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A particle performs simple harmonic moti...

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

A

`A/3sqrt41`

B

`3A`

C

`Asqrt3`

D

`(7A)/3`

Text Solution

Verified by Experts

The correct Answer is:
D
Velocity of particle executing SHM,
`v=omegasqrt(A^(2)-x^(2))" "[omega=` angular velocity of particle]
At `x=(2A)/3,v.=omegasqrt(A^(2)-(4A^(2))/9)=omegasqrt((5A^(2))/9)`
At `x=(2A)/3`, the speed of the particle is trebled.
`therefore" "omegasqrt(A.^(2)-x^(2))=3omegasqrt((5A^(2))/9)` [A. = new amplitude]
or, `A.^(2)-x^(2)=5A^(2)`
or, `A.^(2)-5A^(2)+x^(2)=5A^(2)+((2A)/3)^(2)=5A^(2)+(4A^(2))/9`
= `(49A^(2))/9`
`therefore" "A.=(7A)/3`
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