

Question 2.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
(a) the rotation of earth about its axis.
(b) motion of an oscillating mercury column in a 17-tube.
(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowermost point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.
Solution:
(b) and (c) are SHMs; (a) and (d) are periodic, but not SHMs
Explanations :
(a) During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and fro motion about its axis.

(b) An oscillating mercury column in a [/-tube is simple harmonic. This is because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time.

(c) The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic.

(d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.

Question 3.
Figure depicts four x-t plots for linear motion of a particle. Which of the plots represents periodic motion? What is the period of motion (in case of periodic motion)?
(a)

(b)

(c)

(d)

Solution:
(b) and (d) are periodic
Explanation :
(a) It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition of motion in this case.
(b) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s.
(c) It is not a periodic motion. This is because the particle repeats the motion in one position only. For a periodic motion, the entire motion of the particle must be repeated in equal intervals of time. In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s.

Question 4.
Which of the following functions of time represent (a) simple ‘ harmonic, (b) periodic but not simple harmonic, and (c) non¬periodic motion? Give period for each case of periodic motion (a is any positive constant):
(a) sin ωt – cos ωt
(b) sin 3ωt
(c) 3cos($$\pi / 4$$ -2ωt)
(d) cos ωt +cos 3 ωt+cos 5ωt
(e) exp(-ω2t2)
(f) 1+ ωt+ω2t
Solution:
(a) SHM
The given function is:

This function represents SHM as it can be written in the form: a sin (ωt +Φ)
Its period is: $$\frac{2 \pi}{\omega}$$

(b) Periodic, but not SHM The given function is:
sin3 ωt = $$\frac{1}{4}$$ [3sinωt -sin3ωt] (∵ sin3θ = 3sinθ – 4sin3 θ)
The terms sin cot and sin 3 ωt individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.

Period of$$\frac{3}{4}$$sin ωt = $$\frac{2 \pi}{\omega}$$ = T
Period of$$\frac{1}{4}$$sin3ωt = $$\frac{2 \pi}{3 \omega}$$ = T’ = $$\frac{T}{3}$$
Thus, period of the combination
= Minimum time after which the combined function repeats
= LCM of T and $$\frac{T}{3}$$ = T
Its period is 2 $$\pi / \omega$$

(c) SHM
The given function is:
3 cos $$\left[\frac{\pi}{4}-2 \omega t\right]$$ = 3 cos $$\left[2 \omega t-\frac{\pi}{4}\right]$$
This function represents simple harmonic motion because it can be written in the form:
acos(ωt +Φ)

Its period is :
$$\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}$$

(d) Periodic, but not SHM
The given function is cosωt +cos3ωt +cos5ωt. Each individual cosme function represents SHM. However, the superposition of threc simple harmonic motions is periodic, but not simple harmonic.

cosωt represents SHM with period = $$\frac{2 \pi}{\omega}$$ T (say)
cos 3ωt represents SHM with period = $$\frac{2 \pi}{3 \omega}=\frac{T}{3}$$
cos 5ωt represents SHM with period = $$\frac{2 \pi}{5 \omega}=\frac{T}{5}$$
The minimum time after which the combined function repeats its value is T. Hence, the given function represents periodic function but not SHM, with period T.

(e) Non-periodic motion: .
The given function exp(- ω2t2) is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.

(f) Non-periodic motion
The given function is 1+ ωt + ω2t2
Here no repetition of values. Hence, it represents non-periodic motion.

Question 5.
A particle Is in linear simple harmonic motion between two points, A and B, 10 cm apart Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when It Is
(a) at the end A,
(b) at the end B,
(c) at the midpoint of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
Solution:
The given situation is shown in the following figure. Points A and B are the two endpoints, with AB =10cm. 0 is the midpoint of the path.

A particle is in linear simple harmonic motion between the endpoints
(a) At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at this point.
Its acceleration is positive as it is directed along with AO.
Force is also positive in this case as the particle is directed rightward.

(b) At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at this point.
Its acceleration is negative as it is directed along B.
Force is also negative in this case as the particle is directed leftward.

(c)
The particle is extending a simple harmonic motion. O is the mean position of the particle. Its velocity at the mean position O is the maximum. The value for velocity is negative as the partide Is directed leftward. The acceleration and force of a particle executing SHM is zero at the mean position.

(d)
The particle is moving towards point O from the end B. This direction of motion is opposite to the conventional positive direction, which is from A to R. Hence, the particle’s velocity and acceleration, and the force on it are all negative.

(e)

The particle is moving towards point O from the end A. This direction of motion is from A to B, which is the conventional positive direction. Hence, the value for velocity, acceleration, and force are all positive.

(f)

This case is similar to the one given in (d).

Question 6.
Whi ch of the following relationships between the acceleration a and the displacement x of a particle involves simple harmonic motion?
(a) a=0.7x
(b) a=-200x2
(c) a= – 10 x (d) a=100x3
Solution:
A motion represents simple harmonic motion if it is governed by the force law:
F=-kx
ma’= -kx
∴ a = – $$\frac{k}{m}$$ x

where F is the force
m is the mass (a constant for a body)
x is the displacement
a is the acceleration.
k is a constant
Among the given equations, only equation a = -10 x is written in the
above form with $$\frac{k}{m}$$ =10. Hence, this relation represents SHM.

