Question

A particle is executing SHM of amplitude r. At the distance s from the mean position, the particle receives a blow in the direction opposite to motion which instantaneously doubles the speed. Find the new amplitude.

Answer 1

Here we have to apply the conservation of total energy before blow and after blow.

Before blow

TE= 1/2kr^2 [since r is the amplitude]
PE= 1/2ks^2 [s is the distanec from the mean position]
Then we get KE as TE- PE as shown below

After blow

Here speed increases to double the speed before blow [ v' = 2*v] hence Ke become four times the Ke before blow [Ke'= 4 * Ke]
so we get the TE, PE and KE as


Let X be the farthest point of the particle
Hence TE ' = 1/2kX^2
Hence we have TE'= 1/2kX^2 and TE'= 1/2K [4r^2 -3s^2]
Comparing we get

we get X^2 = 4r^2-3s^2
Hence X= [ 4r^2 -3s^2] ^ 1/2