A load of mass m falls from a height h on the scale pan hung from a...

From the energy conservation principle,

$mgh = \dfrac{1}{2}k{X}_{0}^{2} - mg{X}_{0}$

where, ${X}_{0}$ is maximum elongation in spring.

$⇒ \dfrac{1}{2}k{X}_{0}^{2} - mg{X}_{0} - mgh = 0$

$⇒ {X}_{0}^{2} - \dfrac{2mg}{k}{X}_{0} - \dfrac{2mg}{k}h = 0$

${X}_{0} = \dfrac{2\dfrac{mg}{2} \pm \sqrt{\left(\dfrac{2mg}{k}\right)2 + 4 \times \dfrac{2mg}{k}h}}{2} = \dfrac{mg}{2} + \sqrt{\left(\dfrac{mg}{k}\right)2 + \dfrac{2mg}{k}h}$

Amplitude = elongation in spring for lowest extreme position\left({X}_{0}\right)$~$-$~elongation in spring for equilibrium position\left({X}_{1}\right)$

$= {X}_{0} - {X}_{1} = \dfrac{mg}{k}\sqrt{1 + \dfrac{2hk}{mg}}$,~ & \left[∵ {X}_{1} = \dfrac{mg}{k}\right]