Two bodies M and N of equal mass are hung separtely from two lightweight springs. Force constants of the springs are k_1 and k_2. The bodies are set to vibrate so that their maximum velocities are equal. Find the ratio of the amplitudes of vibration of the two bodies.

Two bodies M and N of equal mass are hung separtely from two lightweig

1.	Two bodies M and N of equal mass are hung separtely from two lightweight springs. Force constants of the springs are k_1 and k_2. The bodies are set to vibrate so that their maximum velocities are equal. Find the ratio of the amplitudes of vibration of the two bodies.
Answer» Solution :A body suspended from a SPRING will ACQUIRE maximum velocity at the equilibriumposition in its path of vibration. At that position potential energy of the body becomes ZERO and its energy becomes totally kinetic. Let the amplitude of vibration of the body M be `x_1` and that of N be `x_2`. Since they are of EQUAL MASS and their maximum velocities are also the same, their maximum kinetic energies will also be equal [`because` Maximum kinetic energy `=1/2timesmasstimes(maxim um velocity)^2`] . This kinetic energy transforms into the potential energy `(1/2kx^2)` of the body at the end of its amplitude. `therefore1/2k_1x_(1^2)=1/2k_2x_(2^2)or,x_1/x_2=sqrt(k_2/k_1)`.