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The maximum velocity of a particle execu...

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

A

18 V

B

6 V

C

12 V

D

3 V

Text Solution

Verified by Experts

The correct Answer is:
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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