A particle performs simple harmonic motion with amplitude A. its speed is double at the instant when it is at distance `(A)/(3)` from equilibrium position. The new amplitude of the motion is `(sqrt(33)A)/(beta)`. Find `beta`

Answer

3

Explanation

`V=omegasqrt(A^(2)-((A)/(3))^(2))`
`V=sqrt((8A^(2)omega)/(9))=(2sqrt(2))/(3)Aomega`.
`V_(new)=2V=(4sqrt(2))/(3)Aomega`
So the new amplitude is given by
`V_(new)=omegasqrt((A_(new))^(2)-((A)/(3))^(2))`
`(32)/(9)A^(2)=(A_(new))^(2)-(A^(2))/(9)`
`A_(new)2=(33A^(2))/(9)`
`A_(new)=(sqrt(33)A)/(3)`