A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance
`(2A)/3` from equilibrium position. The new amplitude of the motion is:

Options

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

Answer

D

Explanation

The velocity of a particle executing SHM at any instant, is defined as the time rate of change of its displacement at that instant. `v=omegasqrt(A^(2)-x^(2))`
where, is angular frequency, A is amplitude and x is displacement of a particle. Suppose that the new amplitude of the motion be
Initial velocity of a particle performs SHM,`v^(2)=omega^(2)[A^(2)-(2A/3)^(3)]`
where, A is initial amplitude and is angular frequency.
final velocity , `(3v)^(2)=omega^(2)[A^(2)-(2A/3)^(3)]`
form Eqs.(i)and(ii),we get `1/9=((A^(2)-4A^(2))/(9))/(A^(2)-4A^(2)/(9))impliesA'(7A)/3`