First slide

Simple hormonic motion

Question

A body of mass M is released from a height h to a scale pan hung from a spring as shown in fig (9). The spring
constant of the spring is k, the mass of the scale pan is negligible and the body does not bounce relative to the pan, then the amplitude of vibration is

Moderate

A

$\frac{m g}{k} \sqrt{(1 - \frac{2 h k}{m g})}$
B

$\frac{mg}{k}$
C

$\frac{mg}{k} [1 + \sqrt{(1 + \frac{2 hk}{mg})}]$
D

$\frac{mg}{k} - \frac{mg}{k} \sqrt{(1 - \frac{2 hk}{mg})}$

Solution

Let y be the displacement of the pan when the mass m hits it. Decrease in potential energy of mass $= mg (h + y), y = amplitude.$
Increase in elastic potential energy of spring
$= \frac{1}{2} {ky}^{2} ∴ mg (h + y) = \frac{1}{2} {ky}^{2} or y^{2} - (\frac{2 mg}{r}) y - \frac{2 mgh}{k} = 0 y^{2} - (\frac{2 mg}{r}) y - \frac{2 mgh}{k} = 0 y = (\frac{(2 mg / k) \pm \sqrt{(2 mg / k)^{2} + 4 (2 mgh / k)}}{2}) Reiecting negative sign, the total extension is y = \frac{mg}{k} + \frac{mg}{k} [\sqrt{(1 + \frac{2 hk}{mg})}] = \frac{mg}{k} [1 + \sqrt{(1 + \frac{2 hk}{mg})}]$

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