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:

Options

A. `(Asqrt(41))/(3)`
B. 3A
C. `Asqrt(3)`
D. `(7A)/(3)`

Answer

D

Explanation

As `upsilon=omegasqrt(A^(2)-x^(2))`
At `x=2A//3, upsilon=omegasqrt(A^(2)=4A^(2)//9)=(sqrt(5))/(3)omega`
When velocity is trebled, then velocity
`upsilon^(')=3upsilon=3xx(sqrt(5))/(3)omega=sqrt(5)omega`
Let `A^(')` be the new amplitude of oscillation, then
`upsilon^(')sqrt(5)omega=omegasqrt(A'^(2)-4A^(2)//9)`
or `5omega^(2)A^(2)=omega^(2)(A'^(2)-(4A^(2))/(9))`
or `A'^(2)=5A^(')+(4A^(2))/(9)=(49)/(9)A^(2)`
or `A'=(7)/(3)A`