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Q.

The maximum velocity of the particle executing S.H.M 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

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a

3V

b

6V

c

12V

d

18V

answer is C.

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Detailed Solution

Case-I :Vmax=Aω=A2πT=V
Case-II : Vmax=(2A)2πT/3=6V

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