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The maximum velocity of a particle execu...

The maximum velocity of a particle executing simple harmonic motion is v. If the amplitude is doubled and the time period of oscillation decreased to 1/3 of its original value the maximum velocity becomes

A

18 v

B

12 v

C

6 v

D

3 v

Text Solution

Verified by Experts

The correct Answer is:
C
`v_(max)=Aomega=(2piA)/(T) implies vprop(A)/(T)`
`(v_(1))/(v_(2))=(A_(1))/(A_(2))xx(T_(2))/(T_(1))=(1)/(2)xx(1)/(3)`
`implies v_(2)=6v_(1)=6v`.
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