Question 7.
The motion of a particle executing simple harmonic motion is described by the displacement function,
x(t) = A cos (ωt+Φ)
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is πs-1. If instead of the cosine function, we choose the sine function to describe the SHM:x = B sin(ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Solution:
Initially, at t = 0:
Displacement, x = 1 cm Initial velocity, ν = ω cm/s.
Angular frequency, ω = π rad s-1
It is given that:
x(t) = Acos (ωt+Φ)
1 = Acos(ω x 0 +Φ) = AcosΦ
AcosΦ =1 ……………………………….. (i)

Velocity, ν = $$\frac{d x}{d t}$$
ω = -Aω sin(ωt +Φ)
1 = -Asin(ω x 0 +Φ) = -AsinΦ
Asin Φ = -1 ………………………… (ii)
Squaring and adding equations (i) and (ii), we get
A2(sin2Φ +cos2Φ) = 1+1
A2 = 2
∴ A = $$\sqrt{2}$$ cm

Dividing equation (ii) by equation (i), we get
tanΦ = -1

SHM is given as
x = Bsin(ωt + α)
Putting the given values in this equation, we get 1 =B sin (ωt + α)
B sin α =1 …………………………… (iii)

Velocity, ν = $$\frac{d x}{d t}$$
ω =(ωB cos (ωt + a)
1 =B cos (ω x 0+α ) = B cos α …………………………………… (iv)
Squaring and adding equations (iii) and (iv), we get
B2 [sin2α +cos2 α] =1+1
B2 =2
B = $$\sqrt{2}$$ cm
Dividing equation (iii) by equation (iv), we get

Question 8.
A spring balance has a scale that reads from 0 to 50kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Solution:
Maximum mass that the scale can read, M = 50 kg
Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m Time period, T =0.6s
Maximum force exerted on the spring, F = Mg where,
g = acceleration due to gravity = 9.8 m/s2
F = 50 × 9.8 = 490 N
∴ Spring constant , K = $$\frac{F}{l}=\frac{490}{0.2}$$ = 2450Nm-1

Mass m, is suspended from the balance,
Time period, T = $$2 \pi \sqrt{\frac{m}{k}}$$
∴ m = $$\left(\frac{T}{2 \pi}\right)^{2} \times k=\left(\frac{0.6}{2 \times 3.14}\right)^{2} \times 2450$$ = 22.36 kg
∴ Weight of the body = mg = 22.36 x 9.8 = 219.167N
Hence, the weight of the body is about 219 N.

Question 9.
Aspringhavingwith aspiring constant 1200Nm-1 is mounted on a horizontal table as shown in figure. A mss of 3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0cm and released.

Determine (i) the frequency of oscillations,
(ii) maximum acceleration of the mass, and
(iii) the maximum speed of the mass.
Solution:
Spring constant, k = 1200 Nm-1
mass,m = 3 Kg
Displacement,A = 2.0 cm = 0.02 m
(i) Frequency of oscillation y, is gyen by the relation
V = $$\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$$
where, T is the time period
∴ v = $$\frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}}$$ = 3.18 s-1
Hence, the frequency of oscillations is 3.18 s-1.

(ii) Maximum acceleration a is given by the relation:
a = ω2A
where,
ω = Angular frequency = $$\sqrt{\frac{k}{m}}$$
A = Maximum displacement
∴ a = $$\frac{k}{m} A=\frac{1200 \times 0.02}{3}$$ = 8 ms-2
Hence, the maximum acceleration of the mass is 8.0 ms2

(iii) Maximum speed, νmax = Aω
= $$A \sqrt{\frac{k}{m}}=0.02 \times \sqrt{\frac{1200}{3}}$$ = 0.4 m/s
Hence, the maximum speed of the mass is 0.4 m/s.

Question 10.
In question 9, let us take the position of mass when the spring is unstretched as x =0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating massif at the moment we start the stopwatch (t = 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude br the initial phase?
Solution:
(a) The functions have the same frequency and amplitude, but different initial phases.
Distance travelled by the mass sideways, A = 2.0 cm
Force constant of the spring, k =1200 N m-1
Mass, m =3kg
Angular frequency of oscillation,
ω = $$\sqrt{\frac{k}{m}}=\sqrt{\frac{1200}{3}}$$ = $$\sqrt{400}$$ = 20 rad s-1
When the mass is at the mean position, initial phase is 0.
Displacement,
x = A sinωt = 2 sin20 t

(b) At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is $$\frac{\pi}{2}$$
Displacement, x = Asin $$\left(\omega t+\frac{\pi}{2}\right)$$
=2sin$$\left(20 t+\frac{\pi}{2}\right)$$
= 2 cos 20t

(c) At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase is $$\frac{3 \pi}{2}$$
Displacement, x = A sin $$\left(\omega t+\frac{3 \pi}{2}\right)$$
= 2sin $$\left(20 t+\frac{3 \pi}{2}\right)$$ = -2cos 20t
The functions have the same frequencyl $$\left(\frac{20}{2 \pi} \mathrm{Hz}\right)$$ land amplitude (2cm),
but different initial phases $$\left(0, \frac{\pi}{2}, \frac{3 \pi}{2}\right)$$

