Feynman Lectures Exercise Challenge

[codelieb]

Hello, all.

My colleagues and I are currently working on an exercise book for The Feynman Lectures on Physics (FLP). This book will include about 1000 exercises from the original Feynman Lectures course as taught at Caltech, covering pretty much the entire range of topics in all three volumes of FLP. Almost all of these exercises have been published before by Caltech, however the old Caltech exercise books are long out of print, and the exercises for FLP Vols. II and III, most of which originated with Feynman himself, were unfortunately never published with answers. We will be publishing them with answers, and worked-out solutions in many cases.

As a "teaser" and special challenge to Feynman physics buffs everywhere, I am posting one of the problems from the new exercise book.

In one of the Review lectures Feynman gave to his freshman students, just before their first big exam, he advised them as follows (copied from Feynman's Tips on Physics, a problem-solving supplement to The Feynman Lectures on Physics):

"Now, all these things you can feel. You don’t have to feel them; you can work them out by making diagrams and calculations, but as problems get more and more difficult, and as you try to understand nature in more and more complicated situations, the more you can guess at, feel, and understand without actually calculating, the much better off you are! So that’s what you should practice doing on the various problems: when you have time somewhere, and you’re not worried about getting the answer for a quiz or something, look the problem over and see if you can understand the way it behaves, roughly, when you change some of the numbers."​

The challenge is to solve the problem given below (originally homework for FLP Vol. I, chapter 23) in the spirit of Feynman's advice, above. It must be solved without using any calculus or differential equations or integral equations or difference equations, etc., without iterative numerical methods, nor any such other fancy mathematical tricks! You may use only algebra, geometry, trigonometry, dimensional analysis, and Newtonian mechanics, in your solution, which should be guided by your physical intuition (however note: all intuitions used in solutions must be justified)! Your answer does not have to be exact, but it should at least be a very close approximation. Here is the problem:

The pivot point of a simple pendulum having a natural period of 1.00 second is moved laterally in a sinusoidal motion with an amplitude 1.00 cm and period 1.10 seconds. With what amplitude should the pendulum bob swing after a steady motion is attained?​

The first person to solve the problem correctly, within the above-described constraints on their solution, will win a free copy (one of my author's copies) of the FLP exercise book when it is published (which we hope will happen late this year or early next year). The problem will also be posted on The Feynman Lectures Website with the winner's solution.

The Feynman Lectures Website will be the sole and final judge of the acceptability and correctness of all submitted solutions. You may email your solutions to me directly (if you want to keep them private) or post them in this thread, so that other people can discuss them.

One other thing: If you are one of the people with whom I have discussed this problem, or you have heard about it indirectly through me, then you are disqualified from this competition - please recuse yourself - sorry!

Good luck!Mike Gottlieb
mg@feynmanlectures.info

www.feynmanlectures.info
www.basicfeynman.com

 

[codelieb]

The outcome of the contest will be announced on January 1, 2012.

 

[K^2]

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.

 

[Dickfore]

So, we solve the problems for you and you have the credit for publishing. Is this how it goes?

 

[codelieb]

So, we solve the problems for you and you have the credit for publishing. Is this how it goes?

Thanks very much for offering, but we don't need anyone to solve problems for us at Caltech. (So, are you from MIT, or what? ). This particular problem has been solved for a long time. It was first published by Caltech in 1963. It was published again in "Exercises for Introductory Physics" (EIP) by Leighton & Vogt in 1969.

Our new exercise book will include all the exercises in EIP, including this one, for which a numeric answer will be provided, but no solution. However, in the books (past and future), where this problem appears, there are no constraints on the form of the solution. I added the "no highfalutin math" constraint for this challenge, in the spirit of Feynman's philosophy of solving physics problem using physical reasoning and intuition wherever possible.

The person who wins the contest, in addition to getting a free book, will have their solution posted with the problem at http://www.feynmanlectures.info" , so if you notice any, please let us know.)


