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# Homework Help: Short question

1. Aug 2, 2004

### TopCat

A drag racer experiences air resistance equal to -cv². Assuming the racer is designed for maximum acceleration, I am to find v[t] and v[x]. If I can just get the equation of motion I'll be set.

I think it is F = ma - μmg - cv² = m dv/dt. According to the answer, the book thinks it is F = - μmg - cv².

Of course, when I evaluate our answers with the assigned numbers for μ, v, and vt (terminal velocity), we both get imaginary answers. :( I have no clue where we are wrong, but I suppose the books values could be the problem there.

2. Aug 2, 2004

### AKG

It's definitely not F = ma - μmg - cv². a = dv/dt, so you'd have:

ma - μmg - cv² = m dv/dt
ma - μmg - cv² = ma
-μmg - cv² = 0

Fnet = ma, as you know, so your equation would also be wrong because it's saying:

Fnet = Fnet - μmg - cv²

And you run into the same problem. ma is not one of the forces acting on the body, it is the NET FORCE, so you can't add it with other forces to get the net force.

Anyways, what is it that you need to find?

3. Aug 2, 2004

### TopCat

The way I see it, the net force, m dv/dt, is equal to the driving force of the engine (ma) minus the frictional force (μmg) minus the air resistance (cv²). Since the racer is designed for optimal acceleration, I assumed that "ma" was this maximal driving force, a constant.

Anyway, the problem is asking to obtain v[t] and v[x]. Then I am to "eliminate the coefficient of friction and solve the resulting equation numerically for the terminal velocity that can produce the 1988 world record of v=129.1 m/s, with t=4.99s for x=0.4 km."

Funny thing is that the v[t] that the book lists has the term ln[(u+v)/(u-v)] where u is the terminal velocity, which is given as 123.6 m/s in the back. Using the values of v and u, the log is imaginary.

Oh, thanks for your help, as well.

4. Aug 2, 2004

### AKG

Okay, well that's just confusing. In that way, "a" is not the acceleration of the vehicle at all, just some constant optimal acceleration. You should replace ma with "Fapplied" or "m*a_optimal" to avoid the confusion.
Just to clarify, you mean speed (v) as a function of time (t) and distance (x)? I would expect to see this as v(t) and v(x), just not sure exactly what it is. I'll look over that question to see if I can get v(t) and v(x).

5. Aug 2, 2004

### TopCat

Yeah, that was confusing. Sorry 'bout that.

v[t] and v[x] are indeed velocity as a function of time and distance. I use the brackets since I've been using mathematica a lot. What I have obtained is:

t = u/(a-g μ) ArcTanh(v/u)

x = -u²/(a-g μ) Ln(v²-u²).

6. Aug 2, 2004

### AKG

Okay, let A represent the optimal acceleration
Let B represent $\mu g$
Let C represent $\frac{c}{m}$

$$m\frac{dv}{dt} = mA - mB - cv^2$$

$$\frac{dv}{dt} = A - B - Cv^2$$

$$\frac{dv}{dx}\frac{dx}{dt} = A - B - Cv^2$$

$$v\frac{dv}{dx} = A - B - Cv^2$$

$$\frac{dv}{dx} = \frac{A - B - Cv^2}{v}$$

$$\frac{v}{A - B - Cv^2} dv = dx$$

$$\int _{v(x=0)} ^{v(x=x)} \frac{v}{A - B - Cv^2} dv = x$$

Let $u = A - B - Cv^2$

$$\int _{A - B - C[v(x=0)]^2} ^{A - B - C[v(x=x)]^2} \frac{-1}{2C} \frac{du}{u} = x$$

$$-\frac{1}{2C} \ln u |_{A - B - C[v(x=0)]^2} ^{A - B - C[v(x=x)]^2} = x$$

$$-\frac{1}{2C} \ln \left ( \frac{A - B - C[v(x=x)]^2}{A - B - C[v(x=0)]^2} \right )= x$$

$$\frac{A - B - Cv(x)^2}{A - B - Cv_0^2} = e^{-2Cx}$$

Isolate v(x), and you should be okay from there, I think.

Last edited: Aug 2, 2004
7. Aug 2, 2004

### AKG

Assuming what you have is right:

v(t) = u Tanh [t(a - gμ)/u]

Also, now that you have v in terms of t, you can replace the v's in the second equation, "x = -u²/(a-g μ) Ln(v²-u²)" with a function of t. You'll have x in terms of t. Isolate t, to get t in terms of x. Take that equation and plug it into your equation for v in terms of t, so you'll have v in terms of x, v(x).


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Determining whether a binary relation is reflexive, symmetric, transitive and antisymmetric

Determine whether the binary relation R on N given in each of the cases below is reﬂexive (r), symmetric (s), transitive (t) or antisymmetric (a), and state whether it is an equivalence relation, an order relation or neither of those.

• (a) $$a\operatorname Rb \iff a < b$$;
• (b) $$a\operatorname Rb \iff b ≤ a$$;
• (c) $$a\operatorname Rb \iff a ≤ b + 1$$;
• (d) $$a\operatorname Rb \iff 3^m a = 3^n b$$ for some $$m,n ∈\Bbb N∪\{0\}$$;

Hi, i'm a bit confused on question D on the worksheet above, The indices are throwing me off a bit, How would I go about in determining whether or not the relation is Reflexive, symmetric, transitive and antisymmetric? Thank you.

• Linking to a worksheet is not an appropriate way to ask a question. Please at least write the question out yourself.
– Matt
Oct 16, 2019 at 18:39
• Welcome to MathStackExchange! Please avoid posting pictures to the problem statements, it is important for you to put the problem in your own words here and also, to improve future usage or the questions answered here. Oct 16, 2019 at 18:40
• Consider $a=4$ and $b=5$. Are there any natural numbers $m$ and $n$ that you can pick that makes $3*4m=3*5n$? Oct 16, 2019 at 19:23

Hint: $$3^m a= 3^nb$$ means $$\frac {3^m a}{3^nb}=1$$ which means $$\frac ab = \frac{3^n}{3^m} = 3^{n-m}$$.

So this is just say that you can get from $$a$$ to $$b$$ multiplying or dividing by $$3$$ some number of times. Example $$4R 36$$ because $$3^2*4 = 3^0*36$$. But $$24 \not R 36$$ becuse $$24 = 8*3$$ while $$36=4*3^2$$ and there is no way you can make $$8*3^k = 4*3^j$$.

So reflexive.

Are there $$m,n$$ so that $$3^n a = 3^m a$$ for all $$a$$? Well, whenever $$m=n$$ that is true.

Symmetric.

If $$3^ma = 3^nb$$ does that mean there are $$m',n'$$ so that $$3^{m'}b = 3^{n'} b$$. Well, of course! Just use the same values. $$3^nb =3^ma$$.

(Or you could use any values where $$n'-m' = m-n$$)

Transitive.

If $$3^ma = 3^nb$$ and $$3^jb =3^kc$$ does that mean there are $$w,v$$ so that $$3^wa = 3^vc$$?

Well, yes use my dividing/multiply by three to get from one to the other.

$$3^ma = 3^nb$$ so

$$3^{m+j} = 3^{n}(3^jb)= 3^{n}(3^kc)= 3^{n+k}$$. Yep.

And anti-symmetric.

Does $$3^n a= 3^mb$$ simultaneously $$3^kb =3^ja$$ imply $$a=b$$? Of course not. $$a = 3^{m-n} b=3^{j-k} b$$ and it's certainly possible to have $$m-n=j-k$$ be other values than $$0$$.

format: plain


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# using the derivative definition how do I find f(x) = -6(square root over x).

Question

using the derivative definition how do I find f(x) = -6(square root over x).

1 Rating


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# zbMATH — the first resource for mathematics

Structural equivalence and ET0L grammars. (English) Zbl 0794.68093
Ésik, Zoltán (ed.), Fundamentals of computation theory. 9th international conference, FCT ’93, Szeged, Hungary, August 23-27, 1993. Proceedings. Berlin: Springer-Verlag. Lect. Notes Comput. Sci. 710, 430-439 (1993).
Summary: For a given context-sensitive grammar $$G$$ we construct ET0L grammars $$G_ 1$$ and $$G_ 2$$ that are structurally equivalent if and only if the language generated by $$G$$ is empty, which implies that structural equivalence is undecidable for ET0L grammars. This is in contrast to the decidability result for the E0L case. In fact, we show that structural equivalence is undecidable for propagating ET0L grammars even when the number of tables is restricted to be at most two. A stronger notion of equivalence that requires the sets of syntax trees to be isomorphic is shown to be decidable for ET0L grammars.
For the entire collection see [Zbl 0875.00104].

##### MSC:
 68Q42 Grammars and rewriting systems


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# [tex4ht] latex -> mediawiki

Sat Jul 23 11:08:15 CEST 2011

On 07/23/2011 06:23 AM, Victor Ivrii wrote:
> It would be interesting and useful to have latex->mediawiki converter
>
> MathJax (which work almost perfectly with mediawiki) should take care
> of math but converter is supposed to convert (say)
>
> \section{My section} to =My section=
> \subsection{My subsection} to ==My subsection==
> \subsubsection{My subsubsection} to ===My subsubsection===
> \paragraph{My paragraph} to ===={My paragraph}====
> \subparagraph{My subparagraph} to ====={My subparagraph}=====
>
> also it should take care of theorem-like envirements and placing (but
> definitely not producing) graphics
>
> Also \texbf{...} should become '''...''' and \textit{...} and
> \emph{...} and \textsl{...} should become ''...'' (all quotes are
> single)

It is not overly difficult to write a package. Anybody who knows the
basics of configuring TeX4ht can easily write a custom configuration
file (say, mediawiki.cfg) that can go with a htlatex run to generate
output for mediwiki.

However, I don't make any promise to write one as I am struggling to
finish html5 and mathml3 packages for TeX4ht for months now. Couldn't
complete yet owing to the pressures at the day job.

> With math also some fix is needed: between two single ' should space
> be inserted (otherwise mediawiki decides that it is its own formatting
> command).

Those things are relatively easier to implement.

> I am not sure that this belongs to tex4ht (may be to some spin-off
> project). As another discussion has shown there is a certain interest
> in different formats (html, off, epub) and there are different ways of
> dealing with math (should be math converted to images, or left as it
> is, so either MathJax takes care or, for example, internal to
> mediawiki image generation should take care).
>
> This is not a feature request but just an invitation to discussion.

