Question

A horizontal spring block system of mass M executes simple harmonic motion. When the block is passing through its equilibrium position, an object of mass m is put on it and the two move together. Find the new amplitude and frequency of vibration. Given, k is the spring constant of the system.

Answer 1

`T_(0) = 2 pi sqrt((M)/(k))` (i)
Let `v_(0)` be the velocity of mass M as it passes the equilibrium position. If `a_(0)` is the initial amplitude, then
`(1)/(2) Mv^(2) = (1)/(2) ka_(0)^(2)`
`implies v = (sqrt((k)/(M))) a_(0)`
If V is the combined velocity of `(M + m)` system in equalibrium possition, then by low of conservation of linear momentum, we have
`Mv = (M + m) V`
`implies V = ((M)/(M + m)) v`
Let a be the new amplitude, then
`(1)/(2) ka = ^(2) (1)/(2) (M + m) V^(2)`
`= (1)/(2)(M + m) ((M)/(M + m))^(2) v^(2)`
`= (1)/(2)(M + m) (M^(2))/((M + m)^(2)) ((k)/(M)) a_(0)^(2)`
`implies a = a_(0) sqrt((M)/(M + m))`
New time period is
`T = 2 pi sqrt((M + m)/(k))`
`= sqrt((M + m)/(M)) T_(0)`
`{{:. T_(0) = 2 pi sqrt((M)/(k))}`