Mike Gottlieb
www.feynmanlectures.info

 

[codelieb]

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

Of course I don't mind you taking a shot, K^2. However, you only get one!

In general, I don't plan to comment on people's solutions to this problem, because I don't want to risk inadvertently giving away clues. So, I will not comment on your method of solution, but, since you posted the problem here, I will comment on a couple of things I notice:

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.

You have (implicitly) defined v as the speed of the bob, a as the amplitude of the bob (which is a distance), and ω as the natural (angular) frequency of the pendulum (= 2π radians/second). Yet you write "v=a/ω," in which the left-hand side has units of distance/time, while the right-hand side has units of distance*time. Similarly, you have "the rate of increase of a=x/ω=x/2π" which has units of distance/time on the left, and distance*time on the right. These equations can not be correct.

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.

There seems to be a small problem with your arithmetic, but perhaps you meant (11*2π) instead of (11/2π).

 

[K^2]

You have (implicitly) defined v as the speed of the bob, a as the amplitude of the bob (which is a distance), and ω as the natural (angular) frequency of the pendulum (= 2π radians/second). Yet you write "v=a/ω," in which the left-hand side has units of distance/time, while the right-hand side has units of distance*time. Similarly, you have "the rate of increase of a=x/ω=x/2π" which has units of distance/time on the left, and distance*time on the right. These equations can not be correct.

(...)

There seems to be a small problem with your arithmetic, but perhaps you meant (11*2π) instead of (11/2π).

You're right on both counts. It never ceases to amaze me how often one can make two mistakes in one problem and still get the right answer. I guess I should have checked the units at every step of the way.

I do feel like there should be a more elegant way to go about it than using energy flow. (Seemed like a good idea at the time, but it got messy.) So hopefully somebody will come up with a better method.

 

[codelieb]

K^2,

You never cease to amuse!

You're right on both counts.

Thank you.

It never ceases to amaze me how often one can make two mistakes in one problem and still get the right answer.

In order for two mistakes to occur in a (otherwise correct) solution, which amazingly produces the right answer, the mistakes have to be "opposite" in the sense that one undoes the other. That is not so in the case of your solution to this problem, however. One of your mistakes is that you are using an equation "v=a/ω" which is simply nonsense. It doesn't mean anything and can not be true any more than apples can be oranges. (Speeds can not have units of time*distance!). Your other nonsense equation, "a=x/ω=x/2π" followed from this one - so those two mistakes did not "cancel": one implied the other. The third (apparent) mistake I pointed out in your solution was in your arithmetic, which I suggested correcting by changing one of the terms, but that does not appear to "undo" your previous (dimensional) mistake. Your answer (with my arithmetic correction) is "A=(x/2π)*(11*2π) =11*x = 11cm." Where x/2π=x/ω has units of distance*time, and 11 has units of time ("The envelope of bob's oscillation must have a period of 11 seconds") So in order for your answer to have units of distance, the "2π" in the term "(11*2π)" would have to have units of time^-2. I am not sure where this "2π" comes from, but if it is merely radians/cycle (as it appears to be) then it is dimensionless and your answer has units of distance*time^2, whereas if it is ω (which equals 2π sec^-1 in your solution), then your answer has units of distance*time. So, in my opinion, if your answer is numerically correct (and I am not saying that it is), it is not because you have made mistakes that cancel each other.

I guess I should have checked the units at every step of the way.

It would have sufficed to check the units of your answer.

I do feel like there should be a more elegant way to go about it than using energy flow. (Seemed like a good idea at the time, but it got messy.) So hopefully somebody will come up with a better method.

With this, I agree completely!

Just to be clear (in case anyone other than K^2 and I are reading this): I am, in general, not commenting on people's solutions or answers to this problem because I don't want to give away any clues. So, if someone claims they have the right answer, and I don't make any comment about that, my silence is not a silent agreement, it is merely silence.​

 

[K^2]

Oh, I saw what happened with that 2π as soon as you mentioned it. What amazes me is how these things happen by random chance to give you identical factors, to within a change of power, in different places of the problem. Of course, what stacks the odds is the choice of "simple" numbers for the problem. In this case, if the period was anything other than 1s, I'd spot the difference immediately. (Not that I'm blaming the problem. This one's great.)

