Question

A particle starts SHM from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is
A. `A/2`
B. `A/sqrt2`
C. `(Asqrt3)/(2)`
D. `(2A)/(sqrt3)`

Answer 1

Correct Answer - C
`v=omegasqrt(A^2-x^2)`
`(v_(max))/(2)=omegasqrt(A^2-x^2)`
`(Aomega)/(2)=omegasqrt(A^2-x^2)`
`A^2/4=A^2-x^2impliesx^2=(3A^2)/(4)`
`x=(sqrt3A)/(2)`