Question

A particle executes simple harmonic motion with an amplitude of `4cm` At the mean position the velocity of tge particle is `10` earth distance of the particle from the mean position when its speed `5` point is
A. `sqrt(3)cm`
B. `sqrt(5)cm`
C. `21(sqrt(3))cm`
D. `21(sqrt(5))cm`

Answer 1

Correct Answer - C
`v_(max) = a omega rArr omega = (v_(max))/(a) = (10)/(4)`
Now `v = omega sqrt(a^(2) - y^(2)) rArr v^(2) = omega^(2)(a^(2) - y^(2))`
`rArr v^(2) = a^(2) - (v^(2))/(omega^(2))`
`rArr T prop sqrt(m) = 2 sqrt(3) cm`