Animated Solution for Physics - Oscillations: A mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period $T$. If the mass is increased by $m$, the time period becomes $\frac{5T}{3}$, then the ratio of $\frac{m}{M}$ is

This problem is a classic example of how changing the physical parameters of an oscillating system affects its time period. Let's take a deep dive into the mechanics and the algebra behind this spring-mass system.

Imagine a block of mass $M$ hanging from a vertical spring with a spring constant $k$. When this block is pulled down slightly and released, it begins to execute Simple Harmonic Motion (SHM). The restoring force provided by the spring is directly proportional to the displacement, which leads us to the fundamental formula for the time period of a spring-mass system.

The time period $T$ is given by the equation:

This equation tells us a beautiful story: the time period depends only on the mass $M$ and the stiffness of the spring $k$. It is completely independent of how far you pull the spring (the amplitude), as long as the elastic limit is not exceeded.

Now, the problem introduces a twist. We add an extra mass $m$ to the original mass $M$. The total mass of the system is now $M+m$. Because the system is heavier, it has more inertia, and therefore, it will oscillate more slowly. This means the new time period $T^{\prime }$ will be greater than the original time period $T$.

The problem states that the new time period $T^{\prime }$ is $\frac{5T}{3}$. We can write the equation for the new time period as:

In physics, whenever you have two states of a system governed by the same formula, taking a ratio is almost always the most elegant way to solve it. By dividing the initial time period by the new time period, we can eliminate the constants $2\pi$ and $k$, which we don't know anyway.

Let's divide the first equation by the second equation:

The $T$ on the left side cancels out, leaving us with $\frac{3}{5}$. On the right side, the $2\pi$ and the $\sqrt{k}$ cancel out perfectly. We are left with a much simpler equation:

To get rid of the square root, we square both sides of the equation:

Now, we just need to perform a simple cross-multiplication to isolate the masses:

Subtracting $9M$ from both sides, we get:

Finally, rearranging the terms to find the ratio of $m$ to $M$:

And there we have it! The ratio of the added mass to the original mass is exactly $\frac{16}{9}$. This matches option (c). By systematically setting up the equations and using ratios to eliminate unknowns, we turned a seemingly complex physics problem into a straightforward algebraic exercise.