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

Answer

C

Explanation

`v=omega sqrt(a^2-((2a)/3)^(2))`
`v=omega sqrt((5)/(9)a^2)`
`v=(sqrt(5))/3 a omega `
so the new amplitude is given by
`v_("new")= omega sqrt(a_("New")^(2)-x^2)`
`sqrt(5)a omega=omega sqrt(a_("New")^(2)-(4a^2)/(9))`
`5a^(2)=a_("New")^(2)-(4a^2)/(9)`
`a_("New")^(2)-(4a^2)/(9)`
`a_("New")^(2)=a^(2)[5+(4)/(9)]`
`a_("New")=a sqrt((49)/(9))=((7a)/(3))`