Question
A particle is doing simple harmonic motion. If the velocity of the particle at distances $x_1$ and $x_2$ from the mean position are $v_1$ and $v_2$ respectively, then prove that its time period will be
$ T =2 \pi \sqrt{\frac{x_2^2-x_1^2}{ v _1^2- v _2^2}}. $
$ T =2 \pi \sqrt{\frac{x_2^2-x_1^2}{ v _1^2- v _2^2}}. $