Question 11.
Figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i. e., clockwise or anti-clockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
solution:
Time period,T =2s
Amplitude, A = 3cm
At time, t = O, the radius vector OP makes an angle $$\frac{\pi}{2}$$ with the positive x -axis, i.e., phase angle Φ = + $$\frac{\pi}{2}$$
Therefore, the equation of simple harmonic motion for the x —projection of OP, at time t, is given by the displacement equation

(b) Time period, T = 4s
Amplitude, a =2 m
At time t = 0, OP makes an angle ir with the x-axis, in the anticlockwise direction. Hence, phase angle, Φ = +π
Therefore, the equation of simple harmonic motion for the x -projection of OP, at time t, is given as

Question 12.
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).
(a) x= – 2sin(3t+$$\pi / \mathbf{3}$$ )
(b) x=cos ($$\pi / 6$$ – t)
(c) x=3 sin (2πt + $$\pi / 4$$ )
(d) x=2 cos πt
Solution:
(a) x = -2 sin $$\left(3 t+\frac{\pi}{3}\right)=+2 \cos \left(3 t+\frac{\pi}{3}+\frac{\pi}{2}\right)=2 \cos \left(3 t+\frac{5 \pi}{6}\right)$$

If this equation is compared with the standard SHM equation,
x =A cos $$\left(\frac{2 \pi}{T} t+\phi\right)$$ then we get
Amplitude, A = 2cm
Phase angle, Φ = $$\frac{5 \pi}{6}$$ =150°
Angular velocity, ω = $$\frac{2 \pi}{T}$$ =3 rad/sec
The motion of the particle can be pokted as shown in the following figure.

(b) x=cos $$\left(\frac{\pi}{6}-t\right)$$ =cos $$\left(t-\frac{\pi}{6}\right)$$
If this equation is compared with the standard SHM equation,
x = A cos $$\left(\frac{2 \pi}{T} t+\phi\right)$$ then we get
Amplitude, A = 1 cm
Phase angle, Φ = $$-\frac{\pi}{6}$$ = – 30°
Angular velocity, ω = $$\frac{2 \pi}{T}$$ =1 rad/s
The motion of the particle can be plotted as shown in the following figure.

(c) x =3sin $$\left(2 \pi t+\frac{\pi}{4}\right)$$

If this equation is compared with the standard SHM equation
x = Acos $$\left(\frac{2 \pi}{T} t+\phi\right)$$ then we get
Amplitude, A = 3cm
Phase angle, Φ = $$-\frac{\pi}{4}$$
Angular velocity, ω = $$\frac{2 \pi}{T}$$ = 2π rad/s
The motion of the particle can be plotted as shown in the following figure.

(d) x=2cosπt
If this equation is compared with the standard SHM equation,
x = A cos $$\left(\frac{2 \pi}{T} t+\phi\right)$$ then we get
Amplitude, A = 2cm
Phase angle, Φ = 0
Angular velocity, ω = π rad/s
The motion of the particle can be plotted as shown in the following figure.

Question 13.
Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in figure (b) is stretched by the same force F.

(a) What is the minimum extension of the spring in the two cases?
(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?
Solution:
(a) For the one block system:
When a force F, is applied to the free end of the spring, an extension l, is produced. For the maximum extension, it can be written as:
F=kl
where k is the spring constant
Hence, the maximum extension produced in the spring, l = $$\frac{F}{k}$$
For the two blocks system:
The displacement (x) produced in this case is:
x = $$\frac{l}{2}$$
Net force, F = +2kx =2k $$\frac{l}{2}$$
∴ l = $$\frac{F}{k}$$

(b) For the one blocks system:
For mass (m) of the block, force is written as
F = ma = m $$\frac{d^{2} x}{d t^{2}}$$
where, x is the displacement of the block in time t
∴ m $$\frac{d^{2} x}{d t^{2}}$$ = -kx
It is negative because the direction of elastic force is opposite to the direction of displacement.

where, ω is angular frequency of the oscillation
∴ Time period of the oscillation,
T= $$\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\frac{k}{m}}}=2 \pi \sqrt{\frac{m}{k}}$$

For the two blocks system:
F=m $$\frac{d^{2} x}{d t^{2}}$$
m $$\frac{d^{2} x}{d t^{2}}$$ =-2kx

It is negative because the direction of elastic force is opposite to the direction of displacement.
$$\frac{d^{2} x}{d t^{2}}$$ = $$-\left[\frac{2 k}{m}\right] x$$ = – ω2x
where, Angular frequency, ω = $$\sqrt{\frac{2 k}{m}}$$
∴ Time period T = $$\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{2 k}}$$

Question 14.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 in. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?
Solution:
Angular frequency of the piston, ω = 200 rad/min.
Stroke =1.0 m
Amplitude, A = $$\frac{1.0}{2}$$ = 0.5m
The maximum speed (νmax) of the piston is given by the relation
νmax =Aω = 200 x 0.5=100 m/min

