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A particle executes simple harmonic moti...

A particle executes simple harmonic motion with an amplitude of 4 cm . At the mean position the velocity of the particle is 10 cm/ s . The distance of the particle from the mean position when its speed becomes 5 cm/s is

A

`sqrt3`

B

`sqrt5` cm

C

`2(sqrt(3))cm`

D

`2(sqrt(5))cm`

Text Solution

Verified by Experts

The correct Answer is:
C
`v_(max)=Aomegaimplies omega=(v_(max))/(A)=(10)/(4)`
Now, `v=omega sqrt(A^(2)-y^(2))`
`implies (5)^(2)=omega^(2)(A^(2)-y^(2))`
`implies (5)^(2)=((10)/(4))^(2) (4^(2)-y^(2))implies y=2sqrt(3)` cm.
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