TeX4ht is an ingenious piece of software ever written in TeX macro
language. It is infinitely complex since it is infinitely configurable
and flexible except for its incompatibility with unicode input. TeX4ht
can generate any kind of html, XML, MathML, custom formats like
mediawiki, mathjax, jsmath, etc. Hence, all the above formats can fit
well into the scheme of TeX4ht.

Best
--


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# Essentials III

In the movie about Stephen Hawking’s life, The Theory of Everything, there is talk about a single unifying equation that would explain everything in the universe. I must assume the real Stephen Hawking is familiar with Feynman’s unworldliness equation: U = 0, which – as Feynman convincingly demonstrates – effectively integrates all known equations in physics. It’s one of Feynman’s many jokes, of course, but an exceptionally clever one, as the argument convincingly shows there’s no such thing as one equation that explains all. Well… What Feynman proves is that one can actually ‘hide‘ all the equations you want in a single equation, but it’s just a trick. As Feynman puts it: “When you unwrap the whole thing, you get back where you were before.” Having said that, some equations in physics are obviously more fundamental than others. Besides Maxwell’s equations – on which I shall come back in a moment – these are the equations one should probably think of as being more fundamental than others:

1. Einstein’s mass-energy equivalence: m = E/c2
2. The wavefunction: ψ = ψ(x, t) = ei(ωt − kx) = ei(kx − ωt)
3. The two de Broglie relations that determine the spatial and temporal frequency of the wavefunction:

ω = E/ħ and k = p/ħ

And, of course, Schrödinger’s equation:

The wavefunction, and Schrödinger’s equation, look the most formidable, of course. However, the mathematical symbols describe rather simple things. First look at the wavefunction: ψ = ei(kx − ωt). That’s Euler’s formula:

eiθ = cosθ + isinθ = cos(kx − ωt) + isin(kx − ωt)

So it’s just a two-dimensional equivalent of writing y = cos(kx – ωt) or, more generally, y = F(x−vt).

Huh? Yes. The y = F(x−vt) function describes any wave that’s traveling in the positive x-direction. In case you forgot, let me remind you of how we’d find the same value for F by simultaneously (a) adding the distance Δx = vΔt to, and (b) subtracting the time vΔt from, the argument. We wrote:

F[x+Δx−v(t+Δt)] = F(x+vΔt–ct−vΔt) = F(x–ct)

What we do here is converting something that’s expressed in time units into something that’s expressed in distance units, and vice versa, as both dimensions are related through the wave velocity: = x/t = Δx/Δt. You’ll say: how do you get from F(x−vt) to cos(kx – ωt)? That’s simple: the argument of a sinusoidal function is expressed in radians, rather than in meter or in seconds. So, using the v = λf formula (the velocity of a regular wave is its frequency times its wavelength), we’d write y = cos(2π(x–ct)/λ). We divide x–ct by the wavelength λ here, so the (x–ct)/λ factor is something that’s measured not in meter but in wavelengths, and then we multiply the thing with 2π because we know that one cycle (i.e. one wavelength) corresponds to 2π radians—and we need an argument expressed in radians in our sinusoidal function, because it’s an angle! It has to be. Now, k = 2π/λ and ω = 2πf, and so we substitute and we get what we wanted to get: cos(2π(x–vt)/λ) = cos[(2π/λ)x–(2πv/λ)t) = cos[kx–(2πf)t) = cos(kx – ωt). I managed to explain this to my 16-year old son, Vincent, so you should have no difficulty in understanding this.

So ψ = ei(kx − ωt) is the same old wavefunction, really. The only thing is that this ei(kx − ωt) function gives you two functions for the price of one. 🙂 To be precise, a sine and a cosine function are one and the same function, really, but with a phase difference of 90 degrees, so that’s π/2 radiansThat’s illustrated below: cosθ = sin(θ+π/2).

So all you need to understand is this:

eiθ = cosθ + isinθ = sin(θ+π/2) + isinθ

What does this say, really? This formula tells us that our ψ(x, t) = ei(ωt − kx) wavefunction is two-dimensional indeed. Now, the term ‘dimension’ is one of those ubiquitous terms in physics and in mathematics that means different things in different contexts, so I should be more precise here but… Well… I can’t. The wavefunction consists of two parts. One part is referred to as ‘real’ (I am talking about the cosine here) while the other (the sine) is referred to as ‘imaginary’. However, both are equally real, of course, because… Well… Both are equally essential: the one bit comes with the other, and vice versa.

Now, we could keep track of the real and imaginary part of our wavefunction in a different way. Instead of writing cosθ + isinθ, we could write our wavefunction as a vector function:

ψ(x, t) = [cos(kx−ωt), sin(kx−ωt)] = [cos(kx−ωt+π/2), sin(kx−ωt)]

Why don’t we do that? We could, but it turns out the imaginary unit is something more than just a delimiter: it’s not just some symbol separating values in some sequence. We use it in calculations. For example, we can move it to the other side in Schrödinger’s equation by multiplying both sides of the equation with its multiplicative inverse 1/i = i1 = −i. In fact, if we also move ħ, then we get an expression which looks somewhat simpler:

So why is it that physicists never write it that way? I am not sure. It’s probably because of that expression using the Hamiltonian operator (H), which is used in the equation that’s engraved on Schrödinger’s bust, and on his grave:

The simpler expression, with i and ħ on the right-hand side of the equation, is interesting though, because the expression makes us think of the geometry involved. Huh? Yes. You should think of two things here:

1. Multiplying some vector quantity with the imaginary unit (i) amounts to a (counter-clockwise) rotation by 90 degrees.
2. When you see a Laplacian in physics – as we do in Schrödinger’s equation (∇· = ∂2/∂x2  + ∂2/∂y+ ∂2/∂z2) − it will usually represent the flux density of the (gradient) flow of a function.

Huh? What? Well.. Yes. Let me first explain the rotation, as that’s the easiest thing to explain. The following illustration should do the trick:

[By the way, I did some posts on complex numbers myself, but here’s a link to an author who does a really good job at explaining them in a very intuitive way. Have fun with it!] As for the geometrical interpretation of the Laplacian in Schrödinger’s equation, I should refer you to my posts on vector analysis, but let’s review the basics here, using the example of heat diffusion in some material body, which was our first example of a vector equation:

h = –κT

We encountered the same equation elsewhere. For example, if the electric potential is Φ, then we can immediately calculate the electric field using the following formula:

E = –Φ

It’s the same thing—mathematically speaking, that is. Now we need a little bit of math to understand these equations in the way we want to understand them, and that’s in some kind of ‘physical‘ way (as opposed to just the math side of it). So… Well… Let me spell it out:

1. The heat flow is represented by the h vector, and its magnitude is the thermal energy that passes, per unit time and per unit area, through an infinitesimally small isothermal surface, so we write: h = |h| = ΔJ/ΔA.
2. The direction of the heat flow is opposite to the direction of the gradient vector T, so we’ve got a minus sign in our equation. That’s simple enough to understand: heat flows from higher to lower temperature, i.e. ‘downhill’, so to speak. So that’s the minus sign. Note that the gradient operator () acts on a scalar-valued function (T), aka a scalar field, but yields a vectorT = (∂T/∂x, ∂T/∂y, ∂T/∂z). [Always remember: boldface symbols represent vectors, while symbols in regular font (e.g. T) are scalars or scalar functions: T = T(x, t) = T(x, y, z, t).]
3. The magnitude of h is proportional to the magnitude of the gradient T, with the constant of proportionality equal to κ (kappa), which is called the thermal conductivity: |h| = h = κ|T|.
4. The gradient of T, i.e. T = (∂T/∂x, ∂T/∂y, ∂T/∂z), measures the rate of change of T, but in a particular direction. In fact, the the rate of change of T in any direction will be the component of the T vector in that direction. Now, the magnitude of a vector component is always smaller than the magnitude of the vector itself, except if it’s the component in the same direction as the vector, in which case the component is the vector. [If you have difficulty understanding this, read what I write once again, but very slowly and attentively.] Therefore, the direction of T is the direction in which the (instantaneous) rate of change of T is largest. In Feynman’s words: “The gradient of T has the direction of the steepest uphill slope in T.”

But so we were talking the Laplacian, which is equal to ∇2T = ·T = equation. The little dot in the middle here matters—and very much so! We’re multiplying the vector differential operator ∇ (i.e. the gradient) with itself here, so we write:

=· = (∂/∂x, ∂/∂y , ∂/∂z)·(∂/∂x, ∂/∂y , ∂/∂z) = ∂2/∂x2  + ∂2/∂y+ ∂2/∂z2

Note that the Laplacian operator (∇2) would, once again, act on a scalar-valued function here (T), aka a scalar field, but that it yields another scalar-valued function, unlike the gradient operator  = (∂/∂x, ∂/∂y, ∂/∂z). Now, the · operator is referred to as the divergence. We wrote:

·h = div h = the divergence of h

Now we can really start talking about what I wrote above: if you see a Laplacian in physics – as we do in Schrödinger’s equation − it will usually represent the flux density of the (gradient) flow of a function. That’s because the physical significance of the divergence is related to the so-called flux of a vector field: it measures the magnitude of a field’s source or sink at a given point. Continuing our example with temperature, consider, for example, air as it is heated or cooled. The relevant vector field will now be the velocity of the moving air at a point. If air is heated in a particular region, it will expand in all directions such that the velocity field points outward from that region. Therefore the divergence of the velocity field in that region would have a positive value, as the region is a source. If the air cools and contracts, the divergence has a negative value, as the region is a sink. Of course, in a block of metal, there’s nothing that moves, except for the heat flow itself, as illustrated below.

The less intuitive but more accurate definition of the divergence is the following: the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. Now, heat is conserved, and this conservation law is expressed as follows:

I am tempted to show you how we get this result, but I don’t want to make this post too long, so I’ll just refer you to Feynman on this. Just note that q is the heat per unit volume (as opposed to Q, which is the total heat). The point is that we can now combine the equation above with our h = –κT expression so as to get the following:

The last step in this rather long development is the following: it is rather logical to assume that the rate of change of the heat per unit volume (i.e. q) will be proportional to the rate of change of the temperature (T), so we can introduce another proportionality constant cv, which, in physics, is referred to as the specific heat (per unit volume) of the material (aka the (volume) heat capacity of the substance), so we write: dq/dt = cv·(dT/dt). Combining all gives us the diffusion equation we wanted to find:

This equation, which – as we demonstrated – models the flow of something (it modeled the flow of heat, or thermal energy, in our example) is very similar to Schrödinger’s equation:

Both equations model a flow, and involve a flux density of something. Let’s do a dimensional analysis here. On the left-hand side, we have a quantity (∂ψ/∂ψ) that’s expressed per second, while on the right-hand side, we have ∇2ψ = ·ψ, which is expressed per square meter, so it is a flux density alright! Now, in our heat diffusion equation, the dimensions came out alright because of the dimension of the diffusion constant, i.e.  D = κ/cv. The specific heat (cv) is expressed in J/(m3·K), while the thermal conductivity, κ, is expressed in W/(m·K) = J/(m·s·K). Dividing the latter by the former gives us a diffusion constant with dimension m2/s, so it works out: we get something expressed per second on both sides of the equation.

