How to prove that motion is periodic but not simple harmonic?

[vcsharp2003]

So it seems. But are the two the same kind of zero in the sense that $\sin(\omega t)$ crosses the axis going the same way? That's why I suggested that you plot this thing.

I have tried to plot the 3 graphs of $y=sin{x}$ in bold black, $y= sin{2x}$ in normal black and $y=sin{4x}$ in bold blue as shown below.

It seems that the combination of these 3 graphs will repeat every 360° or 2π radians. We can't say the combination will repeat every 180° or π radians since the graph of $y = sin x$ is inverted after 180° even though the patterns of $y= sin{2x}$ and $y= sin{4x}$ graphs are repeating every 180°. The superposition of multiple graphs depends entirely on the pattern of individual graphs; so, if patterns of graphs do not change then the superposition of these graphs will not change.
From above explanation, we can say that motion is periodic with a time period equal to the time period of $y= sin {\omega t}$ which is $T= \dfrac {2\pi}{\omega}$.

 

[haruspex]

Can't we just put $t=1$ in the equation obtained for $a$, which would result in $a(1)= -w^2 x(1) + c$ where $c$ is some non-zero number and so clearly $a$ is not proportional to $x$?

No, that does not follow. If you plug in a value for t (ω being some constant) then x and a are also just numbers. To check for proportionality, you need to compare two things that are varying.

 

[vcsharp2003]

No, that does not follow. If you plug in a value for t (ω being some constant) then x and a are also just numbers. To check for proportionality, you need to compare two things that are varying.

I see. Then, do you have any hint on how to go about this and prove that $a$ is not proportional to $x$? I have zero background in differential equations.

 

[haruspex]

I see. Then, do you have any hint on how to go about this and prove that $a$ is not proportional to $x$? I have zero background in differential equations.

In $\ddot x+\Omega^2x=0$, what does doubling x do to a? In your equation in post #18, does doubling x do the same to a?

 

[vcsharp2003]

In $\ddot x+\Omega^2x=0$, what does doubling x do to a? In your equation in post #18, does doubling x do the same to a?

If $x$ is doubled then so should $a$ since $a$ is directly proportional to $x$.

The acceleration equation mentioned in post#18 would not double $a$ if $x$ was replaced by $2x$, due to the term appearing after $-\omega^2 x$.

 

[kuruman]

I have tried to plot the 3 graphs of $y=sin{x}$ in bold black, $y= sin{2x}$ in normal black and $y=sin{4x}$ in bold blue as shown below.

It seems that the combination of these 3 graphs will repeat every 360° or 2π radians. We can't say the combination will repeat every 180° or π radians since the graph of $y = sin x$ is inverted after 180° even though the patterns of $y= sin{2x}$ and $y= sin{4x}$ graphs are repeating every 180°. The superposition of multiple graphs depends entirely on the pattern of individual graphs; so, if patterns of graphs do not change then the superposition of these graphs will not change.
From above explanation, we can say that motion is periodic with a time period equal to the time period of $y= sin {\omega t}$ which is $T= \dfrac {2\pi}{\omega}$.

View attachment 322760

Although a plot generated by computer (there are many free online plotting apps) would have been better, I think you have grasped the idea and expressed it in your own words which is good.

You have found that the period of this function is $T=2\pi/\omega$. Now you have to show that the function is not harmonic. Others have suggested to show that the equation $\dfrac{d^2x}{dt^2}=-\Omega^2 x$ where $\Omega$ is constant does not hold for all values of $t$. That's a good way to do it. However you might also consider how many harmonic functions there are that have the same period $T=2\pi/\omega$ (or frequency $\omega$), make a list with their mathematical expressions and see if $x$ is on the list. If it's not, then $x$ is not harmonic.

 

[vcsharp2003]

Although a plot generated by computer (there are many free online plotting apps) would have been better,

I used an online plotter after reading your answer and I get a graph as below, which clearly is not sinusoidal and therefore, not SHM but it's periodic since a pattern in the graph is being constantly repeated.

Others have suggested to show that the equation d2xdt2=−Ω2x where Ω is constant does not hold for all values of t.

Ok. I'll try to substitute an appropriate value $\dfrac {3\pi}{2\omega}$ in the equation of $x$ as well as of $a$, and then verify if the differential equation is satisfied.

 

[TSny]

I get the following, after taking, second derivative of $x$ wrt $t$.

$a= -\omega ^2 x -\omega ^2 (3 sin{(2\omega t)} + 15sin {(4\omega t)})$

I don't see how you are getting the factors of 3 and 15.