Question 15.
The acceleration due to gravity on the surface of moon is 1.7 ms-2. What is the time period of a simple pendulum on the surface of moon If Its time period on the surface of earth is 3.5 s? (gon the surface of earth is 9.8 ms-2)
Solution:
Acceleration due to gravity on the surface of moon, g’ = 1.7m s-2
Acceleration due to gravity on the surface of earth, g = 9.8 ms-2
Time period of a simple pendulum on earth, T = 3.5 s
T= $$2 \pi \sqrt{\frac{l}{g}}$$

where l is the length of the pendulum
∴ l = $$\frac{T^{2}}{(2 \pi)^{2}} \times g=\frac{(3.5)^{2}}{4 \times(3.14)^{2}} \times 9.8 \mathrm{~m}$$

The length of the pendulum remains constant.
On Moon’s surface, time period,
T’ = $$2 \pi \sqrt{\frac{l}{g^{\prime}}}=2 \pi \sqrt{\frac{(3.5)^{2}}{\frac{4 \times(3.14)^{2}}{1.7}} \times 9.8}$$ = 8.4 s
Hence, the time period of the simple pendulum on the surface of Moon is 8.4 s.

Question 16.
(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:
T = $$2 \pi \sqrt{\frac{m}{k}}$$ A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

(b) The motion of a simple pendulum is approximately simple harmonic for small-angle oscillations. For larger angles of oscillation a more involved analysis shows that T is greater than $$2 \pi \sqrt{\frac{l}{g}}$$ Think of a qualitative argument to appreciate this result.

(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?

(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
Solution :
(a) The time period of a simple pendulum, T = $$2 \pi \sqrt{\frac{m}{k}}$$
For a simple pendulum, k is expressed in terms of mass m, as
k ∝ m
$$\frac{m}{k}$$ = Constant
Hence, the time period T, of a simple pendulum is independent of the mass of the bob.

(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as
F = -mg sinθ
where, F = Restoring force; m = Mass of the bob; g = Acceleration due to
gravity; θ = Angle of displacement
For small θ, sinθ ≈ θ
For large 0,sin0 is greater than 0.
This decreases the effective value of g.
Hence, the time period increases as
T = $$2 \pi \sqrt{\frac{l}{g}}$$
where, l is the length of the simple pendulum

(c) The time shown by the wristwatch of a man falling from the top of a tower is not affected by the fall. Since a wristwatch does not work on the principle of a simple pendulum, it is not affected by the acceleration due to gravity during free fall. Its working depends on spring action.

(d) When a simple pendulum mounted in a cabin falls freely under gravity, its acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.

Question 17.
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small f oscillations in a radial direction about its equilibrium position, what will be its time period?
Solution:
The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.
Acceleration due to gravity = g
Centripetal acceleration = $$\frac{v^{2}}{R}$$
where, v is the uniform speed of the car R is the radius of the track
Effective acceleration (aeff) is given as
aeff = $$\sqrt{g^{2}+\left(\frac{v^{2}}{R}\right)^{2}}$$

Time period, T = $$2 \pi \sqrt{\frac{l}{a_{e f f}}}$$
where, l is the length of the pendulum
∴ Time period, T = $$2 \pi \sqrt{\frac{l}{g^{2}+\frac{v^{4}}{R^{2}}}}$$

Question 18.
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρl. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
T = $$2 \pi \sqrt{\frac{\boldsymbol{h} \rho}{\rho_{\boldsymbol{l}} \boldsymbol{g}}}$$
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Solution:
Base area of the cork = A
Height of the cork = h
Density of the liquid = ρl
Density of the cork = ρ

In equilibrium:
Weight of the cork = Weight of the liquid displaced by the floating cork Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.
Up-thrust = Restoring force, F = Weight of the extra water displaced
F = -(Volume x Density x g)
Volume = Area x Distance through which the cork is depressed Volume = Ax
∴ F = -Ax ρlg ………………………… (i)
According to the force law,
F = kx
k = $$\frac{F}{x}$$

where k is a constant
k = $$\frac{F}{x}$$ = -Aρlg ………………………………. (ii)
The time period of the oscillations of the cork,
T = $$2 \pi \sqrt{\frac{m}{k}}$$ …………………………………… (iii)

where,
m = Mass of the cork
= Volume of the cork x Density
= Base area of the cork x Height of the cork x Density of the cork = Ahρ
Hence, the expression for the time period becomes
T = $$2 \pi \sqrt{\frac{A h \rho}{A \rho_{l} g}}$$ = 2$$\pi \sqrt{\frac{h \rho}{\rho_{l} g}}$$

Question 19.
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Solution:
Area of cross-section of the U-tube = A
Density of the mercury column = ρ
Acceleration due to gravity = g
Restoring force, F = Weight of the mercury column of a certain height
F = -(Volume x Density x g)
F = -(A x 2h x ρ x g) = -2Aρgh = -k x Displacement in one of the arms (h)

where, 2h is the height of the mercury column in the two arms
k is a constant, given by k = $$-\frac{F}{h}$$ = 2Aρg
Time period = $$2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{m}{2 A \rho g}}$$
where, m is the mass of the mercury column
Let l be the length of the total mercury in the U-tube.
Mass of mercury, m = Volume of mercury x Density of mercury = Alρ
∴ T = $$2 \pi \sqrt{\frac{A l \rho}{2 A \rho g}}=2 \pi \sqrt{\frac{l}{2 g}}$$

Hence the mercury column executes simple harmonic motion with time period $$2 \pi \sqrt{\frac{l}{2 g}}$$

Question 20.
An air chamber of volume V has a neck area of cross-section a into which a ball of mass m just fits and can move up and down without any friction (see figure). Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal.