The diffusion constant in Schrödinger’s equation is iħ/2m, whose dimension is [J·s]/[J/(m2/s2] = m2/s, so that works out too! We get something expressed per second on both sides of our equation… Except… Well… We conveniently forgot about i! Is that part of the ‘dimension’ of the right-hand side?

I’d say: it is and it isn’t. Remember that the wavefunction (and, hence, Schrödinger’s equation) gives us two functions for the price of one: we’ve got a real part, and we’ve an imaginary part. Now, multiplying both by amounts to a rotation by 90 degrees, so what’s happening here is that the real part of ∇2ψ becomes the imaginary part of dψ/dt, and the imaginary part of ∇2ψ becomes the real part of dψ/dt. Well… And there’s also this scaling factor ħ/2m, of course.

Huh? Yes. Think of it. If we write dψ/dt as dψ/dt = a + ib, and ∇2ψ as ∇2ψ = c + id, then Schrödinger’s equation tells us that a + ib = (iħ/2m)·(c + id) = (ħ/2m)·(–d + i·c). Hence, a = –(ħ/2m)·d and b = (ħ/2m)·c, or:

1. Re(dψ/dt) = −(ħ/2m)·Im(∇2ψ)
2. Im(dψ/dt) = (ħ/2m)·Re(∇2ψ)

This is very deep. First the minus sign. Does that reflect the clock- or counter-clockwise direction of rotation of the θ argument in the wavefunction matter? No. Time goes in one direction only and, in quantum math, that’s reflected in the argument going clockwise with time. If we’d replace by j = −i, we’d have a complex-number system that works exactly the same (j= (−i)= i= −1). We would just write dψ/dt as dψ/dt = a + jb, and ∇2ψ as ∇2ψ = c + jd, but so we’d still find that a + jb = (jħ/2m)·(c + jd) = (ħ/2m)·(–d + j·c), so we’d write the same: Re(dψ/dt) = −(ħ/2m)·Im(∇2ψ) and Im(dψ/dt) = (ħ/2m)·Re(∇2ψ).

What these bizarre equations above actually mean is not immediately clear. We had an equation that established a relation between (1) a quantity changing in time (i.e. a quantity expressed per second) and (2) its flux density in space (i.e. a quantity expressed per square meter). To be precise, on the left-hand side of Schrödinger’s equation we had the time rate of change of an amplitude, while on the right-hand side we have its flux density. […] Well… We should be precise here: that Laplacian gives us the flux density of the gradient flow of ψ. But… Well… What?

I think of two things right now. First, the famous words that Hermann Minkowski used as introduction to his famous talk at the 80th Assembly of German Natural Scientists and Physicians on September 21, 1908:

“The views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth, space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.”

As you know, Minkowski used these words as an introduction to his theory on four-dimensional space-time, which was nothing but a mathematical formalization and generalization of Einstein’s special relativity theory of relativity. So these words have nothing to do with quantum physics, although Planck had already introduced the fundamental quantum-mechanical postulate (quantization of energy) a decade before. Nevertheless, the Re(dψ/dt) = −(ħ/2m)·Im(∇2ψ) and Im(dψ/dt) = (ħ/2m)·Re(∇2ψ) equations do lead us to thinking that the time rate of change of the wavefunction and the flux density of its gradient flow are both the manifestation of a deeper, underlying, reality. In the case of the heat diffusion equation, that underlying reality is the flow of (heat) energy. Now, as our wavefunction describes some particle moving through spacetime, and remembering Einstein’s mass-energy equivalence relation, it’s obvious we should be able to find a similar interpretation for Schrödinger’s equation.

That brings me to the second point, which I mentioned in other posts too: there is that striking mathematical similarity between (a) the real and imaginary part of the traveling amplitude, and (b) the E and B vectors that characterize the traveling electromagnetic field. Look at the two illustration below.

1. The first one visualizes the real and imaginary part of the ψ(x) = eikx function (so that’s the ψ(x, t) = ei(ωt − kx) at t = 0, or at some other fixed point in time). So we’re missing the time dimension here.
2. The second illustration shows how the electric field vector E varies in space as well as in time, but it does not show the magnetic field vector B.

It’s easy to imagine how an animated version of the ψ(x, t) = ei(ωt − kx) wavefunction would look like: exactly the same as that animation which shows E! Indeed, note that – how convenient ! – E actually rotates clockwise, just like the argument (θ) of our wavefunction! Hence, it is tempting to think we could de-construct E, i.e. we could think of E as consisting of a ‘real’ and ‘imaginary’ part as well! Now that is an interesting idea, isn’t it? 🙂

But so we don’t have B. How would B look like? The animation above shows the electric field vector (E) for a so-called circularly polarized electromagnetic wave. You know the magnetic field vector B is always orthogonal to E, as shown below, with a magnitude that’s equal |B| = B = |E|/c = E/c. The animation below shows how a linearly polarized electromagnetic wave (i.e. a plane wave, really) is supposed to travel through space: the red oscillation is the electric field vector, and the blue oscillation is the magnetic field vector (never mind the 1/c factor here).

The mathematical description of how an electromagnetic wave travels through spacetime is given by Maxwell’s Laws:

I highlighted the linkages between E and B, and I talked about how that works in detail before, so I won’t repeat myself here too much. The point is: Maxwell’s Laws are also structured around time derivatives (∂E/∂t and ∂B/∂t) and flow densities—even if we’re talking circulation densities here, because the left-hand side of the equations with the time derivatives involve the curl operator (×), which is a vector operator that describes the (infinitesimal) rotation of the E and B vector fields respectively. In fact, making abstraction of the electric current (j) and charge densities (i.e. the electric field caused by static charges, which is captured by the first equation), as well as choosing our time and distance units such that c = 1, we get:

1. ∂B/∂t =×E
2. E/∂t =×B

That’s structurally very similar to what we wrote above:

1. Re(dψ/dt) = −(ħ/2m)·Im(∇2ψ)
2. Im(dψ/dt) = (ħ/2m)·Re(∇2ψ)

The only difference between the two mathematical descriptions is that phase shift: Euler’s formula incorporates a phase shift—remember: sinθ = cos(θ − π/2)—and so you don’t have that with the E and B vectors. But then that’s why bosons and fermions are different, of course! This brings me to the final thought I want to make here. You’ll remember we found an equation that was very similar to the dq/dt = −·h equation when we were talking about the energy of fields and the Poynting vector:

Indeed, the similarity triggers the following reflection: if there’s a ‘Poynting vector’ for heat flow (h), and for the energy of fields (S), then there must be some kind of ‘Poynting vector’ for amplitudes too! I don’t know which one, but it must exist! It’s going to be some complex vector, no doubt! But it should be out there. 🙂

Post scriptum 1: In my post on gauges, in which I introduced much more advanced stuff than what you need here (like the concept of the vector potential, usually denoted by A), we derived two equations to describe the traveling electromagnetic field. They were the following:

To see these equations are really structurally similar to Schrödinger’s equation, it suffices to just write out the second one:

It’s again the same thing: a time derivative on one side, and a Laplacian on the other. [Just equate ρ with zero, because we’re interested in the propagation mechanism of the wave only.] As mentioned, the two equations are so similar that we might want to think of how to ‘de-construct’ E in a ‘real’ and an ‘imaginary’ part. But so we’ve got two equations above—one involving the electric potential (ϕ) and one involving the vector potential A—and only one wave equation (i.e. Schrödinger’s equation), so how does that work. Well… As I showed above, Schrödinger’s equation is actually two equations as well: one for the real and one for the imaginary part of the wavefunction. So, yes, “same-same but different”, as they say in Asia. 🙂

Not convinced? Well… There was a quote of Feynman which I never quite understood—till now, that is: “The vector potential A (together with the scalar potential ϕ that goes with it) appears to give the most direct description of the physics. This becomes more and more apparent the more deeply we go into the quantum theory. In the general theory of quantum electrodynamics, one takes the vector and scalar potentials as the fundamental quantities in a set of equations that replace the Maxwell equations.”

I think I finally start to understand what he’s saying here, although I still have a lot to learn. For starters, the E and B vectors have a physical significance that is totally lacking – at our (limited) level of understanding of quantum physics here, for those real and imaginary parts of the wavefunction. Indeed, we’d associate forces with E and B. More specifically, the force on a moving charge would be equal to:

F = qE + qv×B

I guess we’ll learn all about it once we start studying quantum electrodynamics, or quantum field theory. In the meanwhile, it’s nice we can already see what it’s going to look like. As Feynman puts it: “The same equations have the same solutions.” And he’s explicit about what he means with that: “The equations for many different physical situations have exactly the same appearance. Of course, the symbols may be different—one letter is substituted for another—but the mathematical form of the equations is the same. This means that having studied one subject, we immediately have a great deal of direct and precise knowledge about the solutions of the equations of another.” Well… That says it all, doesn’t it? 🙂

Post scriptum 1: In my post on the de Broglie relations, I mentioned this funny E = m·vformula that comes out of them. Just multiply them:

1. f·λ = (E/h)·(h/p) = E/p
2. v = f·λ  ⇒ f·λ = v = E/p ⇔ E = v·p = v·(m·v)

⇒ E = m·v2

That’s weird, isn’t it? E = m·v2? It reminds one of the kinetic energy formula of course—K.E. = m·v2/2—but… Well… That factor 1/2 should not be there. Now I won’t repeat all of my reflections on that here (just check that post of mine if you’d would to freewheel a bit). I just want to show you that we do not find that weird formula when substituting ψ for ψ = ei(kx − ωt) in Schrödinger’s equation, i.e. in:

To keep things simple, we’ll just assume one-dimensional space, so ∇2ψ = ∂2ψ/∂x2. The time derivative on the left-hand side is ∂ψ/∂t = −iω·ei(kx − ωt). The second-order derivative on the right-hand side is ∂2ψ/∂x2 = (ik)·(ik)·ei(kx − ωt) = −k2·ei(kx − ωt) . The ei(kx − ωt) factor on both sides cancels out and, hence, equating both sides gives us the following condition:

iω = −(iħ/2m)·k2 ⇔ ω = (ħ/2m)·k2

Substituting ω = E/ħ and k = p/ħ yields:

E/ħ = (ħ/2m)·p22 = m2·v2/(2m·ħ) = m·v2/(2ħ) ⇔ E = m·v2/2

So now it’s OK!? The only logical conclusion is that we must be doing something wrong when multiplying the two de Broglie equations. To be precise: our v = f·λ equation is probably wrong. Why? Not sure. It’s just something one shouldn’t apply to our complex-valued wavefunction. The ‘correct’ velocity formula for the complex-valued wavefunction should have that 1/2 factor, so we’d write 2·f·λ = v to make things come out alright. But where would this formula come from? The period of cosθ + isinθ is the period of the sine and cosine function: cos(θ+2π) + isin(θ+2π) = cosθ + isinθ, so T = 2π and f = 1/T = 1/2π do not change.

