Feynman Lectures Exercise Challenge

[K^2]

Ok. Given solution x(t)=Acos(ωt) at what t is x(t)=0 and x'(t)=0.

It's a one thing to find a steady solution to a differential equation. It's another to find a steady state attained by a physical system.

 

[Dickfore]

If you use my ODE and you assume that both x(t) and x_{p}(t) are sinusidal motions with the circular frequency of the pivot \omega_{p} (steady solution), then:
<br /> \ddot{x} = -\omega^{2}_{p} \, x<br />
<br /> \ddot{x}_{p} = -\omega^{2}_{p} \, x_{p}<br />
and you get:
<br /> -\omega^{2}_{p} \, x + \omega^{2} \, x - \omega^{2}_{p} \, x_{p} = 0<br />
<br /> x = \frac{\omega^{2}_{p} \, x_{p}}{\omega^{2} - \omega^{2}_{p}} = \frac{x_{p}}{\left(\frac{\omega}{\omega_{p}}\right)^{2} - 1} = \frac{x_{p}}{\left(\frac{T_{\mathrm{pivot}}}{T_{ \mathrm {pend.} }}\right)^{2} - 1}<br />
So, I don't know why codelib says my solution is wrong.

 

[K^2]

Go into a non-inertial reference frame that is moving with the pivot.

Now transform your solution back to inertial frame. That's the step you forgot about. You should get the same answer.

 

[Dickfore]

Ok. Given solution x(t)=Acos(ωt) at what t is x(t)=0 and x'(t)=0.

It's a one thing to find a steady solution to a differential equation. It's another to find a steady state attained by a physical system.

The general solution is:
<br /> x(t) = C_{1} \cos{(\omega t)} + C_{2} \, \sin(\omega t) + A \, \sin{(\omega_{p} t)}<br />

Then, you take:
<br /> x(0) = C_{1}<br />
<br /> \dot{x}(0) = \omega C_{2} + \omega_{p} \, A<br />
If you want C_{1} = C_{2} = 0 (steady solution), you must have:
<br /> x(0) = 0, \; \dot{x}(0) = \omega_{p} \, A<br />

 

[Dickfore]

Now transform your solution back to inertial frame. That's the step you forgot about. You should get the same answer.

Why?

 

[K^2]

I'm well aware of that. You are still forcing the initial condition that forces steady state solution. You cannot start with arbitrary initial condition and attain steady state, as implied by the problem. You have to do this via limits.

Why?

Because you are interested in motion of the bob relative to some fixed frame. You got x(t) in an accelerated frame, relative to pivot. If you transform to inertial frame, you should get the same answer.

 

[codelieb]

So, I don't know why codelib says my solution is wrong.

Compare our solutions.

BTW, my handle is "codelieb" not "codelib."

 

[K^2]

Let me change the names of variables. I'll call x(t) the position of bob relative to ground. I'll call y(t) position of bob relative to pivot. y(t)=x(t)-x_p(t)

Dickfore's equation is for y(t). It is correctly defined.

y''(t)+ω²y(t)+x''_p(t)=0

Convert that to the x(t) frame.

x''(t)-x''_p(t)+ω²(x(t)-x_p(t))+x''_p(t)=0

x''_p(t) simply cancels.

All 3 equations posted by each of us are exactly the same.

The reason Dickfore's steady state amplitude is different is because y(t) has different amplitude from x(t). You have to transform back from y(t) to x(t) coordinates, and you'll have the same answer.

 

[Dickfore]

Compare our solutions.

BTW, my handle is "codelieb" not "codelib."

https://www.physicsforums.com/showpost.php?p=3475333&postcount=25"

The steps that you used to deduce Eq.(2) in that post and Eq.(3) are wrong.

 

[Dickfore]

Let me change the names of variables. I'll call x(t) the position of bob relative to ground. I'll call y(t) position of bob relative to pivot. y(t)=x(t)-x_p(t)

Dickfore's equation is for y(t). It is correctly defined.

y''(t)+ω²y(t)+x''_p(t)=0

Convert that to the x(t) frame.

x''(t)-x''_p(t)+ω²(x(t)-x_p(t))+x''_p(t)=0

x''_p(t) simply cancels.

All 3 equations posted by each of us are exactly the same.

The reason Dickfore's steady state amplitude is different is because y(t) has different amplitude from x(t). You have to transform back from y(t) to x(t) coordinates, and you'll have the same answer.

Question: Assume that the y(t) = 0. Are you calling that swinging of the pendulum or not?

 

[K^2]

Dickfore, they aren't wrong. (See above post). They are merely poorly reasoned. See my derivation from Lagrangian to see how it is properly handled in fixed frame. Your solution for accelerated frame is fine. You just forgot (or didn't think necessary) to convert back. No big deal.

