Question

A particle performs $S.H.M.$ whose velocity is $v_1$ at a distance $x_1$ from mean position and velocity is $v_2$ at distance $x_2$ . The time period and amplitude will be.

Answer 1

Let the amplitude of the oscillation be$A$and$\omega$be the angular frequency then,

$v_1^2={\omega }^2\left(A^2-x_1^2\right)$

$v_2^2={\omega }^2\left(A^2-x_2^2\right)$

Eliminating$A$from the above equations by subtracting we get

${\omega }^2=\left(\frac{v_1^2-v_2^2}{x_2^2-x_1^2}\right)$

We know

$\omega =\frac{2\pi }{T}$

$T=2\pi \sqrt{\left(\frac{x_2^2-x_1^2}{v_1^2-v_2^2}\right)}$

Hence, the time period is $T=2\pi \sqrt{\left(\frac{x_2^2-x_1^2}{v_1^2-v_2^2}\right)}$