But so that’s a mathematical point of view. From a physical point of view, it’s clear we got two oscillations for the price of one: one ‘real’ and one ‘imaginary’—but both are equally essential and, hence, equally ‘real’. You may think the answer must lie in the distinction between the group and the phase velocity when we’re combining waves. Indeed, the group velocity of a sum of waves is equal to vg = dω/dk. In this case, we have:

vg = d[E/ħ]/d[p/ħ] = dE/dp

We can now use the kinetic energy formula to write E as E = m·v2/2 = p·v/2. Now, v and p are related through m (p = m·v, so = p/m). So we should write this as E = m·v2/2 = p2/(2m). Substituting E and p = m·v in the equation above then gives us the following:

dω/dk = d[p2/(2m)]/dp = 2p/(2m) = v= v

However, for the phase velocity, we can just use the v= ω/k formula, which gives us that 1/2 factor:

v= ω/k = (E/ħ)/(p/ħ) = E/p = (m·v2/2)/(m·v) = v/2

Riddle solved! 🙂


text:
Online Education In this page, we are providing Work, Power And Energy Class 9 Extra Questions and Answers Science Chapter 11 pdf download. NCERT Extra Questions for Class 9 Science Chapter 11 Work, Power And Energy with Answers will help to score more marks in your CBSE Board Exams.

## Online Education for Class 9 Science Chapter 11 Extra Questions and Answers Work, Power And Energy

Extra Questions for Class 9 Science Chapter 11 Work, Power And Energy with Answers Solutions

### Work, Power And Energy Class 9 Extra Questions Very Short Answer Type

Work And Energy Class 9 Extra Questions Question 1.
Define the following terms.
(a) Work was done
(b) Energy
(c) Mechanical energy
(d) Kinetic energy
(e) Potential energy
(f) Power
(g) Commercial unit of energy.
(a) Work done: Work done by a force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force.
(b) Energy: Energy of a body is defined as the capacity or ability of the body to do work.
(c) Mechanical energy: Mechanical energy includes kinetic energy and potential energy.
(d) Kinetic energy: The energy possessed by a body by virtue of its motion.
(e) Potential energy: The energy possessed by a body due to its position or configuration.
(f) Power: Power is defined as the rate of doing work or the rate of transfer of energy.
(g) Commercial unit of energy: The energy used in households, industries, and commercial establishment are usually expressed in kilowatt-hour.
1 kWh 1 unit = 3.6 x 106J

Work And Energy Class 9 Extra Questions Numericals Question 2.
Write down the type of energy stored in
(a) spring of a watch
(b) flowing water
(c) rolling stone
(d) raised hammer
(e) running athlete
(a) potential energy
(b) kinetic energy
(c) kinetic energy
(d) potential energy
(e) kinetic energy.

Work And Energy Class 9 Important Questions And Answers Question 3.
What will be the kinetic energy of a body when its mass is made four-time and the velocity is doubled?
Initial kinetic energy,
$$E_{K_{i}}=\frac{1}{2} m v^{2}$$
Final kinetic energy,
$$E_{K_{f}}=\frac{1}{2}(4 m) \times(2 v)^{2}$$
= 16 x $$\frac {1}{2}$$mυ2
$$E_{K_{f}}=16 E_{K_{i}}$$

Class 9 Science Chapter 11 Extra Questions Question 4.
If we lift a body of 7 kg vertically upwards to a height of 10 m, calculate the work done in lifting the body.
Given, m = 7 kg
s = 10m
Workdone, W = F x s
E = mg x s
W = 7 x 10 x 10 J
w = 7000 J

Class 9 Work And Energy Extra Questions Question 5.
State the transformation of energy that takes place when

• Green plants prepare their food.
• Head of a nail hammered hard and it becomes hot.

• Solar energy of sun into chemical energy.
• The kinetic energy of the hammer into heat energy.

Extra Questions Of Work And Energy Class 9 Question 6.
How much work is done by a man who tries to push the wall of a house but fails to do so?
W = Fs = 0
As there is no displacement.

Work And Energy Class 9 Extra Questions And Answers Question 7.
Establish a relationship between SI unit and commercial unit of energy.
SI unit of energy is joule and the commercial unit of energy is the joule.
1kWh = 1000 W x 3600 s = 3.6 x 106J

Questions On Work Energy And Power For Class 9 Question 8.
Write down the energy transformation taking place
(a) In electric bulb
(b) In torch
(c) In the thermal power station
(d) In solar cell
(e) Electric heater
(a) Electricity into light energy
(b) The chemical energy of the cell into light and heat energy
(c) The chemical energy of fuel into electricity
(d) Solar energy into electricity
(e) Electricity into heat energy.

Class 9 Science Chapter 11 Extra Questions And Answers Question 9.
A body of mass m is moving in a circular path of radius r. How much work is done on the body?
Zero. This is because the centripetal force acting on the body is perpendicular to the displacement of the body.

Extra Questions On Work And Energy Class 9 Question 10.
A horse of mass 200 kg and a dog of mass 20 kg are running at the same speed. Which of the two possesses more kinetic energy? How?
The kinetic energy of the horse is more as kinetic energy is directly proportional to mass.

Class 9 Science Ch 11 Extra Questions Question 11.
What is the condition for work done to be positive?
For positive work, the angle between force and displacement should be acute.

Class 9 Science Work And Energy Extra Questions Question 12.
Write down the relation between kinetic energy and momentum of a body.
$$E_{K}=\frac{p^{2}}{2 m}$$
Ek = kinetic energy of a body
p = momentum of the body
m = mass of the body.

Work Energy And Power Class 9 Extra Questions Question 13.
A cyclist comes to a skidding stop at 50 m. During this process, the force on the cycle due to the road is 1000 N and is directed opposite to the motion. How much work does the road do on the cycle?
Given,
Displacement, s = 50 m
Force, F = – 1000 N
Workdone, W = F x s
W = -1000 x 50J
W = 50000J

Class 9 Physics Work And Energy Extra Questions Question 14.
A boy pushes a book by applying a force of 40 N. Find the work done by this force as the book is displaced through 25 cm along the path.
Given, Force (F) = 40 N
Displacement (s) = 25 cm = x 10-2 m
Workdone, W = F x s
= 40 x 25 x 10-2
= 10J
∴ W = 10J

### Work, Power And Energy Class 9 Extra Questions Short Answer Type 1

Work And Energy Class 9 Questions Question 1.
State law of conservation of energy and law of conservation of mechanical energy.
Law of conservation of energy: Energy can neither be created nor be destroyed, it can only be transformed from one form to another.
Conservation of mechanical energy: If there is no energy, then the mechanical energy of a system is always constant.

Work And Energy Class 9 Important Questions Question 2.
Define (a) 1 joule (b) 1 watt.
(a) 1 joule is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.
(b) 1 watt is the power of an agent, which does work at the rate of 1 joule per second.

Ch 11 Science Class 9 Extra Questions Question 3.
Write down SI unit of the following quantities.
(a) work
(b) kinetic energy
(c) potential energy
(d) power
(a) joule (J)
(b) joule (J)
(c) joule (J)
(d) watt (W).

Work And Energy Class 9 Extra Numericals Question 4.
What is the sequence of energy change that takes place in the production of electricity from adam?
The potential energy of stored water is converted into the rotational kinetic energy of turbine blades. The rotational kinetic energy of turbine blades is finally converted into electric energy by the generator.

Work Power And Energy Class 9 Extra Questions Question 5.
A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has larger kinetic energy?
The relation between kinetic energy and momentum
Given,

Question 6.
Why a man does not do work when he moves on a level road while carrying a box on his head?
When a man carries a load on his head, the angle between displacement (s) and force (F) is 900. Therefore work done is zero.

Work, Power And Energy Class 9 Extra Questions Numericals

Question 7.
If an electric iron of 1200 W is used for 30 minutes every day, find electric energy consumed in the month of April.
Given,
Power, P = 1200 W
time, t = 30 minutes
Power, p = $$\frac {W}{t}$$ = $$\frac {E}{t}$$
E = P x t
Energy consumed, E = 1200 x 30 x 60
= 2.16 x 106J = 2.16MJ

Question 8.
What is work done by a force of gravity in the following cases?
(a) Satellite moving around the Earth in a circular orbit of radius 35000 km.
(b) A stone of mass 250 g is thrown up through a height of 2.5 m.
(a) Zero
(b) Given,
mass (m) = 250 g = 0.25 kg
height (h) = 2.5 m
Workdone, W = Fs = mgh
= 0.25 x 10 x 2.5
= 6.25 J
W = 625 J

Question 9.
A car and a truck have kinetic energies of 8 x 10 J and 9 x 10 J respectively. If they are brought to a halt at the same distance, find the ratio of the force applied to both the vehicles.
Given,
Kinetic energy of car, $$\frac {1}{2}$$ = 8 x 10 J
Kinetic energy of truck, $$\frac {1}{2}$$ = 9 x 10 J
W = Fs
Work done,

Question 10.
A bus of mass 10,000 kg is moving with a velocity 90 km/h. Calculate the work done to stop this bus.
Given,
mass of object, m = 1000 kg
Initial velocity of object, u = 90 km/h

= 25 m/s
Work done = $$\frac {1}{2}$$ mυ2 – $$\frac {1}{2}$$ mu2 = 0 – $$\frac {1}{2}$$ x 10000 x (25)2 = 312.5 x 104 J

### Work, Power And Energy Class 9 Extra Questions Short Answer Type 2

Question 1.
The velocity of a body moving in a straight line is inereásed by applying constant force F, for some distance in the direction of motion. Prove that the increase ¡n the kinetic energy of the body is equal to the work done by the force on the body.
Let us consider an object lying on a frictionless surface having mass ‘m’.