 

[codelieb]

Oh, I saw what happened with that 2π as soon as you mentioned it. What amazes me is how these things happen by random chance to give you identical factors, to within a change of power, in different places of the problem. Of course, what stacks the odds is the choice of "simple" numbers for the problem. In this case, if the period was anything other than 1s, I'd spot the difference immediately. (Not that I'm blaming the problem. This one's great.)

Well, I don't see "what happened with that 2π." So far as I can see, your answer is given in the wrong units. Can you please enlighten us by explaining how "random chance [gave] you identical factors, to within a change of power, in different places of the problem" (specifically what places, what powers, etc.) and how that, when accounted for properly by suitable adjustments to your original solution, leaves the answer (11) unchanged and in the correct units (distance)?

 

[K^2]

(1/2)ka²=(1/2)mv² yields v=aω, not a/ω. That's me dividing by 2π instead of multiplying by it. And naturally, changing period of sinusoidal curve from 2π to 11 changes slope by factor of 11/2π, rather than 11*2π. That's me multiplying by 2π instead of dividing by it. Effectively, I include an extra factor of (2π)² and (2π)^-2 in two different places, which canceled each other out in the final result. The physics there is solid. It's only the algebra that's messed up.

 

[codelieb]

(1/2)ka²=(1/2)mv² yields v=aω, not a/ω. That's me dividing by 2π instead of multiplying by it. And naturally, changing period of sinusoidal curve from 2π to 11 changes slope by factor of 11/2π, rather than 11*2π. That's me multiplying by 2π instead of dividing by it. Effectively, I include an extra factor of (2π)² and (2π)^-2 in two different places, which canceled each other out in the final result. The physics there is solid. It's only the algebra that's messed up.

K^2,

You know... authors of academic texts don't make much money, and they don't get many (free) author's copies, so before I make any "decisions" I would like to make sure I understand your solution.

Call the natural period of the pendulum P, the period of the pivot point's motion p, and assume (as per the original problem) that p>P and that p/P <= 1.10 . Call the amplitude of the pivot point's motion a, and the amplitude of the bob A. So that, for example, in the problem as originally stated, P=1.00 sec, p = 1.10 sec, a = 1.00 cm, and we are asked to find A.] Please express A as a function of the variables p, P, and a, and show that it produces your answer of 11 cm.

 

[K^2]

Sure. I called amplitude of the pivot's motion x, so that becomes a. Relative period (period of the envelope) is T=1/(1/P-1/p). ω is 2π/P. With these substitutions:

A = (a*ω)*(T/2π) = a(T/P) = a/(1-P/p)

For p/P = 1.1, 1/(1-P/p)=11. This should work for any ratio p/P, with divergence at p=P corresponding to resonance.

 

[codelieb]

Sure. I called amplitude of the pivot's motion x, so that becomes a. Relative period (period of the envelope) is T=1/(1/P-1/p). ω is 2π/P. With these substitutions:

A = (a*ω)*(T/2π) = a(T/P) = a/(1-P/p)

For p/P = 1.1, 1/(1-P/p)=11. This should work for any ratio p/P, with divergence at p=P corresponding to resonance.

Thanks, K^2.

 

[Dickfore]

That is not the correct answer. The correct answer should read (for small oscillations of the pendulum, small meaning the amplitude of oscillaiton measured by the maxium angle of deflection from the vertical is much smaller than 1 rad):

$$A = \frac{a}{\left(\frac{T_{\mathrm{pivot}}}{T_{ \mathrm{pend.} }}\right)^{2} - 1}$$

When we substitute the numerical values:

$$A = \frac{1.00 \, \mathrm{cm}}{\left(\frac{1.10 \, \mathrm{s}}{1.00 \, \mathrm{s}}\right)^{2} - 1} = \frac{1.00 \, \mathrm{cm}}{1.21 - 1} = \frac{1.00 \, \mathrm{cm}}{0.21} = 4.8 \, \mathrm{cm}$$

 

[K^2]

{Mathematica code follows.}

ω = 2 Pi;

a = 1.0;

s = NDSolve[{x''[t] == -ω^2 (x[t] - a Sin[ω t/1.1]), x[0] == 0, x'[0] == 0}, x[t], {t, 0, 11}];

Plot[x[t] /. s, {t, 0, 11}]

{Output and notebook attached.}


Just saying.

