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A particle executes simple harmonic moti...

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

A

`sqrt(3)` cm

B

`sqrt(5)` cm

C

`2sqrt(3)` cm

D

`2sqrt(5)` cm

Text Solution

Verified by Experts

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