A particle performs simple harmonic motion wit 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

Options

A. `sqrt(11)A`
B. `(sqrt(22)A)/(3)`
C. `(sqrt(33)A)/(2)`
D. none of these

Answer

B

Explanation

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