1.

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 beA. 18 uB. 6 uC. 12 uD. 3 u

Answer» Correct Answer - B
`(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`


Discussion

No Comment Found

Related InterviewSolutions