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. (a) `Asqrt3`
B. (b) `(7A)/3`
C. (c ) `A/3sqrt(14)`
D. (d) `3A`

Answer

D

Explanation

(b) We know that `V=omegasqrtA^(2)=x^(2)`
Initislly `V=omegasqrt(A^(2)-((2A)/3)^(2))`
Fimd `3v=omegasqrt(A'2-((2A)/3)^(2))`
Where `A'=finad amplitude (Given at `x=(2A)/3` veocity
to trebled)
On dividing we get 3/1=sqet((A'^(2)-((2A)/3)^(2))/sqrt((A^(2)-((2A)/3)^(2))`
`9[A^(2)-(4A^(2))/9]=A'2-(4A^(2)/9` :. `A'=(7A)/3`.