The maximum velocity of a particle executing S.H.M. is u. If the amplitude is doubled and the time period of oscillation decreases to (1/3) of its original value, then the maximum velocity will be

Options

A. 18 u
B. 6 u
C. 12 u
D. 3 u

Answer

B

Explanation

`(v_(m_(2)))/(v_(m_(1)))=(A_(2)omega_(2))/(A_(1)omega_(1))=(2T_(1))/(T_(2))=(2T_(1))/((1)/(3)T_(1))=3xx2`
`v_(m_(2))=6v_(m_(1))`
`=6xxu=6u`