Solution:
Volume of the air chamber = V
Area of cross-section of the neck = a
Mass of the ball = m
The pressure inside the chamber is equal to the atmospheric pressure. Let the ball be depressed by x units. As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber.
Decrease in the volume of the air chamber, ΔV = ax
Volumetric strain =
⇒ $$\frac{\Delta V}{V}=\frac{a x}{V}$$

Bulk Modulus of air, B = $$\frac{\text { Stress }}{\text { Strain }}=\frac{-p}{\frac{a x}{V}}$$
In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume.
p = $$\frac{-B a x}{V}$$
The restoring force acting on the ball,
F = p × a = $$\frac{-B a x}{V} \cdot a=\frac{-B a^{2} x}{V}$$ ……………………………. (i)
In simple harmonic motion, the equation for restoring force is
F = -kx …………………………………….. (ii)
where, k is the spring constant Comparing equations (i) and (ii), we get
k = $$\frac{B a^{2}}{V}$$
Time period, T = $$2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{V m}{B a^{2}}}$$

Question 21.
You are riding in an automobile of mass 3000kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (6) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
Solution:
Mass of the automobile, m = 3000 kg
Displacement in the suspension system, x = 15cm = 0.15 m
There are 4 springs in parallel to the support of the mass of the automobile.
The equation for the restoring force for the system:
F = -4 kx = mg

where, k is the spring constant of the suspension system
Time period, T = $$2 \pi \sqrt{\frac{m}{4 k}}$$
and, k = $$\frac{m g}{4 x}=\frac{3000 \times 10}{4 \times 0.15}$$ = 50000 = 5 x 10 4N/m
Spring constant, k = 5 x 104 N/m

Each wheel supports a mass, M = $$\frac{3000}{4}$$ = 750 kg
For damping factor b, the equation for displacement is written as:
x = x0e-bt/2M

The amplitude of oscillation decreases by 50%

where, Time petiod,t =$$2 \pi \sqrt{\frac{m}{4 k}}=2 \pi \sqrt{\frac{3000}{4 \times 5 \times 10^{4}}}$$ =0.7691s
∴ b =$$\frac{2 \times 750 \times 0.693}{0.7691}$$ =1351.58kg/s
Therefore, the damping constant of the spring is 1351.58 kg/s.

Question 22.
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Solution:
The equation of displacement of a particle executing SHM at an instant t is given as
x = Asinωt
where,
A = Amplitude of oscillation
ω = Angular frequency = $$\sqrt{\frac{k}{M}}$$
The velocity of the particle is
ν = $$\frac{d x}{d t}$$ = Aωcosωt

The kinetic energy of the particle is
Ek = $$\frac{1}{2} M v^{2}=\frac{1}{2} M A^{2} \omega^{2} \cos ^{2} \omega t$$
The potential energy of the particle is
Ep = $$\frac{1}{2} k x^{2}=\frac{1}{2} M \omega^{2} A^{2} \sin ^{2} \omega t$$

For time period T, the average kinetic energy over a single cycle is given as

And, average potential energy over one cycle is given as

It can be inferred from equations (i) and (ii) that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.

Question 23.
A circular disc of mass 10 kg is suspended by a wire attached to its center. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = -αθ, where J is the restoring couple and 0 the angle of twist).
Solution:
Mass of the circular disc, m = 10 kg
Radius of the disc, r = 15cm = 0.15 m
The torsional oscillations of the disc has a time period, T = 1.5 s The moment of inertia of the disc is
I = $$\frac{1}{2}$$ mr2 = $$\frac{1}{2} \times(10) \times(0.15)^{2}$$ = 0.1125kg-m2

Time period, T = $$2 \pi \sqrt{\frac{I}{\alpha}}$$
where, α is the torsional constant.
An21 4 x(3.14)2x 0.1125 , M
α = $$\frac{4 \pi^{2} I}{T^{2}}=\frac{4 \times(3.14)^{2} \times 0.1125}{(1.5)^{2}}$$ = 1.972 N-m/rad
Hence, the torsional spring constant of the wire is 1.972 N-m rad-1.

Question 24.
A body describes simple harmonic motion with amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm.
Solution:
Amplitude, A = 5 cm = 0.05m
Time period, T = 0.2 s
(a) For displacement, x = 5 cm = 0.05m
Acceleration is given by
a = -ω2x = $$-\left(\frac{2 \pi}{T}\right)^{2} x=-\left(\frac{2 \pi}{0.2}\right)^{2} \times 0.05$$
Velocity is given by
ν = ω $$\sqrt{A^{2}-x^{2}}=\frac{2 \pi}{T} \sqrt{(0.05)^{2}-(0.05)^{2}}$$ = 0
When the displacement of the body is 5 cm, its acceleration is -5π2 m/s2 and velocity is 0.

