The maximum velocity of a particle executing SHM is v. If its amplitude of oscillation is doubled and its time period is made half of its initial value, the maximum velocity of it will be

Options

A. v
B. 2 v
C. 4 v
D. 8 v

Answer

C

Explanation

`V=Aomega=(2piA)/(T),V'=((2A)xx2pi)/(T//2)`
`=(4A)/(T)xx2pi`
`=4V`.