A block of mass M attached to the free end of a spring of force constant k is nounted on a smooth horizontal table as shown in figure.

The block executes SHM with amplitude A and frequency f . If an object of mass m is put on it, when the block is passing through its equilibium position and the two move together ,then what is the new amplitude and frequency of vibration?

Explanation

Initially the mass of osciallting system is M
`:. f=(1)/(2pi) sqrt((k)/(M))`
Let f be the new frequency and A' be the new amplitude of vibration.
When block of mass m block M, then the total mass of osciallting system becomes `M+m`
`:' f' = (1)/( 2pi ) sqrt((k)/( ( m+M)))`
Now `(l')/( f )= sqrt((M)/( (m+M)))`
`:. f'= fsqrt((M)/( (m+ M)))` .....(i)
The new frequency of oscillaiton f' is less than f
When mass M passes through its mean position, it has maximum speed.
Let v and V be the speed of block of mass M and `M + m`respectively at equilibrium position.
By conservation of linear momentum for the collision.
`Mv=( m+ M) V`
But at equilibrium position
`v= A omega =A xx 2 pi f`
` :. Ma ( 2pi f ) = ( m+ M) A' 2pif`
`implies (A')/( A) = (M)/( (m+M))xx(f)/(f')`
Using equation (i), we have
`A'= A sqrt((M)/((m+M)))`