A body of mass `0.5kg` is performing `SHM` with a time period `.^(pi)//_(2)` seconds. If its velocity at mean position is `1ms^(-1)`, then restoring force acting on the body at a phase angle `60^(@)` from exterme position is

Options

A. 0.5 N
B. 1 N
C. 2 N
D. 4 N

Answer

B

Explanation

`omega = (2pi)/(((pi)/(2)))=4`
`A omega = 1 rArr A = (1)/(4)m`
`x = A cos omega t = A cos 60^(@) = (A)/(2) = (1)/(8)`
`a = omega^(2) x = 4^(2) xx (1)/(8) = 2m//s^(2)`
`F = ma = (0.5)(2) = 1N`