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A particle performs simple harmonic miti...

A particle performs simple harmonic mition with amplitude A. Its speed is trebled at the instant that it is at a destance
`(2A)/3` from equilibrium position. The new amplitude of the motion is:

A

`A/3sqrt(41)`

B

3A

C

`Asqrt(3)`

D

`7/3A`

Text Solution

Verified by Experts

The correct Answer is:
D
The velocity of a particle executing SHM at any instant, is defined as the time rate of change of its displacement at that instant. `v=omegasqrt(A^(2)-x^(2))`
where, is angular frequency, A is amplitude and x is displacement of a particle. Suppose that the new amplitude of the motion be
Initial velocity of a particle performs SHM,`v^(2)=omega^(2)[A^(2)-(2A/3)^(3)]`
where, A is initial amplitude and is angular frequency.
final velocity , `(3v)^(2)=omega^(2)[A^(2)-(2A/3)^(3)]`
form Eqs.(i)and(ii),we get `1/9=((A^(2)-4A^(2))/(9))/(A^(2)-4A^(2)/(9))impliesA'(7A)/3`
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