(b) For displacement, x =3 cm = 0.03 m
Acceleration is given by
a = – ω2x = – $$\left(\frac{2 \pi}{T}\right)^{2}$$x = $$\left(\frac{2 \pi}{0.2}\right)^{2}$$ 0.03 = -3π2 m/s2
Velocity is given by

When the displacement of the body is 3 cm, its acceleration is -3π m/s2 and velocity is 0.4π m/s.

(c) For displacement, x = 0
Acceleration is given by
a = – ω2x = 0
Velocity is given by

When the displacement of the body is 0, its acceleration is 0, and velocity is0.5π m/s.

Question 25.
A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x0 and pushed towards the center with a velocity v0 at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0, and v0. [Hint: Start with the equation x = a cos(ωt + θ) and note that the initial velocity is negative.]
Solution:
The displacement equation for an oscillating mass is given by
x = Acos(ωt + θ) …………………………… (i)
where A is the amplitude
x is the displacement
θ is the phase constant

Squaring and adding equations (iii) and (iv), we get

Hence, the amplitude of the resulting oscillation is $$\sqrt{x_{0}^{2}+\left(\frac{v_{0}}{\omega}\right)^{2}}$$


text:
# Forward Exchange Rate

Forward exchange rate is the exchange rate at which a party is willing to enter into a contract to receive or deliver a currency at some future date.

Currency forwards contracts and future contracts are used to hedge the currency risk. For example, a company expecting to receive €20 million in 90 days, can enter into a forward contract to deliver the €20 million and receive equivalent US dollars in 90 days at an exchange rate specified today. This rate is called forward exchange rate.

Forward exchange rates are determined by the relationship between spot exchange rate and interest or inflation rates in the domestic and foreign countries.

## Formula

Using the relative purchasing power parity, forward exchange rate can be calculated using the following formula:

 f = s × 1 + Id n 1 + If

Where,
f is forward exchange rate in terms of units of domestic currency per unit of foreign currency;
s is spot exchange rate, in terms of units of domestic currency per unit of foreign currency;
Id domestic inflation rate;
If is foreign inflation rate; and
n is number of time periods

Using the covered interest rate parity, forward exchange rate is calculated using the following formula:

 f = s × 1 + id n 1 + if

Where,
f, s and n stand for the same as stated above;
Id domestic interest rate; and
If is foreign interest rate.

## Example

Exchange rate between US$and British £ on 1 January 2012 was$1.55 per £. This is our spot exchange rate. Inflation rate and interest rate in US were 2.1% and 3.5% respectively. Inflation rate and interest rate in UK were 2.8% and 3.3%.

Estimate the forward exchange rate between the countries in $/£. Solution Using relative purchasing power parity, forward exchange rate comes out to be$1.554/£

 f = $1.5507/£ × 1 + 3.3% n =$1.554/£ 1 + 3.3%

Using the interest rate parity, forward exchange rate is

 f = $1.5507/£ × 1 + 2.1% n =$1.5401/£ 1 + 2.8%

Actual exchange rate was $1.6244/£. US$ has depreciated more than predicated by the relative purchasing power parity and interest rate parity.

Written by Obaidullah Jan


text:
# O. G. Storozh

Search this author in Google Scholar

Articles: 3

### On a criterion of mutual adjointness for extensions of some nondensely defined operators

Methods Funct. Anal. Topology 20 (2014), no. 1, 50-58

In the paper the role of initial object is played by a pair of closed linear densely defined operators $L_0$ and $M_0$, where $L_0 \subset M_0^{\ast}:= L,$ acting in Hilbert space. A criterion of mutual adjointness for some classes of the extensions of finite-dimensional (non densely defined) restrictions of $L_0$ and $M_0$ are established. The main results are based on the theory of linear relations in Hilbert spaces and are formulated in the terms of abstract boundary operators.

### The criteria of maximal dissipativity and self-adjointness for a class of differential-boundary operators with bounded operator coefficients

Methods Funct. Anal. Topology 14 (2008), no. 4, 372-379

A class of the second order differential-boundary operators acting in the Hilbert space of infinite-dimensional vector-functions is investigated. The domains of considered operators are defined by nonstandard (e.g., multipoint-integral) boundary conditions. The criteria of maximal dissipativity and the criteria of self-adjointness for investigated operators are established.

### On some perturbations changing the domain of proper extension of positively definite operator

Methods Funct. Anal. Topology 11 (2005), no. 3, 257-269


text:
# Keyword:FULLXS

FULLXS

This keyword causes the full, energy-dependent group cross-sections to be stored when the cross section library data are being collapsed. Note that this can be used at any different instance of the GETXS keyword.

Example usage:

< -- Control phase -- >
...
FULLXS
GETXS 1 709
FISPACT
* Title of the simulation
< -- Initial phase -- >
...
< -- Inventory phase -- >
...
FULLXS
GETXS 1 709
...


text:
# Can the running time be reduced to something lower than $O(d^4)$?

