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

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##.
  • #36
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 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}##.

CamScanner 02-23-2023 13.24.jpg
 
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  • #37
vcsharp2003 said:
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.
 
  • #38
haruspex said:
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.
 
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  • #39
vcsharp2003 said:
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?
 
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  • #40
haruspex said:
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##.
 
  • #41
vcsharp2003 said:
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.
 
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  • #42
kuruman said:
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.

Screenshot_20230223-221552.jpg

kuruman said:
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.
 
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  • #43
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)})##

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##.
 
  • #44
TSny said:
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.

CamScanner 02-23-2023 13.24_2.jpg
 
  • #45
vcsharp2003 said:
I tried again and get the following as shown in the screenshot.
1677178463448.png

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##.
 
  • #46
TSny said:
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?
 
  • #47
vcsharp2003 said:
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##.
 
  • #48
TSny said:
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.
 
  • #49
vcsharp2003 said:
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.
 
  • #50
@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.
 
  • #51
TSny said:
@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.
 
  • #52
vcsharp2003 said:
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.
 
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