Animated Solution for Physics - Oscillations: A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency $\omega$. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

Imagine you are standing on an elevator that is moving up and down in a perfectly smooth, rhythmic manner—executing Simple Harmonic Motion (SHM). Now, imagine the elevator starts moving faster and faster, increasing its amplitude. At some point, as the elevator drops down, you might feel a sudden sensation of weightlessness. If it drops fast enough, you might even lose contact with the floor! This is exactly the scenario we are exploring with a coin placed on a horizontally oscillating platform.

We have a coin of mass $m$ resting on a horizontal platform. The platform is undergoing vertical SHM with an angular frequency $\omega$. As the amplitude $A$ of the oscillation is gradually increased, we want to find the exact moment the coin loses contact with the platform.

To understand this, we must look at the forces acting on the coin. There are only two vertical forces at play here: 1. The gravitational force, $mg$, acting downwards. 2. The normal force, $N$, exerted by the platform on the coin, acting upwards.

As long as the coin is in contact with the platform, it shares the platform's acceleration. The condition for the coin to just lose contact is that the normal force becomes zero ($N=0$). This happens when the platform accelerates downwards faster than gravity can pull the coin down.

In SHM, the acceleration of the platform at any displacement $y$ from the mean position is given by $a=-{\omega }^{2}y$. The negative sign indicates that the acceleration is always directed towards the mean position.

The maximum downward acceleration occurs at the highest point of the oscillation, where $y=+A$. At this extreme position, the downward acceleration is $a_{max}={\omega }^{2}A$.

Let's apply Newton's Second Law to the coin at this highest point. Taking the downward direction as positive, the net force is $mg-N$. Therefore, the equation of motion is:

Substituting the maximum acceleration, we get:

We are looking for the critical amplitude where the coin just loses contact. At this exact moment, the normal force $N$ drops to zero. Let's substitute $N=0$ into our master equation:

The mass $m$ cancels out from both sides, beautifully showing that the condition is independent of the coin's mass. Solving for the amplitude $A$, we find:

This is the critical amplitude. If the amplitude is increased beyond this value, the platform will accelerate downwards faster than $g$, and the coin will be left behind in free fall, losing contact with the platform. Thus, the coin leaves contact for the first time for an amplitude of $\frac{g}{{\omega }^2}$.