A force of constant magnitude ‘F’ is acting on the body. Here initial velocity of the body is u and the final velocity is u. As there is no dissipative forces, work on the body will be stored in the form of a change in kinetic energy. It can be proved as
W = Fs …………(1)
and hence from the equation of motion
υ2 – u2 = 2as
s = $$\frac{v^{2}-u^{2}}{2 a}$$ ………….(2)
Asweknowthat F = ma
Using (1), (2) and (3),
W = ma x $$\frac{v^{2}-u^{2}}{2 a}$$
= $$\frac {1}{2}$$ m (υ2 – u2)
= $$\frac {1}{2}$$mυ2 – $$\frac {1}{2}$$mu2
= change in kinetic energy

Question 2.
Derive an expression for potential energy. Write Its SI unit.
When work is done on the body, the work is stored in the form of energy. Consider an object of mass, m. Let it be raised through a height, h from the ground. A force is required to do this. The minimum force required to raise the object is equal to the weight of the object, mg. The object gains energy equal to the work done on it. Let the work done on the object against gravity h be W.
i.e., W = force x displacement = mgh

Since work done on the object is equal to mgh, an energy equal to mgh units is gained by the object. This is the potential energy (Ep) of the object.
∴ Ep = mgh
SI unit of potential energy is the joule (J).

Work, Power And Energy Class 9 Extra Questions Numericals

Question 3.
A girl having a mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 ms1 by applying a force. The trolley comes to rest after traversing a distance of 16 m.
(a) How much work is done on the trolley?
(b) How much work is done by the girl?
Given, mass of girl, m = 35 kg
mass of trolley, m = 5 kg.
initial velocity of trolley, u = 4 m/s

(a) using work done = change in kinetic energy
W = $$E_{K_{f}}-E_{K_{i}}$$
= o – $$\frac {1}{2}$$ x 40 x (4)2 = – 320
∴ W = 320 J_J
(b) Work done by the girl = 0.

Question 4.
Express kilowatt in terms of joule per second. A 150 kg car engine develops 500 W for each kg. What forces does it exert in moving the car at a speed of 20 ms-1?
1 kW = iooo w = 103 J/S
Given, mass (m) = 150 kg
power (P) = 500 W
velocity (u) = 20 m/s
Using P = Fυ
or, 500 = F x 20
= F =$$\frac {500}{20}$$ = 25
F= 25 N

Question 5.
How is the power related to the speed at which a body can be lifted? How many kilogrims will a man working at the power of 100 W, be able to lift at constant speed of 1 mr1 vertically? (g = 10 ms-2)
We know that,

or, p = $$\frac {F.s}{t}$$ P = Fυ = mg.υ [F = mg]
where, F = force
s = displacement
υ = velocity
t = time
Given,
power (P) = 100 W
velocity (υ) = 1 m/s
g = 10 m/s2
or, 100 = m x 10 x 1
∴ m = 10 kg

Question 6.
A car of mass 2000 kg is lifted up a distance of 30 m by a crane in 1 minute. A second crane does the same job in 2 minutes. What is the power applied by each crane?
Given, mass of the car to be lifted, m = 2000 kg
height through which the car is to be lifted, h = 30 m
Time taken by first crane, t1 = 1 minute = 60 s
time taken by second crane, t1 = 2 minutes = 120 s
Amount of work done by each crane,
W = mgh = 2000 x 10 x 30 J
W= 6 x 106 J
Power of first crane, ‘
P1 = $$\frac{W_{1}}{t_{1}}=\frac{6 \times 10^{5} \mathrm{J}}{60 \mathrm{s}}$$ = 10kW
Power of second crane
P2 = $$\frac{W_{2}}{t_{2}}=\frac{6 \times 10^{5} \mathrm{J}}{120 \mathrm{s}}$$ = 10kW

Question 7.
Calculate the electricity bill amount for a month of June, if 6 bulbs of 100 W for5 hours, 4 tube lights of 60 W for 5 hours, a TV of 50 W for 6 hours are used per day. The cost per unit is ₹ 5.
Total energy used in a day = (6 x 100 x 5 + 4 x 60 x 5 + 1 x 50 x 6) Wh
= (3000+ 1200 +300)Wh = 4500 Wh
Total energy = 4.5 kWh = 4.5 unit
Total energy used in 30 days = 4.5 x 30 = 135 units
Bill amount = 135 units x ₹ 5 unit = ₹ 675.

### Work, Power And Energy Class 9 Extra Questions Long Answer Type

Question 1.
State the conditions for positive, negative, and zero work. Give at least one example of each.
1. Zero work: If the angle between force and displacement is 90°, then work done is said to be zero work.
Example: When a man carries a load on his head and moves on a level road. Work done by the man on the load is zero.

2. Positive work: Work done is said to be positive if the force applied on an object and displacement are in the same direction.
W = Fs

Example: Work done by the force of gravity on a falling body is positive.
3. Negative work: Work done is said to be negative if the applied force on an object and displacement is in opposite direction.
W = -Fs
Her displacement is taken to be negative (- s).

Example: Work done by friction force applied is negative on a moving body.

Question 2.
Give a reason for the following:
(a) A bullet is released on firing the pistol.
(b) An arrow moves forward when released from the stretched bow.
(c) Winding the spring of a toy car makes it to run on the ground.
(d) Falling water from a dam generates electricity.
(e) Winding the spring of our watch, the hands of the watch movement.
(a) The chemical energy of gun powder is converted into kinetic energy of the bullet.
(b) The elastic potential energy in a stretched bow is converted into kinetic energy of the arrow.
(c) The potential energy of a spring is converted into kinetic energy of the toy.
(d) The kinetic energy of water is converted into electric energy.
(e) The potential energy of spring due to its windings is converted into mechanical energy of the watch.

Question 3.
State the law of conservation of energy. Show that the energy of a freely falling body is conserved.
Energy can neither be created nor be destroyed, it can only be transformed ‘ m A from one form to another. The total energy before and after the transformation always remains constant.
Let us consider an object of mass ‘m’ dropped from a height h.
Total energy at point A

$$E_{T_{A}}=E_{K}+E_{P}$$
or, $$E_{T_{A}}=0+m g h$$
∴ $$E_{T_{A}}=m g h$$
Total energy at point B,
$$E_{T_{B}}=E_{T}+E_{p}$$
For finding out velocity at point B
apply υ2 – u2 = 2as
$$v_{\mathrm{B}}^{2}=2 g h$$ = 2gh
Hence, $$E_{T_{B}}=\frac{1}{2} m \mathrm{V}_{\mathrm{B}}^{2}+m g$$
$$E_{T_{B}}=\frac{1}{2} m(2 g h)=m g h$$
Here, $$E_{T_{A}}=E_{T_{B}}$$
Hence if there is no energy loss, total energy is conserved.

### Work, Power And Energy Class 9 Extra Questions HOTS

Question 1.
A running man has half the kinetic energy that a body of half of his mass has. The man speeds up by 1 m/s and then has the same kinetic energy as the boy. What are the original speeds of the man and the boy?
Let us take
mass of boy = m
mass of man = M
velocity of boy = u
velocity of man = υ
Here, m = $$\frac {M}{2}$$
Initially
EKof man = $$\frac {1}{2}$$ EK of boy
$$\frac {1}{2}$$ Mυ2 = $$\frac {1}{2}$$ mu2 x $$\frac {1}{2}$$
and $$\frac {1}{2}$$ Mυ2 = $$\frac {1}{2}$$ $$\frac{1}{2}\left(\frac{M}{2}\right) u^{2} \times \frac{1}{2}$$
∴ υ2 = $$\frac{u^{2}}{4}$$ ………….(1)

Finally
EK of man = EK of boy
$$\frac {1}{2}$$M(υ +1)2 = $$\frac {1}{2}$$ mu2
$$\frac {1}{2}$$M(υ +1)2 = $$\frac{1}{2}\left(\frac{M}{2}\right) u^{2}$$
(υ +1)2 = $$\frac{u^{2}}{2}$$
∴ υ +1 = $$\frac{u}{\sqrt{2}}$$ …………(2)
υ = $${\sqrt{2}}$$ + 1 = 2.41 m/s
and u = 4.82 m/s

Question 2.
Avinash can run with a speed of 8 m/s against the frictional force of 10 N, and Kapil can move with a speed of 3 m/s against the frictional force of 25 N. Who is more powerful and why?
Power can be expressed as, P = Fu
Power of Avinash, P = Fu = 10 x 8 = 80 W
Power of Kapil, P = Fu = 25 x 3 = 75 W
Avinash is more powerful than Kapil.

Question 3.
The weight of a person on a planet A is about half that on the earth. He can jump up to 0.4 m height on the surface of the earth. How high can he can jump on planet A?
For the case, of jump, the energy imparted by the person is converted into potential energy,
Hence,
(mAgA)hA = (mege)he ………(1)
Given, mAgA = $$\frac{m_{e} g_{e}}{2}$$ ………….(2)
Using (1) and (2),
$$\frac{h_{\mathrm{A}}}{2}=h_{e}$$
∴ hA = 2he = 2 x 0.4 = 0.8m

Question 4.
A ball is dropped from a height of 10 m. If the energy of the bal] is reduced by 40% after striking the ground, how much high can the ball bounce back? (g 10 m/s2)
Given,
height, h = 10
If the energy of the ball is reduced by 40%, the remaining energy of the ball is 60% of initial.
Hence ball will rebound to 60% of the initial height
h = $$\frac {60}{100}$$ x 10 m
∴ h = 6m

Question 5.
Four men lift a 250 kg box to a height of 1 m and hold it. Without raising or lowering it
(a) How much work is done by men in lifting the box?
(b) How much work they do in just holding it?
Given, mass of block, m = 250 kg
height, h = 1 m
(a) work done in lifting,
W = Fs = mgh = 250 x 10 x 1
W = 2500 J
(b) work done in holding, W = 0

Question 6.
What is power? How do you differentiate kilowatt from kilowatt-hour? The Jog falls in Karnataka state are nearly 20 m high. 2000 tonnes of water falls from it in a minute. Calculate the equivalent power if all this energy can be utilized. (g = 10 ms-2)
1. Power is rate of doing work.
2. Kilowatt is the unit of power and kilowatt-hour is unit of energy.
3. Given, height, h = 20 m
Mass per unit time,
m/t = 2000 tonnes per minutes = $$\frac{2000 \times 10^{3}}{60}$$ kg/s
Power P = $$\frac{W}{t}=\frac{m g h}{t}=\frac{2000 \times 10^{3}}{60}$$ = 10 x 20
∴ P = 6.67 x 106 W = 6.67 MW

Question 7.
What happens to the kinetic energy when:
1.  the mass of the body is doubled at constant velocity?
2. the velocity of the body is doubled at constant mass?
3. the mass of the body is doubled but velocity is reduced to one fourth?
1. The kinetic energy of body is given by,
Ek = $$\frac {1}{2}$$ mυ6, Ek ∝ m. If the mass of the body is doubled its kinetic energy is also doubled.

