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 Tb/Ta = \sqrt x , where value of x is \underline\hspace2cm . (Round off to the Nearest Integer)

Simple Harmonic Motion PYQ — JEE Main 2021 · 17 Mar · Shift 1 · Q67 with Solution

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Correct answer: 2

For a mass $M$ attached to a single spring of spring constant $k$ , the time period is

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

For two springs, each of spring constant $k$ , connected in series:

\frac{1}{k_{\text{eq}}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}

So,

k_{\text{eq}} = \frac{k}{2}

Now the mass $M$ oscillates with effective spring constant $k/2$ , so

T_b = 2\pi \sqrt{\frac{M}{k/2}} = 2\pi \sqrt{\frac{2M}{k}}

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

Given

\frac{T_b}{T_a} = \sqrt{x}

Hence,

\sqrt{x} = \sqrt{2} \implies x = 2

\boxed{x=2}

The nearest integer is also $2$ .