Question

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

Answer 1

before the blow, we know the total energy of the system was 1/2 k r^2 since when the particle is at maximum amplitude, all the mechanical is in the form of elastic PE which is equal to 1/2 k x^2 where x is the displacement from equilibrium

just before the blow, the PE of the particle was 1/2 k s^2, so its KE at the time was

1/2 k r^2 - 1/2 k s^2 = 1/2 k(r^2-s^2)

if the blow doubled the velocity, the KE quadrupled (since KE depends on v^2), so the KE immediately after the blow was 2 k(r^2 - s^2)

now, the PE is still 1/2 k s^2, so the total energy of the system was

total energy = total PE + total KE = 1/2 k s^2 + 2 k(r^2 - s^2) = 2 k r^2 - 3/2 k s^2

since this is the total energy, it must be equal to the new total energy when the particles is at the farthest point, which we call X. The PE at X is

1/2 k X^2 and this equals total energy

so 1/2 k X^2 = 2 k r^2 - 3/2 k s^2

divide through by 1/2 k and obtain

X^2 = 4 r^2 - 3 s^2

and X = Sqrt[4 r^2 - 3s^2]