2. Kinetic energy, Ek ∝ u2
If velocity of the body is doubled, its kinetic energy becomes four times.

3. Initial kinetic energy,
$$E_{K_{i}}$$= $$\frac {1}{2}$$ mu2
Final kinetic energy,
$$E_{K_{f}}$$ = $$\frac {1}{2}$$(2m) $$\left(\frac{v}{4}\right)^{2}=\frac{m v^{2}}{16}$$
Kinetic Energy becomes one eighth.

Question 8.
Why is the water at the bottom of a waterfall warmer than the water on the top?
When waterfalls, its potential energy is converted into kinetic energy of molecules, and the kinetic energy of molecules is converted into heat energy.

### Work, Power And Energy Class 9 Extra Questions Value Based (VBQs)

Question 1.
Aman is a student of class IX. He saw an old man trying to keep his box on the roof of a bus but was unable to do so. Aman picked up his box and placed the box on the roof of the bus.
The old man thanked Aman.
Answer the following questions based on the given paragraph:

• Is the work done by Aman while placing the box on the roof of the bus positive or negative?
• Is the work done by gravity on the box positive or negative?
• What values are shown by Aman?

• Positive
• Negative
• Aman is a kind and helpful person.

Question 2.
In the winter season, John gifted an electric heater to his grandfather. The electric heater uses electricity to increase room temperature.
Answer the following questions based on the above paragraph:

• Write down the energy transformation in the electric heater.
• What values are shown by John?


text:
# Reef knot

Reef knot
Names Reef knot, Square knot, Hercules knot
Category Binding
Category #2 Bend
Origin Ancient
Related Thief knot, Granny knot, Grief knot, Surgeon's knot, Shoelace knot
Releasing Jamming
Typical use Joining two ends of a single line to bind around an object.
Caveat Not secure as a bend. Spills easily if one of the free ends is pulled outward. Does not hold well if the two lines are not the same thickness.
ABoK #74, #75, #460, #1204, #1402, #2096, #2573, #2574, #2577, #2580
Instructions [1]

The reef knot or square knot is an ancient and simple binding knot used to secure a rope or line around an object. Although the reef knot is often seen used for tying two ropes together, it is not recommended for this purpose due to potential instability of the knot.

A reef knot is formed by tying a left-handed overhand knot and then a right-handed overhand knot, or vice versa. A common mnemonic for this procedure is "right over left; left over right", which is often appended with the rhyming suffix "... makes a knot both tidy and tight". Two consecutive overhands of the same handedness will make a granny knot. The working ends of the reef knot must emerge both at the top or both at the bottom, otherwise a thief knot results.

## Naming

The reef knot is at least between 4,000 and 9,000 years old. The name "reef knot" dates from at least 1794[1] and originates from its common use to reef sails,[2][3] that is to tie part of the sail down to decrease its effective surface area in strong winds. To release the knot a sailor could collapse it with a pull of one hand; the sail's weight would make the collapsed knot come apart. It is specifically this behavior which makes the knot unsafe for connecting two ropes together.[4]

The name "square knot" is found in Dana's 1841 maritime compendium A Seaman's Friend, which also gives "reef knot" as an alternative name.[5][6]

## Uses

The reef knot is used to tie the two ends of a single line together such that they will secure something, for example a bundle of objects, that is unlikely to move much. In addition to being used by sailors for reefing and furling sails, it is also one of the key knots of macrame textiles.[7]

The knot lies flat when made with cloth and has been used for tying bandages for millennia. As a binding knot it was known to the ancient Greeks as the Hercules knot (Herakleotikon hamma) and is still used extensively in medicine.[8] In his Natural History, Pliny relates the belief that wounds heal more quickly when bound with a "Hercules knot".[9]

It has also been used since ancient times to tie belts and sashes. A modern use in this manner includes tying the obi (or belt) of a martial arts keikogi.

With both ends tucked (slipped) it becomes a good way to tie shoelaces, whilst the non-slipped version is useful for shoelaces that are excessively short. It is appropriate for tying plastic garbage or trash bags, as the knot forms a handle when tied in two twisted edges of the bag.

The reef knot figures prominently in Scouting worldwide. It is included in the international membership badge[10] and many scouting awards.[11] In the Boy Scouts of America demonstrating the proper tying of the square knot is a requirement for all boys joining the program.[12]

## Misuse as a bend

The reef knot can capsize (spill) when one of the free ends is pulled outward.

The reef knot's familiarity, ease of tying, and visually appealing symmetry conceal its weakness. The International Guild of Knot Tyers warns that this knot should never be used to bend two ropes together.[13] A proper bend knot, for instance a sheet bend or double fisherman's knot, should be used instead. Knotting authority Clifford Ashley claimed that misused reef knots have caused more deaths and injuries than all other knots combined.[14] Further, it is easily confused with the granny knot, which is a very poor knot.

## Physical analysis

An approximate physical analysis[15] predicts that a reef knot will hold if $1 \le 2\mu e^{\mu\pi}$, where μ is the relevant coefficient of friction. This inequality holds if $\mu \gtrsim 0.24$. Experiments show that the critical value of μ is actually somewhat lower.[16]


text:
Differential equations driven by fractional Brownian motion as random dynamical systems: qualitative properties (08frg140)

Arriving Sunday, September 28 and departing Sunday October 5, 2008

Organizers

David Nualart (University of Kansas)
Frederi Viens (Purdue University)

Objectives

Background and relevance: stochastic integration and fractional
Brownian motion

A central mathematical object in Stochastics/Stochastic Processes is the Ito
integral. It plays an important role in many areas of pure and applied
mathematics including: mathematical finance, population dynamics, fluid
dynamics, statistics, signal processing, control, particle systems, to name a
few. The integrator of such an integral is often chosen to be Brownian motion
(the Wiener process) or its semimartingale generalizations. These random
functions are of unbounded total variation, so that their Stieltjes integrals
do not exist. Special properties of the integrators and the integrands are
necessary to generalize the definition of the Stieltjes integral to the Ito
integral, and enable the definition of solutions of differential equations
driven by Brownian motion.

A property of paramount importance to this effect for Brownian motion is the
independence of its increments. To move beyond integrals and processes
constructed using this property is one of the most important tasks in the
theory of Stochastics. We are most interested in using the fractional Brownian
motion (fBm) process $B^{H}$ where $Hin(0,1)$ is fixed. It is a type of
stochastic process which deviates significantly from Brownian motion and
semimartingales, and others classically used in probability theory. As a
centered Gaussian process, it is characterized by the stationarity of its
increments and a medium- or long-memory property which is in sharp contrast
with martingales and Markov processes. Specifically, $B^{H}left( 0right) =0$ and $Varleft[ B^{H}left( tright) -B^{H}left( sright) right] =leftvert t-srightvert ^{2H}$. It also exhibits power scaling and path
regularity properties with Holder parameter $H$, which are very distinct from
Brownian motion. In fact, this single textquotedblleft
Hursttextquotedblright parameter $Hin(0,1)$ is also responsible for the low
decorrelation speed of fBm: for increments that are $n$ time units apart, the
correlation is precisely $c_{H}n^{2H-2}$ where the constant $c_{H}=H(2H-1)$.
Note that the standard Brownian motion is included in this family of models:
it is fBm with $H=1/2$, since $c_{H}$ is indeed null in this case. FBm has
become a popular choice of late for applications where classical processes
cannot model these non-trivial properties; for instance long memory, which is
also known as persistence, and corresponds to the case $Hin(1/2,1)$, is of
fundamental importance for financial data and in internet traffic: see
cite{Mandelbrot}, cite{w1}, cite{w2} . Our research group will have these
applications in mind in all discussions.

The mathematical theory of fBm is currently being developed vigorously by a
number of stochastic analysts, in various directions, using complementary and
sometimes competing tools. Ever since the pioneering works of Z{"{a}}hle
cite{Zah98}, Decreusefond and "{U}st"{u}nel cite{DecUst99}, and Lyons
cite{L}, the main thrust has been to understand how to perform stochastic
integration with respect to fBm in a way which is consistent with some
properties of the classical Ito theory for Brownian motion. In the case of
higher regularity ($H>1/2$), simple trajectorial methods, labelled as
emph{pathwise}, can be used which make it easy to translate one integration
theory into another, as emph{fractional derivatives} allow a pathwise
estimate of the integrals in terms of integrand and integrator using special
norms. Pathwise integrals historically gave the first cases where adequate
solutions to stochastic differential equations (sde) were established, see
Nualart and Rascanu cite{NR02}; infinite-dimensional equations have been
treated with the same success as finite-dimensional ones, e.g. Nualart and
Maslowski cite{MN}, Viens et al. cite{TTV}.

Timeliness, motivation, and importance: stochastic differential
equations driven by fractional Brownian motion, from general theory to random
dynamical systems

Solving sde can be considered a benchmark for testing the adequacy of an
integration theory. Among those methods which allow the use of more irregular
fBm ($H<1/2$), the so-called rough path theory, which uses non-probabilistic
constructions even for very irregular signals, is more efficient at defining
solutions of non-linear sde driven by fBm; it also has the advantage of being
applicable to processes that share some regularity properties with fBm, but
are otherwise very different; its main disadvantage is to not allow the same
ease of calculations as for Ito equations for Brownian motion. The Skorohod
(divergence) integral theory, based on stochastic analysis and Malliavin
calculus, is better at exploiting the Gaussian property of fBm, for instance
to generalize the Ito integral and formula with ease, resulting in mean-zero
stochastic integrals, and other convenient constructions: see Cheridito and
Nualart cite{CN05}, Mocioalca and Viens cite{MV05}. But Skorohod integration
has not produced a way to define solutions to fully non-linear equations; this
is one of the longest-standing open problems in stochastic analysis.medskip

Many mathematicians have yet to move decisively beyond the basic "existence-uniqueness" theory of sde driven by fBm. While some problems of this type, such as for Skorohod sde, remain open and are worthy of study, in this focused research group, we will also delve deeper into the qualitative properties of fBm-driven equations. In particular, we will investigate the equations' asymptotics (e.g. in large time). The two most popular theories dealing with the asymptotic qualitative behavior for general sde are: the theory of emph{random dynamical systems} (RDS) and the theory of existence and uniqueness of emph{invariant measures} for the associated Markov semigroup. However, similarly to fBm itself, equations driven by fBm do not generate a Markov process. This precludes the study of invariant measures using classical tools for fBm-driven systems. It motivates our plan to concentrate on the investigation of fBm-driven sde as RDS, the interface between the two being at the heart of our focused research group.

