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 VB. 6 VC. 12 VD. 3 V

Answer» Correct Answer - B
`(v_(m_(2)))/(v_(m_(1)))=(A_(2)omega_(2))/(A_(1)omega_(1))=(A_(2))/(A_(1))xx(T_(1))/(T_(2))`
`=2xx3=6`
`v_(m_(2))=6v`


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