Imagine I have a weighted complete directed graph $$G$$ with $$d$$ vertices(so $$d(d-1)$$ edges) and I want to do the following:

1. Set $$D$$ to be a DAG with the same set of vertices but without any edges
2. sort the edge weights $$c_{ji}$$ of the original graph by size
3. From the biggest to the lowest edge weight: Add an edge $$(j,i)$$ to the DAG $$D$$ if we do not cause a cycle with it

Now I see two approaches to implement this.the first one is:

1. Set $$D$$ to be a DAG with the same set of vertices but without any edges
2. sort the edge weights $$c_{ji}$$ of the original graph by size
3. From the biggest to the lowest edge weight: Add an edge $$(j,i)$$ to the DAG $$D$$ if there is no path from $$i$$ to $$j$$

Now sorting can be neglected in complexity ( $$O(d^2\log(d))$$)

And number three is performed at most $$d(d-1)$$ times and to check if a path exists we can use BFS or DFS and the number of edges, unfortunately, can be as big as $$d^2$$ so we have $$d^4$$ in total

My second idea: Work with a reachability matrix and update the reachability matrix in every iteration of the loop... But then updating the reachability matrix is again of complexity $$d^2$$ because if we set an edge from $$j$$ to $$i$$ then denoting the ancestors of $$j$$ by $$k_1$$ and the descendants of $$i$$ by $$k_2$$ we need to set $$R(u,w)=1$$, $$R(u,i)=1$$, $$R(j,w)=1$$ for any $$u$$ in $$k_1$$ and $$w$$ in $$k_2$$ .... Assuming that in the worst case that each of $$k_1$$ and $$k_2$$ have $$d/2-1$$ vertices we get a complexity of $$d^2$$ That we need to update the reachability matrix more than $$O(d)$$ times can e.g. be seen by looking at bipartite graphs... So I'm again stuck with the same complexity... Is there actually a more efficient solution?

• The bottommost link in this excellent answer to a similar question promises to make adding edges to a DAG and checking for connectivity both amortised $O(d)$-time operations, bringing the overall complexity down to $O(d^3)$. – j_random_hacker Mar 27 '19 at 15:54
• For undirected graphs, this algorithm is called Kruskal's algorithm and returns the minimum spanning tree. But a different algorithm must be used for MSTs in directed graphs. Is that what you're attempting to find, or do you really want the "almost MST" graph returned by this algorithm? – BlueRaja - Danny Pflughoeft Mar 27 '19 at 16:50
• @Johannes This method does not necessarily produce a MST analogue for directed, weighted graphs, as the edgeweight of the resulting acyclic graph needn't be maximal. Are you sure you don't want it to be maximal? – Sudix Mar 27 '19 at 19:37
• My (and @Sudix's) point was that you appear to be looking for the MST, but this algorithm only does that for undirected graphs. If so, it's not NP-hard. Chu-Liu's algorithm solves it in $O(d^3)$ – BlueRaja - Danny Pflughoeft Mar 27 '19 at 20:37
• You are talking about Optimum branching I assume... of course the solution I am talking about is not a solution for the optimal branching because in my case I can get a complete DAG whereas Optimum branching I have $d-1$ edges - but thank you very much for this remark – Johannes Mar 27 '19 at 21:20


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This week, I’m going to cover the Maximum Binary Tree problem. It’s an interesting tree problem that combines multiple concepts such as searching, traversals, and recursion.

# Problem

Given an integer array with no duplicates, a maximum tree building on this array is defined as follows:

1. The root is the maximum number in the array.
2. The left subtree is the maximum tree constructed from the left part subarray divided by the maximum number.
3. The right subtree is the maximum tree constructed from the right part subarray divided by the maximum number.

Construct the maximum tree by the given array and output the root node of this tree.

For example, the array [3, 2, 1, 6, 0, 5] would give us the following tree:

      6
/   \
3     5
\   /
2 0
\
1

# Solution

When reading the problem statement, the first thing that should go through your mind is recursion. Once steps 1-3 are completed for the root node, the steps can also repeated for the left and right children, assuming they exist. Therefore, the key to solving the problem is solving it for once case correctly so that it can be recursively applied to other cases.

Let’s take a look at the first couple steps of the algorithm. The maximum number in the array is 6, which exists at index 3. The maximum number in the left subarray (from indices 0 -> 2) is 3. The maximum number in the right subarray (from indices 4 -> 5) is 5. The root node is 6 and the left and right children are 3 and 5 respectively. We can repeat the same algorithm on the left and right subarrays until the array is empty.

The process of visiting a root node then recursing on its left and right subtrees should seem familiar. This is a DFS pre-order tree traversal. The only difference is that we have to build our tree as we traverse it; this shouldn’t be a problem because the values are supplied by the maximum number in each recursive step.

Therefore, we are left with the following algorithm:

• Check if nums is empty. If it is, return None. This will presumably form the leaves of our tree.
• Find the index of the maximum number. Let’s say it’s at index i.
• Create a node whose internal element is nums[i].
• Recurse on the left subtree with nums[:i] and the right subtree with nums[i + 1:].
• Return the node.

