A body of mass m falls from a height h on to the pan of a spring balance, figure, The masses of the pan and spring are negligible. The spring constant of the spring is k. The body gets attached to the pan and starts executing S.H.M. in the vertical direction. Find the amplitude and energy of oscillation.

Explanation

When a body of mass m falls from height h on to the pan of spring balance, let the spring of spring balance get compressed by a lengthx. The loss of P.E. of the mass `=mg (h+x)`. The gain in elastic potential energy by the spring due to compression `=(1)/(2)kx^(2)`
Accordin to law of conservation of energy, we have
`mg(h+x)=(1)/(2)kx^(2)`
or `(1)/(2)kx^(2)-mgx-mgh=0` or `x^(2)-(2mg)/(k)x-(2mgh)/(k)=0`
Solving this quadratic equation, we have
`x=((2mg)/(k)-sqrt(((mg)/(k))^(2)+((8mgh)/(k))))/(2)=(mg)/(k)+-(mg)/(k)sqrt(1+(2kh)/(mg))`
In equilibrium position, the spring will be compressed by distance `(mg)/(k)`. If r is the amplitude of oscillation then
`(mg)/(k)+r=(mg)/(k)+-(mg)/(k)sqrt(1+(2kh)/(mg))`
`:.` amplitude, `r=(mg)/(k)sqrt(1+(2kh)/(mg))`
Energy of oscialltion `=(1)/(2)kr^(2)=(1)/(2)k((mg)/(k))^(2)(1+(2kh)/(mg))=mgh+((mg)^(2))/(2k)`