A load of mass m falls from a height h onto a | vedclass

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(B) Let $x_0$ be the equilibrium extension of the spring when the mass $m$ is placed on the pan. At equilibrium,$mg = kx_0$,so $x_0 = \frac{mg}{k}$.Le

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$\frac{mg}{k} \sqrt{1 + \frac{2hk}{mg}}$

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(B) Let $x_0$ be the equilibrium extension of the spring when the mass $m$ is placed on the pan. At equilibrium,$mg = kx_0$,so $x_0 = \frac{mg}{k}$.Let $x_1$ be the maximum downward displacement (maximum elongation) of the spring from its natural length when the mass $m$ falls from height $h$. Applying the principle of conservation of mechanical energy between the initial position (mass at height $h$ above the pan) and the lowest point of the motion:Initial energy (potential energy relative to the lowest point) = Final energy (elastic potential energy of the spring).$mg(h + x_1) = \frac{1}{2} kx_1^2$$2mgh + 2mgx_1 = kx_1^2$$kx_1^2 - 2mgx_1 - 2mgh = 0$Solving this quadratic equation for $x_1$:$x_1 = \frac{2mg \pm \sqrt{(2mg)^2 - 4(k)(-2mgh)}}{2k} = \frac{2mg + \sqrt{4m^2g^2 + 8kmgh}}{2k} = \frac{mg}{k} + \sqrt{\frac{m^2g^2}{k^2} + \frac{2mgh}{k}}$$x_1 = \frac{mg}{k} + \frac{mg}{k} \sqrt{1 + \frac{2hk}{mg}}$The amplitude of vibration $A$ is the distance from the equilibrium position $x_0$ to the extreme position $x_1$:$A = x_1 - x_0 = \left( \frac{mg}{k} + \frac{mg}{k} \sqrt{1 + \frac{2hk}{mg}} \right) - \frac{mg}{k} = \frac{mg}{k} \sqrt{1 + \frac{2hk}{mg}}$

$A$ load of mass $m$ falls from a height $h$ onto a scale pan hung from a spring as shown in the figure. If the spring constant is $k$,the mass of the scale pan is zero,and the mass $m$ does not bounce relative to the pan,then the amplitude of vibration is

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