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.