A body of mass m falls from a height h onto the pan of a spring balance. The masses of the pan and spring are negligible. The force constant of the spring is k. The body sticks to the pan and oscillates simple harmonically. The amplitude of oscillation is

Options

A. `(mg)/(k)`
B. `(mg)/(k)sqrt(1+(2hk)/(mg))`
C. `(mg)/(k)+(mg)/(k)sqrt((1+2hk)/(mg))`
D. None of these

Answer

b

Explanation

From the conservation principle,
`mgh=(1)/(2)kX_(0)^(2)-mgX_(0)`
where , `X_(0)` is maximum elongation in spring.
`rArr(1)/(2)kX_(0)^(2)-mgX_(0)-mgh=0`
`rArrX_(0)^(2)-(2mg)/(k)X_(0)-(2mg)/(k)h=0`
`X_(0)=(2(mg)/(k)+-sqrt(((2mg)/(k))^(2)+4xx(2mg)/(k))h)/(2)`
Amplitude = elongation in spring for lowest extreme position
- elongation in spring for equilibrium position
`=X_(0)-X_(1)=(mg)/(k)sqrt((1+(2hk)/(mg)) " " [becauseX_(1)=(mg)/(k)]`