Question

A body of mass $m$ fell from a height $h$ onto the pan of a spring balance having a mass $M$ . The mass of the spring are negligible, the stiffness of the latter is $x$ . Having stuck to the pan, the body starts performing harmonic oscillations in the vertical direction. Find the oscillation amplitude in this case.

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Accepted Answer

Due to a perfectly inelastic collision with the pan, the kinetic energy of the body $m$ decreases. Obviously, the body $m$ will strike the pan with velocity $v_{0} = \sqrt{2 g h}$ . If $v$ is the common velocity of the "body $m +$ pan" system, because of the collision, then from the conservation of linear momentum,

$m v_{0} = (M + m) v$
or, $v = \frac{m v_{0}}{(M + m)} = \frac{m \sqrt{2 g h}}{(M + m)}$

At the moment the body $m$ strikes the pan, the spring is compressed due to the weight of the pan, by the amount $\frac{M g}{κ}$

If $l$ is the further compression of the spring due to the velocity acquired by the pan-body system, then from the conservation of mechanical energy of the system, in the field generated by the joint action of both the gravity and spring forces,

$\frac{1}{2} (M + m) v^{2} + (M + m) g l = \frac{1}{2} κ {(\frac{M g}{κ} + l)}^{2} - \frac{1}{2} κ {(\frac{M g}{κ})}^{2}$
or, $\frac{1}{2} (M + m) \frac{m^{2} 2 g h}{(M + m)} + (M + m) g l = \frac{1}{2} κ {(\frac{M g}{κ})}^{2} + \frac{1}{2} κ l^{2} + M g l - \frac{1}{2} κ {(\frac{M g}{κ})}^{2}$
$\frac{1}{2} κ l^{2} - m g l - \frac{m^{2} g h}{(m + M)} = 0$

or, $\frac{1}{2} κ l^{2} - m g l - \frac{m^{2} g h}{(m + \frac{M}{2})} = 0$
Thus, $l = \frac{m g \pm \sqrt{m^{2} g^{2} + \frac{2 \times g h m^{2}}{M + m}}}{κ}$

As minus sign is not acceptable,
$l = \frac{m g}{κ} + \frac{1}{κ} \sqrt{m^{2} g^{2} + \frac{2 κ \cdot n^{2} g h}{(M + m)}}$

If the oscillating pan $+$ body $m$ system were at rest, it corresponds to their equilibrium position, i.e., the spring was compressed by $\frac{(M + m) g}{κ}$ . Therefore, the amplitude of oscillation,

$a = l - \frac{m g}{κ} = \frac{m g}{κ} \sqrt{1 + \frac{2 h κ}{m g}}$

The mechanical energy of oscillation, which is only conserved with the restoring forces, becomes $E = U_{extreme} = \frac{1}{2} κ a^{2}$ (Because spring force is the only restoring force not the weight of the body)
Alternately,
$E = T_{mean} = \frac{1}{2} (M + m) a^{2} ω^{2}$

Thus, $E = \frac{1}{2} (M + m) a^{2} (\frac{κ}{M + m}) = \frac{1}{2} κ a^{2}$

I E Irodov Solutions for Chapter: OSCILLATIONS AND WAVES, Exercise 1: MECHANICAL OSCILLATIONS

I E Irodov Physics Solutions for Exercise - I E Irodov Solutions for Chapter: OSCILLATIONS AND WAVES, Exercise 1: MECHANICAL OSCILLATIONS

Questions from I E Irodov Solutions for Chapter: OSCILLATIONS AND WAVES, Exercise 1: MECHANICAL OSCILLATIONS with Hints & Solutions

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