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

[vcsharp2003]

TL;DR Summary: Prove that a sum of trigonometric ratios is periodic but not not simple harmonic.

We need to prove that $x = sin{\omega t} + sin{2\omega t} + sin{4\omega t}$ where $x$ is the displacement from the equilibrium position at time $t$.

I can see that each term is a SHM, but not sure if the sum of these terms would be a simple harmonic motion.
Also, I can see that the time period of third term is one-fourth that of first term and time period of second term is half that of first term.

But after the above analysis, I am confused.

 

[Vanadium 50]

Is there a number I can add to t to get the same x? What if I add 2t? 3t?

 

[kuruman]

Addendum to the question posed by @Vanadium 50 above.
To complete the answer, you also have to prove that $x$ is periodic and find its period $T$.
You might also wish to plot $x$ to guide your thinking.

 

[TSny]

For SHM at angular frequency $\Omega$, $x$ can always be expressed as $x = A\sin(\Omega t + \phi)$.

So, for SHM, how is $\ddot x$ related to $x$?

 

[hutchphd]

It is interesting that the easiest way to do this problem is to graph it for interval τ such that ωτ~10 π. You will immediately see if it repeats and if it looks like a sinusoid. Then you can show the result analytically as required. A lesson well learned

 

[vcsharp2003]

For SHM at angular frequency $\Omega$, $x$ can always be expressed as $x = A\sin(\Omega t + \phi)$.

So, for SHM, how is $\ddot x$ related to $x$?

$a= -\omega ^2 x$

 

[vcsharp2003]

It is interesting that the easiest way to do this problem is to graph it for interval τ such that ωτ~10 π. You will immediately see if it repeats and if it looks like a sinusoid. Then you can show the result analytically as required. A lesson well learned

Did you mean $\omega t = 10 \pi$?

 

[TSny]

$a= -\omega ^2 x$

Or, $\ddot x = -\Omega^2 x$ for the example I gave.

The important thing is that for SHM the acceleration equals a constant times the displacement.

Does your series expression for $x$ have this property?

 

[hutchphd]

Did you mean ωt=10π?

Did I say that awkwardly? The longest period T possible will be T=2π/ω so I want to look for times ~ five times that interval. Brain fog lately

 

[vcsharp2003]

Did I say that awkwardly?

You didn't use the equal sign, so I was confused.

 

[hutchphd]

Brain fog. I could see the = sign!! Apologies

 

[vcsharp2003]

Or, $\ddot x = -\Omega^2 x$ for the example I gave.

The important thing is that for SHM the acceleration equals a constant times the displacement.

Does your series expression for $x$ have this property?

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

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

Since for SHM, $a= -\omega ^2 x$, so it's not SHM after looking at the above equation for $a$. Is that correct reasoning?

 

[vcsharp2003]

Is there a number I can add to t to get the same x? What if I add 2t? 3t?

I get the following, but I cannot see the value of $x$ repeating.

$x(t+2t) = sin3\omega t + sin6\omega t +sin12\omega t$

$x(t+3t) = sin4\omega t + sin8\omega t +sin16\omega t$

 

[erobz]

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

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

That looks like it needs some parenthesis?

 

[vcsharp2003]

That looks like it needs some parenthesis?

You mean parenthesis around $2\omega t$ and $4\omega t$?

 

[erobz]

You mean parenthesis around $2\omega t$ and $4\omega t$?

I meant:

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

 

[berkeman]

I meant:

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

Thanks! I couldn't parse what he posted...

 

[vcsharp2003]

Thanks! I couldn't parse what he posted...

Yes, my mistake. I had the following in my notebook, but wrote it incorrectly in the forums.

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

Also, I've corrected the equation in post#12 for which you reported the error.

 

[kuruman]

I get the following, but I cannot see the value of $x$ repeating.

Hint: Clearly $x=0$ at $t=0$. At what value of $t$ does the next $x=0$ occur? How about the one after that?

 

[vcsharp2003]

Hint: Clearly $x=0$ at $t=0$. At what value of $t$ does the next $x=0$ occur? How about the one after that?

That's not easy to me. Do I have to solve the trigonometric equation $0=sin{\omega t} + sin{2\omega t} + sin{4\omega t}$?

It seems when $t= \dfrac {\pi}{w}$ then also $x=0$.

 

[kuruman]

That's not easy to me. Do I have to solve the trigonometric equation $0=sin{\omega t} + sin{2\omega t} + sin{4\omega t}$?

Not really. Take first term,$~\sin(\omega t)$. At what values of $t$ is it zero? Were you not taught that when you took trig?

 

[erobz]

That's not easy to me. Do I have to solve the trigonometric equation $0=sin{\omega t} + sin{2\omega t} + sin{4\omega t}$?

Further hint: think of certain multiples of a very famous transcendental number.

 

[vcsharp2003]

Not really. Take first term,$~\sin(\omega t)$. At what values of $t$ is it zero? Were you not taught that when you took trig?

Yes, I can see it. First term would be zero when $t = n\pi$ where n is any integer i.e. +ve or -ve or zero.

 

[vcsharp2003]

Further hint: think of certain multiples of a very famous transcendental number.

Hahaha. That's a nice hint.

 

[kuruman]

It seems when $t= \dfrac {\pi}{w}$ then also $x=0$.

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.

 

[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 see. I will try that too.

 

[erobz]

Hahaha. That's a nice hint.

Parity matters as @kuruman points out, which is why I said certain multiples.

 

[erobz]

That's why I suggested that you plot this thing.

First thing I did. We often take for granted the tools we now have at our disposal to answer these types of questions with almost no effort. Then (once you know what you are working toward) worry about the mathematical purist approach!

 

[vcsharp2003]

Parity matters as @kuruman points out, which is why I said certain multiples.

$n=0,1,2,3...$. Is this what you meant by certain multiples?

 

[erobz]

$n=0,1,2,3...$. Is this what you meant by certain multiples?

parity refers to even-odd

 

[vcsharp2003]

parity refers to even-odd

This is getting complex for me. But I'll try to figure it out.

 

[erobz]

This is getting complex for me. But I'll try to figure it out.

Just plot it first like @kuruman suggested.

 

[haruspex]

Is there a number I can add to t to get the same x? What if I add 2t? 3t?

Did you mean,

Is there a number, T, I can add to t to get the same x? What if I add 2T? 3T?

 

[haruspex]

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

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

Since for SHM, $a= -\omega ^2 x$, so it's not SHM after looking at the above equation for $a$. Is that correct reasoning?

It's not quite that simple. As @TSny pointed out in post #8, $\omega$ is a value given in the equation of motion. To show it is not SHM you need to show there is no value $\Omega$ for which $\ddot x+\Omega^2x=0$.

 

[vcsharp2003]

It's not quite that simple. As @TSny pointed out in post #8, $\omega$ is a value given in the equation of motion. To show it is not SHM you need to show there is no value $\Omega$ for which $\ddot x+\Omega^2x=0$.

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 $x$? Also, we could use $\omega = 1$ in the acceleration equation since $\omega$ can be any value other than zero.

 

[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.