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

• vcsharp2003
In summary, we need to prove that the sum of trigonometric ratios ##x = sin{\omega t} + sin{2\omega t} + sin{4\omega t}## is periodic but not simple harmonic. This can be shown by analyzing the time periods of each term and noticing that they are not all equal. Additionally, when taking the second derivative of ##x##, we see that it does not follow the property of SHM, further proving that this sum is not simple harmonic. To find the period, we can plot the function and notice that it repeats at intervals of ##2\pi/\omega##.

#### vcsharp2003

Thread moved from the technical forums to the schoolwork forums
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.

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

vcsharp2003
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.

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##?

nasu and 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

vcsharp2003
TSny said:
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##

hutchphd said:
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##?

vcsharp2003 said:
##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?

vcsharp2003 said:
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
hutchphd said:
Did I say that awkwardly?
You didn't use the equal sign, so I was confused.

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

vcsharp2003
TSny said:
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?

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erobz
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##

vcsharp2003 said:
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?

erobz said:
That looks like it needs some parenthesis?
You mean parenthesis around ##2\omega t## and ##4\omega t##?

vcsharp2003 said:
You mean parenthesis around ##2\omega t## and ##4\omega t##?
I meant:

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

vcsharp2003 and berkeman
erobz said:
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...

erobz
berkeman said:
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.

erobz and berkeman
vcsharp2003 said:
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?

erobz
kuruman said:
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##.

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vcsharp2003 said:
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?

vcsharp2003 said:
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
kuruman said:
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.

erobz said:
Further hint: think of certain multiples of a very famous transcendental number.
Hahaha. That's a nice hint.

vcsharp2003 said:
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.

erobz and vcsharp2003
kuruman said:
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
vcsharp2003 said:
Hahaha. That's a nice hint.
Parity matters as @kuruman points out, which is why I said certain multiples.

vcsharp2003
kuruman said:
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!

erobz said:
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?

vcsharp2003 said:
##n=0,1,2,3...##. Is this what you meant by certain multiples?
parity refers to even-odd

vcsharp2003
erobz said:
parity refers to even-odd
This is getting complex for me. But I'll try to figure it out.

vcsharp2003 said:
This is getting complex for me. But I'll try to figure it out.
Just plot it first like @kuruman suggested.

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?

vcsharp2003 said:
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, ChiralSuperfields and TSny
haruspex said:
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.

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