 

[codelieb]

K^2,

That's an interesting graph. It shows that the amplitude of the pendulum grows and diminishes periodically, every 11 seconds. (I have attached another graph generated using your notebook but for t=0 to 88 instead of t=0 to 11.)

I have a suggestion for you: make a pendulum from a piece of string with a weight tied to one end, hold the other end in you hand, and move it slowly back and forth with a steady motion, a little slower than the natural period of the pendulum. Keep it up until the bob attains a steady motion (as per the stated problem). Then compare the amplitude of the bob as a function of time to your graph.

 

[K^2]

The solution I show assumes no damping. That results in a periodic envelope. Nothing that can be done about that.

If you take a real pendulum, there is going to be damping. The envelope, then, has a rather complex shape. Try running the same notebook with the following equation:

x''[t] == -ω^2 (x[t] - a Sin[ω t/1.1]) - 2 ω c x'[t]

For different values of c, there will be entirely different solutions.HOWEVER, one may ask what the amplitude is as c->0 and t->Inf. That is closer in spirit to the question asked, and it does have a solution. I did not consider that. Neither mine nor Dickfore's equations have anything with a solution to THAT problem. If that's what you want solved, that one's still open.

 

[Dickfore]

K^2, the equation you are solving does not correspond to the situation described in the problem.

 

[K^2]

K^2, the equation you are solving does not correspond to the situation described in the problem.

Really? Would you like to write out the diff-eq that does?

 

[codelieb]

Really? Would you like to write out the diff-eq that does?

K^2,

Would you please show us how you found your differential equation,

x''[t] == -ω^2 (x[t] - a Sin[ω t/1.1]).

Mike

 

[K^2]

I found it by simply adding a driving term to a simple harmonic oscillator. But I can easily verify it.

$$L = \frac{m}{2}(\dot{x}^2+\dot{y}^2) - mgy$$
$$f(x,y) = (x-x_p)^2 + y^2 - L^2 = 0$$

Where x_p is the pivot location and L is the length of the arm.

$$m\ddot{x}-2\lambda (x-x_p)= 0$$
$$m\ddot{y} + mg - 2\lambda y = 0$$
$$\frac{d^2}{dt^2}(x-x_p)^2 + \frac{d^2}{dt^2}y^2 = 0$$

That last equation is simply the differential form of the constraint, and reduces to the following.

$$(\dot{x}-\dot{x}_p)^2 + (x-x_p)(\ddot{x}-\ddot{x}_p) + \dot{y}^2 + y\ddot{y} = 0$$

Second equation, in the mean time, can be solved for 2*lambda.

$$2\lambda = \frac{m\ddot{y}+mg}{y}=-\frac{m\ddot{y}+mg}{\sqrt{L^2-(x-x_p)^2}}$$

Note the minus sign. That comes from the fact that there are two solutions for y, and we are interested in one where y~-L.

Cutting board time. First simplification that has to be made is (dy/dt)²->0. Velocity in y should be very small for a pendulum in linear mode to begin with. Square of that term can be safely dropped if we are hoping to linearize the equation. More terms will follow, but dropping this here makes things easier.

