The displacement of a particle executing S.H.M. is given by y = 0.25 sin 200t cm. The maximum speed of the particle is

Options

A. `200 cm s^(-1)`
B. `100 cm s^(-1)`
C. `50 cm s^(-1)`
D. `5.25 cm s^(-1)`

Answer

C

Explanation

Given : `x=0.25sin200t`
Comparing it with standard SHM equation `x=Asinomegat`, we get
`A=0.25cm, omega=200s^(-1)`
`therefore` Maximum speed, `v_(m)=Aomega=(0.25cm)(200s^(-1))=50cms^(-1)`