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?
I tried by finding the second order differential of the given equation.
$a = \dfrac {d^2 x}{d t^2} = – A \omega ^2 \cos (\omega t + \phi)$
Comparing it with the general equation for acceleration $a = – \omega ^2 x$, we can find $\omega$ from here.
But that is where the problem is coming. It makes no sense if I write $\omega = \omega \sqrt {A}$.
What is the correct method to find the time period of the SHM? What am I missing?
Best Answer
There is a very simple mistake in your math. Notice $A$ is part of $x$, it is factored so you'll get to $\omega=\omega$ again. If you want to find a meaning to $\omega T = 2\pi$, consider the fact that $\cos$ (or $\sin$) are periodic functions with period $2\pi$. Hence, every time you have a time difference such that $\omega(t_1− t_2)=2 \pi$ you are back at the same point. Hence the period is given by $\omega T = 2\pi$.