Animated Solution for Physics - Oscillations: Consider two identical springs each of spring constant $k$ and negligible mass compared to the mass $M$ as shown. Fig.1 shows one of them and Fig.2 shows their series combination. The ratios of time period of oscillation of the two SHM is $\frac{T_b}{T_a}=\sqrt{x}$, where value of $x$ is ......... . (Round off to the nearest integer)

Imagine you are in a physics lab, looking at two different setups. In the first setup (Fig. 1), a block of mass $M$ is hanging from a single spring with a force constant $k$. In the second setup (Fig. 2), the same mass $M$ is hanging, but this time it's supported by a chain of two identical springs, each with a constant $k$, connected end-to-end. This end-to-end connection is what we call a series combination.

Our goal is to find the ratio of their time periods of oscillation, specifically $\frac{T_b}{T_a}$, and compare it to $\sqrt{x}$ to find the value of $x$.

For any simple harmonic oscillator consisting of a mass attached to a spring, the time period $T$ is governed by the beautiful and fundamental equation:

Here, $k_{eq}$ represents the equivalent spring constant of the entire system.

For the first system (Fig. 1), there is only one spring. So, the equivalent spring constant is simply $k$. Therefore, its time period $T_a$ is straightforward:

Now, let's tackle the second system (Fig. 2). When springs are connected in series, the overall system becomes less stiff or more "stretchy". Think of it like tying two rubber bands together; the resulting mega-band stretches much easier than a single one.

Mathematically, the equivalent spring constant $k_{eq}$ for springs in series is found by adding their reciprocals:

Since both of our springs are identical ($k_1=k_2=k$), we can substitute this into our formula:

Inverting both sides gives us the equivalent spring constant for the series combination:

As expected, the effective stiffness is exactly half of a single spring.

Armed with the equivalent spring constant, we can now write the time period $T_b$ for the second system:

The problem asks for the ratio $\frac{T_b}{T_a}$. Let's divide the two expressions we've derived:

Notice how elegantly the terms cancel out. The $2\pi$, the mass $M$, and the spring constant $k$ are present in both the numerator and the denominator.

The problem states that this ratio is equal to $\sqrt{x}$.

By direct comparison, it is crystal clear that the value of $x$ must be exactly 2.