A block of mass M rests on a friction less horizontal table and is connected to two fixed posts by springs haveing spring constants k₁ and k₂ respectively. Suppose the block is vibrating with amplitude A and that at the instant that it is passing through its equilibrium position, a mass m is dropped vertically on to the block and sticks to it. Find the new amplitude of vibration.

 as the 2nd  body is dropeed w(omega )changes to w_new=(k₁+k₂/M+m)^1/2

but the velocity at that point would be the same Aw_initial

let the new amp be A_new

A_new w_new=Aw_initial

whr w_initial =(k₁+k₂/M)^1/2