Question

The amplitude of a executing `SHM` is `4cm` At the mean position the speed of the particle is `16 cm//s` The distance of the particle from the mean position at which the speed the particle becomes `8 sqrt(3)cm//s` will be
A. `2sqrt(3) cm`
B. `sqrt(3) cm`
C. `1 cm`
D. `2 cm`

Answer 1

Correct Answer - D
At mean position velocity is maximum
`i.e. v_(max) = omega a rArr omega rArr omega = (v_(max))/(a) = (16)/(4) = 4`
`:. V = omega sqrt(a^(2) - y^(2)) rArr 8 sqrt(4^(2) - y^(2))`
`rArr 192 = 16(16 - y^(2)) rArr 12 = 16 - y^(2) rArr y = 2 cm`