The theory of RDS, developed by L. Arnold and coworkers (see cite{Arn98}),
can be used to describe the asymptotic and qualitative behavior of systems of
random and stochastic differential/difference equation in terms of stability,
Lyapunov exponents, invariant manifolds, and attractors. A RDS consists of two
parts. The first part is a model for the noise path $omega$, leading to a
emph{metric dynamical system}. In particular, it is known that the fractional
Brownian motion forms an emph{ergodic} metric dynamical system, see Maslowski
and Schmalfu{ss } cite{MS04}. The second part of a RDS is the dynamics of an
sde: it is given by the solution mapping $phi(t,omega,x)$, which describes a
solution at time $t$ starting at time zero with initial condition $x$ in some
phase space $E$ which is driven be a noise path $omega$. This mapping $phi$
satisfied a generalized (semi)-group property, called emph{cocycle}%
--property, see Arnold cite{Arn98}. In other contexts, some answers have
already been given to basic questions such as stationarity and ergodicity
using fBm: see for instance Hairer cite{Hm05}, Hairer and Ohashi cite{HO07},
where conditions are given so that results familiar to the Brownian case also
hold for fBm. In Maslowski and Schmalfu{ss } cite{MS04},
infinite-dimensional RDS driven by fBm are exhibited as stochastic evolution
systems, and are shown to have unique, exponentially attracting fixed points.

Specific objectives for fractional-Brownian-driven random dynamical
systems: stability, infinite systems, invariant manifolds, Lyapunov
exponents

For many infinite dimensional Brownian-driven sde with non-trivial diffusion
coefficients, it is not known whether these equations generate a RDS. The
reason is that typically stochastic differential equations are only defined
$omega$-almost surely because the exceptional set is related to the
definition of an Ito integral as a limit of random variables emph{in
probability}. However, such a family of exceptional sets does not allow to use
the theory of RDS's full power. One advantage of sde driven by an fBm with
$H>1/2$ is that one can consider pathwise integrals, which avoid exceptional
sets. The case of $H>1/2$ is also interesting from the physical standpoint
since it is the case where path memory (persistence) is the longest; in that
sense, it is most different from standard Brownian motion.

On the other hand, when $H<1/2$, as we alluded to above in the relation
between pathwise and Skorohod integration, there does not exist a universally
accepted way of looking at stochastic integration or sde's, even finite
dimensional, and the problem of solving sde's in the Skorohod sense is
entirely open. Our focused research group will address the issue by trying to
understand whether any synergy can exist between the various approaches, which
have thus far been largely disjoint. In particular, it is possible that the
difficulties being experienced by stochastic analysts are related to the
non-existence of a RDS interpretation of sdes driven by irregular fBm. Since
non-linear equations driven by irregular fBm can be solved via the rough path
method (see recent progress by Nualart and Hu cite{NuaHu06}, and also the
original works of Lyons and Qian cite{LioQui98} or cite{LioQia02}), finding
even a single counter-example where such a solution occurs in a semi-explicit
way, but does not allow a RDS to be defined, would help understand the open
problem.

A key concept describing the dynamics of RDS generated by fBm-driven sde is
the so-called global attractor, which is an invariant random set attracting
other bounded random sets; it is a hallmark of the concept of emph{stability}%
. The essential dynamics take place in a neighborhood of the attractor. Even
for infinite dynamical systems the attractor often has a finite (fractal)
dimension which allows to describe the dynamics by finitely many parameters.
To show that there is an attractor, the main point is to show the existence of
an absorbing set (see Chueshov and Schmalfu{ss } cite{r-8}). In the
classical theory, it can usually can be constructed via Gronwall's lemma. This
leads to the question of finding a Gronwall lemma with respect to the special
fractional norms used for estimating stochastic integrals of fBm; this forces
one to try and modify the phase space of the RDS accordingly. Some ideas for
this kind of Gronwall lemma can be found for finite dimensional systems in
Garrido, Maslowski and Schmalfu{ss } cite{GarMasSchm08}. Our focused
research group will discuss these findings, their scope, and their extensions
to infinite systems driven by fBm.

Our plan to discuss stability for fBm-driven sde will continue with the
existence of stable and unstable manifolds and Lyapunov exponents, see Lu and
Schmalfu{ss } cite{r-9}. Such smooth manifolds are invariant under the
dynamics of the systems, and on them, the states are attracted or repelled by
a steady state. Standard methods to obtain these manifolds are the
Lyapunov-Perron method, based on fixed point arguments. We plan to discuss how
to obtain these fixed points with respect to the special fractional norms for
fBm integration, which will again force us to use an appropriate phase
space.

Our focused research groups also intends to deal with the interesting question
of stability expressed in terms of pointwise Lyapunov exponents. The recent
work Viens and Zhang cite{VZ} considers the stochastic heat equation with
multiplicative infinite-dimensional noise which is fBm in time. If the spatial
behavior of the noise is non trivial, it is shown that when $Hleq1/2$, the
main Lyapunov exponent is finite and positive, but when $H>1/2$, it is
infinite, and in fact the solution explodes no slower than $expleft( ct^{2H}/log tright)$ and no faster than $expleft( ct^{2H}/log tright)$. Surprisingly, the case of $Hleq1/2$ is the easiest one to
analyze, and leads to familiar results. In the case of long memory ($H>1/2$),
the behavior of the solution is yet unclear. Is the memory length the main
factor in this difficulty for $H>1/2$? Or are there scaling arguments which
would help us show that the magnitude of the noise is the main driving force?
We will discuss these questions which could have important consequences in
statistical physics.

Specific objectives for long memory: memory preservation and
non-linear time series, effect on climate models, delay equations, and
statistical estimation

Further physically intriguing results regarding memory length are provided to
us by engineering and sociological practitioners. In Jumarie cite{Jg02},
arguments are given indicating that the introduction of random decision times
can lead to fractal behavior in dynamical systems, specifically with long
memory. In the opposite direction, numerical analysis on some linear dynamical
systems driven by fBm indicate they may lose the long memory property:
Grigoriu cite{Gm07}. The mathematical issue here is whether fBm's long memory
is preserved when passed through a RDS. As a first step to elucidate the
truths behind the numerical indications, we will discuss the use of
discrete-time RDS driven by long-memory time series. Indeed, it has been known
for a few years that some linear and nonlinear quadratic ARCH($infty$) time
series exhibit precisely fractional Brownian memory: Giraitis et al.
cite{GLRS04}. Here we will focus the discussion on how long memory may effect
products of non-independent matrices; an ARCH($infty$) model may be
considered as a product of infinite-dimensional matrices based on IID noise
terms, but it might be more efficient to model the discrete version of an
fBm-driven system as a product of $2times2$ matrices whose components are
correlated over long ranges. It is possible that for such a problem, using
continuous time and fractional stochastic calculus may provide a distinct
avenue of attack.

FBm is a good candidate to model random long-time influences in climate
systems, see Palmer et al. cite{Palmer2005}, cite{Palmer}. Here, the first
issues to address are existence questions for sde of from fluid dynamics when
they are driven by an fBm. For instance, although several authors appear
interested by the issue of driving a Navier-Stokes equation by fBm, no
definitive publications have yet appeared. The stochastic heat equation
mentioned above can be considered a toy model for this problem, and could
help, via Feynman-Kac representation as in cite{hv} for the corresponding
vorticity equation, understand the full non-linear equation driven by fBm.
After the existence question is better understood, our group will discuss the
dynamics of larger climate systems with long memory.

Stochastic differential delay equations and their asymptotic behavior have
received much attention in recent years (see for instance work of Caraballo
cite{Ca90}, Garrido-Atienza and collaborators cite{CaGaRe03}, cite{GaRe03},
Lisei cite{Lisei01}, Taniguchi et al. cite{TaLiuTru02}): such delayed
problems often appears in applications in physics, biology, engineering,
finance, etc. They offer an alternative to assuming that the driving noise is
fBm, while preserving some of fBm's memory features; but by combining the
effect of delay with the long memory of fBm, more realistic models will
emerge; to our knowledge these have not been considered. During our stay at
BIRS, we plan to discuss whether such stochastic delay equations with long
memory generate RDS, and if so, whether there exist random fixed points.
Judging for instance by the work of Garrido-Atienza et al. cite{CGS}
(stationary solutions for delayed sde driven by Brownian motion), our sde
coefficients will have to satisfy particular conditions.

Stochastic analysis has recently proved useful in tackling statistical
estimation issues for fBm. $H$ can be estimated consistently, and in some
cases with asymptotic normality, using simple power variation statistics, see
Tudor and Viens cite{TV07}. Similar constructs with weights are capable of
giving rise to various Ito integral limits, see for instance Nourdin et al.
cite{NNT07}. The statistics results have also motivated the study of
generalizations of fBm, such as the Rosenblatt process, and higher-order
Hermite processes, which share fBm's covariance structure, but can be
arbitrarily highly non-Gaussian. For simple fBm-driven sde with non-linear
drift scale parameter $a$, Gaussian and Malliavin calculus have yielded the
strong consistency and asymptotic normality of $a$'s maximum likelihood
estimator based on discrete observations: cite{TVstat}. Much remains to be
understood in estimating parameters for RDS driven by fBm; our group will
discuss the future research directions which are most likely to be of interest
to statisticians and other practitioners who need to know how to determine and
test their long-memory models.

Conclusion
Stochastic differential equations with fractional Brownian motion have the potential to provide a wealth of new models in many applied areas; our plan to use ideas from random dynamical systems to study their properties will give them the new perspective they deserve.

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text:
# bi-directional half-duplex communication between two AT89C2051's micros without serial port

I'm trying to do a direct two-way communication between two AT89C2051 microcontrollers with only two GPIO pins so that I have other GPIO lines free for other tasks.

P3.7 is the data wire, and P3.2 (INT0) is the control wire, and they are externally pulled high to VCC (of 5 volts) through 10K resistors.

One microcontroller has an 18.432Mhz crystal connected to it so I can connect it to my computer at a clean 9600bps baud. The other microcontroller is connected to a 20Mhz crystal (which is the highest speed lowest-priced compact crystal I ordered from the internet) for fast memory and wireless communication.