With that in mind, equation for lambda can be re-written to be in terms of x variables only.

$$2\lambda = -\frac{mg}{\sqrt{L^2-(x-x_p)^2}}+\frac{m}{L^2-(x-x_p)^2}\left((\dot{x}-\dot{x}_p)^2 + (x-x_p)(\ddot{x}-\ddot{x}_p)\right)$$

And that gives us the differential equation.

$$m\ddot{x}+(x-x_p)\left(\frac{mg}{\sqrt{L^2-(x-x_p)^2}}-\frac{m}{L^2-(x-x_p)^2}\left((\dot{x}-\dot{x}_p)^2 + (x-x_p)(\ddot{x}-\ddot{x}_p)\right)\right)=0$$

Now the tricky part. This is a crazy non-linear mess, but x/L and x_p/L are very small. That means contributions in higher orders of these terms are small, leaving us with only the following terms.

$$m\ddot{x}+(x-x_p)\frac{mg}{\sqrt{L^2-(x-x_p)^2}}=0$$

Furthermore, the radical can be expanded in powers of (x-x_p), and we only keep the constant term: 1/L.

This results in the equation

$$\ddot{x}+(x-x_p)\frac{g}{L}=0$$

Setting g/L=ω² and substituting x_p=a sin(ω_pt), the final form is obtained.

$$\ddot{x}+\omega^2(x-a sin(\omega_p t))=0$$

Where ω_p=ω/1.1, of course.

 

[Dickfore]

The differential equation corresponding to the problem at hand is:

$$\ddot{x} + \omega^{2} \, x + \ddot{x}_{p} = 0$$

 

[K^2]

Derive it.

 

[codelieb]

Hoo boy!

Guys (I keep saying I am busy, but...),

K^2: Your answer (to this ancient problem at Caltech) is wrong. Your solutions are wrong - both of them. Your differential equation is wrong. Your graph is wrong. And your ideas about the behavior of harmonically forced linear oscillators are wrong.

DickFore: You have the right idea, but your answer is also wrong.

Now, I am going to show you jokers how we solve this problem with differential equations at Caltech. After that, everyone will know the right answer (which is not 11 cm and not 4.8 cm), and how to find it in a conventional way. The challenge to find it in an unconventional way remains. However, K^2 and Dickfore are disqualified from the competition (for having submitted wrong answers). Please excuse me for typing this out in ASCII - I don't have time to make it pretty.

The equation of motion for a forced harmonic oscillator with one degree of freedom is

mx'' + kx = F(t),

or

(1) x'' + (k/m)x = F(t)/m,

where F(t) is the driving force.

(Note: in this problem x is the horizontal position of the bob.)

The frequency of _unforced_ oscillations, whose period is given to be T0 = 1 sec, is

w0 = Sqrt(k/m) = 2Pi/T0 = 2Pi sec^-1,

while the frequency of _forced_ oscillations, whose period is given to be T = 1.1 sec is

w = 2Pi/T = 2Pi/1.1 sec^-1.

Let X be the position of the pivot, such that

X = X0 cos(wt), with X0 = 1 cm.

Then we must have (by Newton's law),

mx'' + k(x-X) = 0,

and thus,

(2) x'' + (k/m)x = (k/m)X0 cos(wt) = w0^2 X0 cos(wt) .

Comparing (2) to (1), we find that the driving force is

F(t) = F0 cos (wt), with F0 = kX0,

thus (1) becomes,

(3) x'' + (w0^2)x = (w0^2)X0 cos(wt).

When steady-state motion is attained we have

(4) x = A cos(wt),

where A is the amplitude we are seeking. Substituting (4) into (3), dividing both sides by the common factor cos(wt), we get

-Aw^2 + Aw0^2 = X0(w0^2),

and thus,

A = X0 (w0^2)/(w0^2 - w^2)

= X0 1/(1-(w/w0)^2)

= 1 cm * 1/(1-(1/1.1)^2)

~= 5.76 cm.

 

[K^2]

1) Your differential equation is exactly the same as mine.

2) Steady state cannot be attained with this differential equation. (Hint: What's the x(0) and x'(0) for your solution?)

So much for Caltech.

P.S. I did explain earlier how you can obtain a steady state solution. The answer will be the same 5.8cm, but solution is different, because you must take certain things in the limit.

