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If the displacement of a particle executing S.H.M. is given by x = 0.24 sin (400 t + 0.5)m, then the maximum velocity of the particle is

A

`24 m//s`

B

`48 m//s`

C

`96 m//s`

D

`72 m//s`

Text Solution

Verified by Experts

The correct Answer is:
C
`x=0.24sin(400t+0.5)m`
`v_(m)=Aomega`
`=0.24xx400=96m//s`
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