Question

How is time period and frequency of simple harmonic motion is related to its acceleration?

Answer 1

Expression for time period is,

$T=\dfrac {2\pi}{\omega}\quad\longrightarrow\quad (1)$

where ω is the angular frequency.

For a simple harmonic motion, the acceleration,

$a=\omega^2y$

where y is the displacement from the mean position. (Magnitude is only considered here.)

From this, we get,

$\omega=\sqrt {\dfrac{a}{y}}$

Then (1) becomes,

$T=\dfrac {2\pi}{\sqrt {\dfrac {a}{y}}}\\\\\\\boxed {T=2\pi\sqrt {\dfrac{y}{a}}}$

We know that frequency is the reciprocal of time period. Hence,

$\boxed {\nu=\dfrac {1}{2\pi}\sqrt {\dfrac{a}{y}}}$