The velocity of a particle in `SHM` at the instant when it is `0.6cm` away from the mean position is `4cm//s`. If the amplitude of vibration is `1cm` then its velocity at the instant when it is `0.8cm` away form the mean postion is (in cm//s)

Options

A. 6 cm `s^(-1)`
B. 3 cm `s^(-1)`
C. 1.5 cm `s^(-1)`
D. 4.5 cm `s^(-1)`

Answer

B

Explanation

`V_(1)omega = sqrt(A_(1)^(2) - y_(1)^(2)), V_(2)omega = sqrt(A^(2) - y_(2)^(2))`
`(V_(1))/(V_(2)) = sqrt((1^(2) - (0.6)^(2))/(1^(2) - (0.8)^(2))) = (0.8)/(0.6) = (4)/(V_(2)) = (4)/(3)`
`rArr V_(2) = 3 ms^(-1)`