You are given $x = sin{\omega t} + sin{2\omega t} + sin{4\omega t}$.

Take the time derivative of both sides. What do you get for $\dot x$?

Repeat to get an expression for $\ddot x$.

 

[vcsharp2003]

I don't see how you are getting the factors of 3 and 15.

You are given $x = sin{\omega t} + sin{2\omega t} + sin{4\omega t}$.

Take the time derivative of both sides. What do you get for $\dot x$?

Repeat to get an expression for $\ddot x$.

I tried again and get the following as shown in the screenshot.

 

[TSny]

I tried again and get the following as shown in the screenshot.

OK. You have $$\ddot x = -\omega^2 \sin \omega t - 4 \omega^2 \sin 2\omega t -16 \omega^2\sin 4 \omega t$$
If the motion is SHM, then there must exist a number $\Omega$ such that $\ddot x = -\Omega^2 x$. That is, $$\ddot x =-\Omega^2 \sin \omega t - \Omega^2 \sin 2\omega t - \Omega^2\sin 4 \omega t$$ So there must be an $\Omega$ such that $$\Omega^2 \sin \omega t + \Omega^2 \sin 2\omega t + \Omega^2\sin 4 \omega t = \omega^2 \sin \omega t + 4 \omega^2 \sin 2\omega t +16 \omega^2\sin 4 \omega t$$ and this must hold for all times $t$.

 

[vcsharp2003]

So there must be an Ω such that Ω2sin⁡ωt+Ω2sin⁡2ωt+Ω2sin⁡4ωt=ω2sin⁡ωt+4ω2sin⁡2ωt+16ω2sin⁡4ωt and this must hold for all times

Can we not say from the equation that there is no such $\Omega$ since the coefficients of sine terms on RHS are not equal?

 

[TSny]

Can we not say from the equation that there is no such $\Omega$ since the coefficients of sine term on RHS are not equal?

OK. But if you've never really proven this sort of thing, then it would be good to provide a proof.
For example, what does the relation tell you when $t$ is such that $\omega t = \pi/2$? Repeat for $\omega t = \pi/4$.

 

[vcsharp2003]

OK. But if you've never really proven this sort of thing, then it would be good to provide a proof.
For example, what does the relation tell you when $t$ is such that $\omega t = \pi/2$? Repeat for $\omega t = \pi/4$.

I did what you suggested and get the following. But, I don't get what it achieves?

When $\omega t = \dfrac {\pi}{2}$ then $\Omega^2 = \omega^2$.

When $\omega t = \dfrac {\pi}{4}$ then $\Omega^2 (\dfrac {1}{\sqrt{2}} + 1) = \omega^2 (\dfrac {1}{\sqrt{2}} + 4)$.

So $\omega$ and $\Omega$ are inconsistent in their relationship.

I hope it could be explained more simply as it's just getting too complex.

 

[TSny]

I did what you suggested and get the following. But, I don't get what it achieves?

When $\omega t = \dfrac {\pi}{2}$ then $\Omega^2 = \omega^2$.

When $\omega t = \dfrac {\pi}{4}$ then $\Omega^2 (\dfrac {1}{\sqrt{2}} + 1) = \omega^2 (\dfrac {1}{\sqrt{2}} + 4)$.

So $\omega$ and $\Omega$ are inconsistent in their relationship.

Yes, that is right.

 

[TSny]

@vcsharp2003
Your answer to @haruspex in post #40 is actually a good way to show that it's not SHM. Sorry that I did not see that sooner.

 

[vcsharp2003]

@vcsharp2003
Your answer to @haruspex in post #40 is actually a good way to show that it's not SHM. Sorry that I did not see that sooner.

I put the second derivative in the following form.

$a= -\omega ^2 x -\omega ^2 (3 sin{(2\omega t)} + 15sin {(4\omega t)})$

Isn't that clearly saying $a$ is simply not $a= -\omega ^2 x$. If it were SHM, then we would have got $a= -\omega ^2 x$, but we didn't. To me that is so obvious. Based on the above fact, clearly it's not SHM.

No need of ultra-complex proofs and other logic.

 

[haruspex]

I put the second derivative in the following form.

$a= -\omega ^2 x -\omega ^2 (3 sin{(2\omega t)} + 15sin {(4\omega t)})$

Isn't that clearly saying $a$ is simply not $a= -\omega ^2 x$.

Yes, but as pointed out, you need to show there does not exist any constant, A, the given $\omega$ or otherwise, for which $\ddot x=-A^2x$. Post #40 does that.