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)

We need to determine the value of $$x$$, which represents the ratio of the time periods of oscillation between a single spring system and a series combination of two identical springs.

The time period ($$T$$) of a simple harmonic motion (SHM) mass-spring system is given by the formula:

$$T = 2\pi \sqrt{\frac{M}{k_{\text{eq}}}}$$

Where $$M$$ is the attached mass and $$k_{\text{eq}}$$ is the equivalent spring constant of the configuration.

1. Time Period for Configuration A ($$T_a$$)

In Fig. 1, a single mass $$M$$ is attached to a single spring of spring constant $$k$$.

• Equivalent spring constant ($$k_{\text{eq}, a}$$) = $$k$$

The time period $$T_a$$ is:

$$T_a = 2\pi \sqrt{\frac{M}{k}}$$

2. Time Period for Configuration B ($$T_b$$)

In Fig. 2, two identical springs, each of spring constant $$k$$, are connected end-to-end in a series combination.

• The formula for the equivalent spring constant of two springs in series is: $$\frac{1}{k_{\text{eq}, b}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} \implies k_{\text{eq}, b} = \frac{k}{2}$$

The time period $$T_b$$ is:

$$T_b = 2\pi \sqrt{\frac{M}{\left(\frac{k}{2}\right)}} = 2\pi \sqrt{\frac{2M}{k}}$$

3. Find the Ratio $$x = \frac{T_b}{T_a}$$

Now, we take the ratio of the two time periods:

$$x = \frac{T_b}{T_a} = \frac{2\pi \sqrt{\frac{2M}{k}}}{2\pi \sqrt{\frac{M}{k}}}$$

Canceling out the common terms ($$2\pi$$ and $$\sqrt{\frac{M}{k}}$$) simplifies the expression to:

$$x =2