How to find time period of SHM from equation of motion?

[Wrichik Basu]

Homework Statement

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Say for example I've got the equation of a SHM as: $$x = A \cos (\omega t + \phi)$$ where ##A## is the amplitude.

How do I find the time period of this motion?

Homework Equations

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Stated above.

The Attempt at a Solution

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I tried by finding the second order differential of the given equation.

## a = \dfrac {d^2 x}{d t^2} = - A \omega ^2 \cos (\omega t + \phi)##

And comparing it with the general equation for acceleration ##a = - \omega ^2 x##.

So, we can compare and find ##\omega## from here. And then the time period.

But is this approach correct? I don't think so, as ##\omega## gets canceled in the process.

 

[kuruman]

So, we can compare and find ##\omega ## from here. And then the time period.

Right.

But is this approach correct? I don't think so, as ##\omega## gets canceled in the process.

What process? It is always true by definition that ##\omega T =2\pi.##

 

[Wrichik Basu]

What process? It is always true by definition that ##\omega T =2\pi.##

No, I don't mean that. If I compare the two equations for ##a##, then I get ##\omega = \sqrt {A \omega ^2}##. So, I've got ##\omega## on both sides of the equation. Isn't that wrong? Because ##\omega ^2## can get canceled from both sides.

 

[kuruman]

What two equations are you comparing? It looks like you start with ##x(t)=A\cos(\omega t+\phi)## and then take a double derivative to find ##a(t)## under the assumption that ##\omega## is an arbitrary constant that may be defined separately. You cannot extract a value for that constant from these equations.

 

[Wrichik Basu]

What two equations are you comparing? It looks like you start with ##x(t)=A\cos(\omega t+\phi)## and then take a double derivative to find ##a(t)## under the assumption that ##\omega## is an arbitrary constant that may be defined separately. You cannot extract a value for that constant from these equations.

OK, I get it. So, I'll do it like this:

Start with $$x = A \cos (\omega t + \phi)$$
Find second differential: ## a = \dfrac {d^2 x}{d t^2} = - A \omega ^2 \cos (\omega t + \phi)##

Comparing it with the general equation for acceleration ##a = - (\omega ') ^2 x##, $$\omega ' = \sqrt {A \omega ^2}$$
Then, time period ##T = \frac {2\pi }{\omega '} = \frac {2 \pi}{\omega \sqrt{A}}##.

This will solve any ambiguity between the two ##\omega##.

 

[kuruman]

OK, I get it.

No, you don't get it.
If ##x(t)=A\cos(\omega t+\phi)##, and ##a = -\omega^2 A\cos(\omega t +\phi)##, then ##a = -\omega^2 x## which means ##\omega'=\omega.##

 

[Wrichik Basu]

No, you don't get it.
If ##x(t)=A\cos(\omega t+\phi)##, and ##a = -\omega^2 A\cos(\omega t +\phi)##, then ##a = -\omega^2 x## which means ##\omega'=\omega.##

Then what is the correct method?

 

[kuruman]

Correct method for what? What do you know and what are you trying to find? Your original statement of the problem implies that you want to find the period ##T##. Based on what you give, the only thing you can say is ##T=2\pi / \omega## as I indicated in post #2.

 

[Nguyen Son]

First of all, I just don't know how you can get ##\omega'=\sqrt{A\omega^{2}}##
You wrote ##a=-(\omega'^{2})x##, it's ok, but when you put ##x=A\cos(\omega t+\phi)## from above into this equation, how did you get ##\omega'=\sqrt{A\omega^{2}}## ?
Secondly, ##T=\frac{2\pi}{\omega}##, that's a definition, but if you still want to find out why ##T=\frac{2\pi}{\omega}##, you can look at my instruction below.
Here, ##k/n## must be 1 because ##T## is "period", it means that time ##T## must be a smallest number (and non-negative, of course)
p/s: sorry for my laziness, I think that writing on the board is faster than typing on the keyboard