A block of mass M is suspended from a ceiling by a spring with spring stiffness constant k. A penny of mass m is placed on top of the block. What is the maximum amplitude of oscillations that will allow the penny to just stay on top of the block? (Assume \(m \ll M\).)

According to the law of equilibrium, all forces must be balanced by the counter forces acting in opposite directions.

The mass of the block is\(M\).

The mass of the penny is\(m\).

The spring stiffness constant is\(k\).

The mass of the penny is very lesser than the mass of the block as \(m \ll M\).

In this situation, for a penny to always stay on top of the block, it must be balanced by force acting on it. According to the symmetry, the net force acting on the system is:

\(\begin{aligned}{l}{F_{{\rm{net}}}} = mg - {F_{\rm{N}}}\\ma = mg - {F_{\rm{N}}}\end{aligned}\)

Here,\({F_{\rm{N}}}\)is a normal reaction and\(a\)is the vertical acceleration and\(g\)is the acceleration due to gravity.

The vertical acceleration experienced by the system must act upwards, as for the penny to stay in stable equilibrium on the top of the block, the penny must be constantly accelerated upwards. Thus, the force acting due to the weight must be greater than the normal reaction.

Now, in this situation, acceleration is acting upwards, and it couldn’t be downwards since alternate would result in a normal negative reaction, which is not possible in this case. This case will result in a limiting condition, which is \({a_{{\rm{down}}}} = g\), which is the maximum applied acceleration on the block-penny system.

Also, for an SHM system, the maximum acceleration experienced is:

\(\begin{aligned}{c}{a_{{\rm{max}}}} = {\omega ^2}A\\ = \frac{k}{{m + M}}A\\ \approx \frac{k}{M}A\end{aligned}\)

Here,\(\omega \)is the angular frequency and A is the amplitude.

On equating the relations of maximum acceleration.

\(\begin{aligned}{c}g = \frac{k}{M}A\\A = \frac{{Mg}}{k}\end{aligned}\)

Hence, the maximum amplitude of the oscillations are\(\frac{{Mg}}{k}\).