A particle is executing S.H.M. along a straight line. Its velocities at distances $ x_1 $ and $ x_2 $ from the mean position are $ v_1 $ and $ v_2 $, respectively. Its time period is: (2015)
$ 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} $
$ 2\pi \sqrt{\frac{v_1^2 - v_2^2}{x_1^2 + x_2^2}} $
$ 2\pi \sqrt{\frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}} $
$ 2\pi \sqrt{\frac{x_1^2 + x_2^2}{v_1^2 + v_2^2}} $
Solution:
Velocity in SHM is $ v^2 = \omega^2(A^2 - x^2) $. So, $ v_1^2 = \omega^2(A^2 - x_1^2) $ and $ v_2^2 = \omega^2(A^2 - x_2^2) $. Subtracting these equations gives $ v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2) $, yielding $ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} $.
Leave a Reply