Question: Assume that the y(t) = 0. Are you calling that swinging of the pendulum or not?

I'm not interested in the argument on which frame we should measure amplitude in. As long as you agree that the differential equations and solutions are identical, simply represented in different frames, I'm fine with whichever you chose.

 

[Dickfore]

Dickfore, they aren't wrong. (See above post). They are merely poorly reasoned. See my derivation from Lagrangian to see how it is properly handled in fixed frame. Your solution for accelerated frame is fine. You just forgot to convert back. No big deal.

There is no converting back. I used Lagrange's formalism to derive the equation as well, but then I justified my solution according to the rules of the competition. y(t) = 0 does not correspond to a swinging pendulum. The pivot and the bob lie on a vertical line and move horizontally then.

 

[K^2]

Like I said, it doesn't really matter, so long as you agree that solutions are identical.

 

[Dickfore]

Like I said, it doesn't really matter, so long as you agree that solutions are identical.

The fact of the matter is that whatever you are describing, it does not correspond to the solution:

That is a nice problem. Hope you don't mind me taking a shot.

First, I'm assuming a linear oscillator. The envelope of bob's oscillation must have a period of 11 seconds, since that's the time it takes for the relative phase between two oscillation to go around. The envelope is sinusoidal, and must pass through the node at the same slope as the envelope of a resonance, since it's effectively at resonance at these points. (The phase of the support's displacement leads or lags bob's displacement by exactly 90° there.)

So the only question is at what rate would amplitude of bob's oscillation grow if both periods were at 1s exactly. The kinetic energy of the bob as it passes through the node is (1/2)mv². The force due to bob displacement is F=kx for some k and x=1cm. We don't know what the k or m is, but k/m=ω², and that's available. Rate of change of pendulum's energy is therefore F*v. With total energy related to current amplitude by E=(1/2)ka², where a is amplitude. v=a/ω. Energy increases at the rate of kxa/ω. So a² is increasing at rate of 2xa/ω. Since a(t)=ct (resonance), a²=c²t² which is area under a triangle y=2c²t, and therefore increases at a rate of 2c²t = 2ac. So the rate of increase of a=x/ω=x/2π.

So the solution is the amplitude of the sinusoidal wave with period of 11 seconds and passes through zero at slope of x/2π. Sin with period 2π will pass through zero at slope 1. So A=(x/2π)*(11/2π) =11*x = 11cm.

that you had posted in post #3.

 

[K^2]

That solution assumes no damping and at-rest initial condition. Exactly the same differential equation, though. Yes, it does not follow the spirit of the problem. I realized that early on page 2. Hence my comments about damping coefficient being needed to turn this into a proper physical problem, which can later be taken to zero as a limit.

 

[codelieb]

Dear all,

The point of this exercise (besides winning a free book) is to find a solution that is correct, physically appealing, and does not depend on the calculus or higher-level maths, in the spirit of Feynman's philosophy of intuitive problem-solving. I would add that, if at all possible, one should try to have fun while doing it. Otherwise, what's the point?

What's going in in this thread now is not fun. (Nor is it going to win anyone a free book.) Richard Feynman would not approve! I am unsubscribing.

If anyone else wants to take a stab at The Feynman Lectures Exercise Challenge, please email your solution to me, or post it on http://www.feynmanlectures.info" Forum, where you can find the problem posted on the Exercises board.


Mike Gottlieb
Editor, The Feynman Lectures on Physics
Author, Feynman's Tips on Physics
mg@feynmanlectures.info

 

[codelieb]

You cannot start with arbitrary initial condition and attain steady state, as implied by the problem.

There is nothing in the problem that defines the initial conditions, limits them in any way, or implies that steady state can be obtained with arbitrary initial conditions. The problem states that a steady state is obtained, which implies that the necessary initial conditions were satisfied.

Your (current) objection to this problem, regarding initial conditions, is remarkably similar to your objections about initial conditions in the bead parabola accelerometer problem, after you gave the wrong answer to it ... and are similarly irrelevant.

 

[codelieb]

See what you did there, Dickfore? You scared the poor computer scientist away with your accelerated frames. :p

K^2,

I did unsubscribe, but for some reason, I got an email about your most recent post.

Dickfore didn't scare me away, K^2. I like Dickfore. It's clear to me that he knows what he is talking about. I left this thread because of you, and your constant blame-laying and bantering about irrelevancies that seem to follow every time you fail to solve an elementary physics problem, as per this https://www.physicsforums.com/showthread.php?t=524137". I just don't have the time to deal with such foolishness.

 

[Borek]

Locked pending moderation.