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The amplitude of a executing SHM is 4cm ...

The amplitude of a executing `SHM` is `4cm` At the mean position the speed of the particle is `16 cm//s` The distance of the particle from the mean position at which the speed the particle becomes `8 sqrt(3)cm//s` will be

A

`2sqrt(3)` cm

B

`sqrt(3)` cm

C

1 cm

D

2 cm

Text Solution

Verified by Experts

The correct Answer is:
D
(d) At mean position , velocity is maximum
`v_("max") =omegaaimpliesomega=(v_("max"))/(a)=(16)/(4)=4`
`therefore v=sqrt(a^(2)-y^(2))implies 8sqrt(3)=4sqrt(4^(2)-y^(2))`
`192=16(16-y^(2))implies12=16-y^(2)impliesy=2 cm`
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