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