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

In summary, if you have the equation of a SHM, you can find the time period by finding the second order differential and comparing it with the general equation for acceleration.
  • #1
Wrichik Basu
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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

:[/B]

Stated above.

The Attempt at a Solution

:[/B]

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.
 
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  • #2
Wrichik Basu said:
So, we can compare and find ##\omega ## from here. And then the time period.
Right.
Wrichik Basu said:
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.##
 
  • #3
kuruman said:
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.
 
  • #4
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.
 
  • #5
kuruman said:
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##.
 
  • #6
Wrichik Basu said:
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.##
 
  • #7
kuruman said:
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?
 
  • #8
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.
 
  • #9
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
 

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

1. How do I determine the time period of SHM from the equation of motion?

The time period of SHM can be determined by taking the inverse of the frequency, which is given by the coefficient of the sine or cosine term in the equation of motion.

2. What is the significance of the angular frequency in determining the time period of SHM?

The angular frequency, given by the coefficient of the sine or cosine term in the equation of motion, is directly related to the time period of SHM. The time period can be calculated by taking 2π divided by the angular frequency.

3. Can the time period of SHM be affected by changes in the amplitude or mass?

No, the time period of SHM is only affected by the spring constant and the mass of the object. The amplitude and mass do not have an impact on the time period.

4. How does the equation of motion for SHM relate to the formula for calculating time period?

The equation of motion, x = A cos(ωt + φ), is directly related to the formula for calculating time period. The coefficient of the cosine term, ω, represents the angular frequency, which can be used to calculate the time period by taking the inverse of the frequency (1/ω).

5. Is there a specific unit for time period in SHM?

Yes, the unit for time period in SHM is seconds (s). This unit is derived from the units of the angular frequency, which is radians per second (rad/s).

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