The maximum velocity of a particle executing SHM is V, If the amplitude is doubled and the time period of oscillation de4creased to `1/3` of its original value, the bmaximum velocity becomes,

Options

A. 14 V
B. 12 V
C. 6 V
D. 3 V

Answer

c

Explanation

Max velocity=V
`V=Aomega`
`V=A(2pi)/T`
New velocity `V'omega=A'(2pi)/(T')`
`=2Axx(2pi)/(T//3)`
`=6(Axx(2pi)/T)`
`therefore V'=6 V`