Below is an example of the algorithm’s implementation in Python. I’m assuming there exists a Node class that can be used to create a binary tree.

def construct_maximum_binary_tree(nums):
if not nums:
return None

i = nums.index(max(nums))
node = Node(nums[i])

node.left = construct_maximum_binary_tree(nums[:i])
node.right = construct_maximum_binary_tree(nums[i + 1:])

return node

# Analysis

In each recursive step, we are doing three things – searching for the maximum number, recursing on the left subarray, and recursing on the right subarray. The recurrence for this algorithm can be set up as $T(n) = 2T(\frac{n}{2}) + O(n)$ since we make two recursive calls and searching requires $O(n)$ time.

Using the Master theorem, this gives us a time complexity of $O(n \log n)$. Because we make around $\log n$ recursive calls, the space complexity is $O(\log n)$.

However, we have not considered the worst-case scenario. Imagine if we had an array that was sorted in ascending or descending order, such as [1, 2, 3, 4, 5, 6] or [6, 5, 4, 3, 2, 1]. Given an array of $n$ elements, we would recurse with $n - 1$ elements since either the left or right subarray would be empty. The recurrence would then turn into $T(n) = T(n - 1) + O(n)$. Because we have to make $n$ recursive calls, the time complexity would be $O(n^2)$ and the space complexity would be $O(n)$.


text:
# The convex hull of a manifold whose cobordism class is trivial

Let $$M$$ be a compact orientable $$n$$ dimensional manifold. Assume that $$M$$ has trivial cobordism class.
Is there an embedding of $$M$$ in some Euclidean space $$\mathbb{R}^m$$ such that the convex hull of $$M$$ is a $$n+1$$ dimensional manifold whose boundary is $$M$$?

Here the image of $$M$$ under the embedding is denoted again by $$M$$.

Note: One can pose the same question in the following geometric manner:

Let $$M$$ be a compact Riemannian manifold with trivial cobordism class. Is there an isometric embedding of $$M$$ in some Euclidean space such that the convex hull of $$M$$ is a manifold whose boundary is $$M$$?

• Generally no. For example, plenty of homotopy spheres do not bound contractible manifolds. Many (?most?) homology 3-spheres do not bound contractible 4-manifolds, yet all 3-manifolds are null cobordant. May 14 at 7:09
• @RyanBudney Snap! May 14 at 7:12
• @RyanBudney I think Freedman proved that every homology 3-sphere bounds a contractible topological 4-manifold. May 14 at 11:33
• @Zerox, I doubt it, but if you can find such a claim somewhere, you will have found a contradiction. For example, the Poincare Dodecahedral Space does not bound a contractible 4-manifold, due to the Rochlin invariant. 2 days ago
• @RyanBudney I think you missed the word topological in Zerox's comment. On the other hand, the original question is a bit ambiguous about the category one is working in. The first part could be in the smooth/PL/topological category, and might have different answers. The second part presumably refers to smooth manifolds (because of the Riemannian metric). 2 days ago

Implicit in the other responses is the fact that if $$M$$ bounds a convex manifold $$W$$, then $$W$$ is contractible and so M has the homology of a sphere. So any null-cobordant manifold that is not a homology sphere is a counterexample. Eg take $$M = X \# -X$$ where $$X$$ is any orientable manifold with non-trivial homology.


text:
# Reference for the 3d Ising model

+ 3 like - 0 dislike
466 views

Before I read the papers on the numerical solutions of the 3d Ising model (such as the work of El Showk et. al) I would like to know a bit more about the correspondende between the 3d Ising model and conformal field theory, in specific the field content, the scaling dimensions, correlation functions etc (I would like to see why they are what they are rather than a list as shown in the work I include above).

Therefore, are there any good introductory references for the 3d Ising model and CFT? I am quite familiar with the 2d case and its solution (also there are many references available for this) so I am only interested in 3d.

+ 4 like - 0 dislike

Section 3.8 of Cardy's book Scaling and Renormalization in Statistical Physics (only the first few pages are publicly online) describes scaling fields and scaling dimensions for the 3D Ising model, giving in particular the relations $\Delta_\sigma=1/2+\eta/2$ and $\Delta_\epsilon=3-1/\nu$ figuring on p.25 of the paper Bootstrapping mixed correlators in the 3D Ising model by Filip Kos, David Poland, David Simmons-Duffin. The analogous relations to the fields in More constraining conformal bootstrap by Ferdinando Gliozzi are $\Delta_\phi=1/2+\eta/2$ and $\Delta_{\phi^2}=3-1/\nu$. Numerical values for comparison are given in [12,13] of the latter paper.

Together with the general outline of the relations between the renormalization flow and CFTs in the last chapter of Cardy's book, this seems enough to make the identifications on a heuristic level. I haven't seen any more definite work on the relations, and would be interested in anything better that you can come up with in your own search.

answered Mar 25, 2016 by (15,608 points)

@ArnoldNeumaier I do not do research on this field, I am just purely interested in learning about it. I have asked for references also some senior faculty in my department and if they provide me with something good I will let you know.

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