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A particle executing simple harmonic mot...

A particle executing simple harmonic motion has an amplitude of 6 cm . Its acceleration at a distance of 2 cm from the mean position is `8 cm//s^(2)` The maximum speed of the particle is

A

`8 cms^(-1)`

B

`12 cms^(-1)`

C

`16 cms^(-1)`

D

`24 cms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`a=omega^(2)yimpliesomega=sqrt((a)/(y))=sqrt((8)/(2))=2"rads"^(-1)`
Now `v_("max")=Aomega=6xx 2=12 cms^(-1)`
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