Initially I was going to reserve 4 GPIO pins for data communication but I wouldn't have any free for other tasks such as memory and LED manipulation.

This is my idea behind the two wires.

The sender loads P3.7 with the one bit and then sends P3.2 low then waits a bit for remote processing. This repeats until the whole byte is sent.

When the receiver receives data, the interrupt is supposed to be executed (from the remote sender making P3.2 low) and the byte received from P3.7 is stored in a temporary location until all bytes are received which a counter tracks. Once all are received, a flag is set.

The sending seems to be ok, but the receiving always stalls.

Do I need to adjust my timings or is there a better approach of doing half-duplex bi-directional communication with only two wires connected to the same pins (with pull-up resistors) on both micros?

Here is my code for reference. I have to remove the first line of code after programming the first chip to ensure the correct code is on each chip.

UNIT equ 1h ;1 = 8051 to interface to PC, 0 = 8051 to interface to radio.
RDATS EQU 40h ;remote data temp
RDAT EQU 42h
RDATCT EQU 41h
RDI EQU 0h ;remote-data-interrupt bit
CT EQU 1h ;carry temp - receive
CTS EQU 2h ;carry temp - send

org 0h
ljmp main
;INT 0 natural address is here (03h)
ljmp intZ
org 002Bh

;called
intZ:
clr EX0
ifdef UNIT
clr P3.4 ;This is supposed to execute on 8051 connected to PC but does not.
endif
mov CT,C
push ACC
mov C,P3.7
mov A,RDATS
RLC A
mov RDATS,A
inc RDATCT
mov A,RDATCT
cjne A,#8h,no8
setb RDI
mov RDAT,RDATS
mov RDATCT,#0h
no8:
jnb P3.2,$pop ACC mov C,CT ifdef UNIT setb P3.4 endif setb EX0 reti sendrdat: clr EX0 mov CTS,C mov R7,#8h push ACC sendrmore: clr C rlc A mov P3.7,C clr P3.2 nop nop setb P3.2 djnz R7,sendrmore pop ACC mov C,CTS setb EX0 ret serialin: clr RI jnb RI,$
mov A,SBUF
ret

serialout:
clr TI
mov SBUF,A
jnb TI,$ret pserial: rewr: clr A movc A,@A+DPTR jz exl acall serialout inc DPTR ajmp rewr exl: ret main: mov RDAT,#0h mov RDATS,#0h mov RDATCT,#0h clr RDI mov P3,#0FFh mov P1,#0FFh mov SP,#50h ;skip most ram so SP doesnt be wrecked mov scon,#50h mov tmod,#020h mov 087h,#80h ;pcon.7 mov TH1,#0F6h ;9600 baud on 18.432Mhz setb EX0 setb EA setb TR1 IFDEF UNIT ajmp PCmode ELSE ajmp RADmode ENDIF ;branch PCmode: clr RI jnb RI,$
mov DPTR,#ptest1
acall pserial
;send 0AAh to remote 8051
mov A,#0AAh
acall sendrdat
mov DPTR,#ptest2
acall pserial
; code stalls here forever.
; It should only stall until remote 8051 calls this one.
rwait:
jnb RDI,rwait
mov A,RDAT
push ACC
mov DPTR,#ptestr
acall pserial
pop ACC
acall serialout
mov A,#0Dh
acall serialout
mov A,#0Ah
acall serialout
mov DPTR,#ptestdone
acall pserial
sjmp $RADmode: ;radio mode... only obey commands at this time nop jnb RDI,RADmode clr RDI mov A,RDAT cjne A,#0AAh,nosig ;AAh=get signature = 41h (A) mov A,#041h acall sendrdat nosig: ajmp RADmode sjmp$

ptest1:
db 'Sending data to unit 2...',0Dh,0Ah,00h
ptest2:
db 'Receiving data from unit 2...',0Dh,0Ah,00h
ptestr:
ptestdone:
db 'This is a test that worked!',0Dh,0Ah,00h
END

• Have you ever heard of I²C? – Dave Tweed Oct 15 '16 at 1:07
• I read up about it and It seems I'm using the protocol to an extent. If possible, I want to enable as much multitasking as possible in the IC's instead of stalling for the full 8 bits. – user116345 Oct 15 '16 at 1:52
• The point is, I²C is a completely static, synchronous protocol that lots of people understand. You'll get a lot more help if you stick to an existing standard rather than inventing your own. You can run I²C as slow or fast as you like, and since it is entirely driven by edges, even a bit-banged implementation can be done entirely in interrupts, with no busy-wait loops at all. Plus, there's no way I'm going to wade through that mess of poorly-formatted, uncommented assembly code. I've written a lot of 8051 code in my time (C and assembly), and I'd be ashamed to show code like that in public. – Dave Tweed Oct 15 '16 at 2:01
• I could write substantially worse code. Anyways, I'm using the interrupt line as a trigger to start the reception of a byte. I'm gonna look for better code examples to see where I'm wrong but I'm still open to answers. – user116345 Oct 15 '16 at 2:05
• So why don't you use the onboard UART? I googled the datasheet for that chip and it seems to come in a 20 pin package... plenty of gpio – crowie Oct 15 '16 at 2:24

• I want to enable as much multitasking as possible

That just means that you will be using interrupts in some fashion.

The alternative, salting your code with short snippets of identical polling code placed strategically to simulate periodic events is as ugly as it gets. So you don't ever do it. I've seen it done before. And it's terrible and will make your code for all practical purposes unmaintainable.

• until all bytes are received which a counter tracks
• half-duplex bi-directional communication

These suggest to me that you actually may want a master/slave relationship, but where either can become the master for the period of the communication. It can be done by having only one master, ever. But the way you write suggests you would either like a protocol without a specific master (such as the asynchronous protocol used in RS-232, for example) or else a means by which either can become a master for the purposes of a burst of communication.

• 18.432Mhz crystal
• 20Mhz crystal

With two different processors, even if they are running at the so-called same clock rate, will start out with some phase difference which also drifts around relative to each other. Nothing is perfect in life. But with two different clock rates, one thing I also look at is the GCD. In this case, that's 32000. This means one option might be to use a divisor of 576 on your $18.432\:\textrm{Mhz}$ micro and a divisor of 625 on your $20.000\:\textrm{Mhz}$. That would get the bit timing, if you used a protocol where that mattered, into the similar ballparks. You'd still have phase error and drifting. But if you chose to use an asynchronous RS-232 like protocol, you could probably manage that after a fashion at a bit rate of $32\:\textrm{kHz}$.

But I also think Dave is right, too. Using something like $\textrm{I}^2\textrm{C}$ has its advantages over dinking around with too much over-sampling (you may need 2X over-sampling or more, regardless, since this is software; but this varies a bit with the hardware interrupt modes you can use, too) and getting your bit timings zeroed in.

The tradeoffs are now something like this:

1. Use multi-master $\textrm{I}^2\textrm{C}$. This takes a little extra effort in the code for master and slave to achieve the multi-master parts. But there is a standard for this too and plenty of example code to examine, I'm sure. The limitation here is that your communications are half-duplex using two lines. But you've already accepted this. And arbitration issues might have you scrambling for those GCD divisors I mentioned above.

2. Use a copy of the asynchronous format (start bit, N data bits, and let's say two stop bits) used by RS-232 and commonly found in UART hardware. The advantage here is that you get full-duplex communications. But you've indicated you don't need that. You'd probably want to use those GCD divisors here, too, to get the bit timing close enough to each other.

3. Use $\textrm{I}^2\textrm{C}$ in simple master-slave mode. Slaves can not push any data to the master. Instead, the master would have to ask the slave if there was something to send back. But you could also go this route. It would spare you arbitration issues, too.

In all cases you will have interrupts involved and either, (A) use state machines; or, (B) use an operating system with threads (cooperative ones, if not pre-emptive.)

It doesn't take much, not even on the 8051, to pony up a simple cooperative set of threads. (Yes, I've done it on a SiLabs C8051F061 and using SDCC and not Keil.) It really cleans up code and makes it very maintainable, because it avoids the chain of state transitions.

But a clearly documented set of state transitions isn't bad, either. Most with any experience using state machines can follow the clear documentation and see which code performs which step. So I think either approach would be fine.

• they are externally pulled high to VCC (of 5 volts) through 10K resistors.

So this is good as $\textrm{I}^2\textrm{C}$ wants that and asynchronous protocols can work with that.

Why don't you do some googling on $\textrm{I}^2\textrm{C}$ state machines, read through them until you feel you understand them, and also read through the documentatation available from this $\textrm{I}^2\textrm{C}$ documentation that is available to you. Similarly, consider an asynchronous protocol, as well.

One thing in your favor is that you do have the ability to get similar bit times on each device. This doesn't fix phase differences or drift. But a little bit of over-sampling can address that. Or you can just leave them free-run and work through the issues of that process.

Chances are, there are many different examples of code readily available on the web to cover you here, too. I'd recommend doing some searching there, as well. You are the one who needs this. So you should be doing the up-front work here. Not asking us to do that for you.

That said, if you get yourself educated about these methods and can narrow down your questions to something specific, I'm sure you'll get some wise advice. But right now, you are just throwing code against a wall to see what sticks. And I think we'd like you to do your own work getting up to speed on existing methods which are tried and true, before you start asking anyone to come up with something new for you or something based only on your own (apparently quite limited) thinking about this. If you are experienced, you can indeed draft up your own protocol and make it work -- first time, too, out of the box. But you aren't experienced. So you should first work from the benefit of others.

• At this rate, I might as well buy the same crystal (same speed) for each microcontroller for best speed and i/o reliability. – user116345 Oct 15 '16 at 4:58
• @Mike You could consider driving them both from the same crystal module. They'd be in phase as well as running at the same clock rate. – jonk Oct 15 '16 at 5:05
• I never knew such a setup is possible. I already made one circuit board where each micro has its own crystal. How do I wire the xtal pins of each microcontroller so both use the same crystal without blowing either part up? – user116345 Oct 15 '16 at 5:31
• @Mike That depends on the specific micro. Many can have one of their clock input pins driven externally (one is actually an output, one an input, in a class-A oscillator scheme.) You should follow guidance, though. Most manufacturers will talk about how this can be done for their product. If not, do a search on the web. If that doesn't work, consider calling up the local rep for the micro and ask them to get their field engineer to call you so you can ask about it. They'll know. – jonk Oct 15 '16 at 5:33
• @Mike Look at the datasheet for the AT89C2051. You will see the two clock lines clearly labeled, one as output and one as input. Figure 5-2 seems to cover the case. – jonk Oct 15 '16 at 5:42