P.P.S. I'm very glad to find that you are a comp-sci major. My respect for Caltech's physics department remains untarnished.

 

[Dickfore]

2) Steady state cannot be attained with this differential equation. (Hint: What's the x(0) and x'(0) for your solution?)

It can. And that's exactly done by adjusting the x(0) and x'(0) in the absence of friction. Otherwise, just make the eigenvalues with a small negative imaginary part so that

$$|e^{-i \, \omega \, t}| = e^{t \, \Im{\omega}} \rightarrow 0, t \rightarrow +\infty$$

This means that you substitute:
$$\omega \rightarrow \omega - i \epsilon \Rightarrow \omega^{2} \rightarrow (\omega - i \, \epsilon)^{2} = \omega^{2} - i \, \epsilon$$

Thus, after a very long time, the solutions of the homogeneous equation will dampen and you will get only the solution of the inhomogeneous equation.

 

[K^2]

Right. But that's exactly what I'm saying. You have to take the limit as damping goes to zero. You cannot find a solution at damping=0.

The problem with adjusting x(0) and x'(0) is that then there is no t such that x(t)=0 and x'(t)=0. There can be no time at which bob was at rest. That's not a physical solution.

Look at the statement of the problem.

With what amplitude should the pendulum bob swing after a steady motion is attained?

This implies starting at rest, or at very least, not from steady state. With differential equation given by codeleib, you have to start with steady state. If you don't, it will never be achieved.

 

[Dickfore]

Derive it.

Go into a non-inertial reference frame that is moving with the pivot. The acceleration of the frame is $\ddot{x}_{p}(t)$. This will cause an extra inertial force in the x direction. The torque that acts on the pendulum gives angular acceleration. If we measure the position of the pendulum by the deflection angle from the vertical $\theta$, then the equation of motion for rotation is:

$$m \, l^{2} \, \ddot{\theta} = - m \, g \, l \, \sin{(\theta)} + (-m \, a_{p}(t)) \, l \, \cos{(\theta)}$$

In the small amplitude approximation $\theta \ll 1$, we have:
$$\sin{(\theta)} \approx \theta$$
$$\cos{(\theta)} \approx 1$$
and the eqauation becomes:
$$\ddot{\theta} + \frac{g}{l} \, \theta + \frac{\ddot{x}_{p}(t)}{l} = 0$$
Now, instead of measuring the declination by an angle, we can measure it by the x-component of the deflection:
$$x = l \, \sin{(\theta)} \approx l \, \theta$$
We then get:
$$l \, \ddot{\theta} + \frac{g}{l} \, (l \, \theta) + \ddot{x}_{p}(t) = 0$$
$$\ddot{x} + \omega^{2} \, x + \ddot{x}_{p}(t) = 0, \; \omega = \sqrt{\frac{g}{l}}$$

 

[Dickfore]

The problem with adjusting x(0) and x'(0) is that then there is no t such that x(t)=0 and x'(t)=0. There can be no time at which bob was at rest.

There is. Your answer was wrong because you don't know what a steady solution is.

 

[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:
$$\ddot{x} = -\omega^{2}_{p} \, x$$
$$\ddot{x}_{p} = -\omega^{2}_{p} \, x_{p}$$
and you get:
$$-\omega^{2}_{p} \, x + \omega^{2} \, x - \omega^{2}_{p} \, x_{p} = 0$$
$$x = \frac{\omega^{2}_{p} \, x_{p}}{\omega^{2} - \omega^{2}_{p}} = \frac{x_{p}}{\left(\frac{\omega}{\omega_{p}}\right)^{2} - 1} = \frac{x_{p}}{\left(\frac{T_{\mathrm{pivot}}}{T_{ \mathrm {pend.} }}\right)^{2} - 1}$$
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:
$$x(t) = C_{1} \cos{(\omega t)} + C_{2} \, \sin(\omega t) + A \, \sin{(\omega_{p} t)}$$

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

 

[Dickfore]

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

Why?