A mass m is attached to two strings as shown in figure. The spring constants of two springs are K_1 and K_2 . For the frictionless surface, the time period of oscillation of mass m is :

1 \over 2\pi \sqrt K_1 + K_2 \over m

1 \over 2\pi \sqrt K_1 - K_2 \over m

Simple Harmonic Motion PYQ — JEE Main 2023 · 6 Apr · Shift 1 · Q55 with Solution

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Correct answer: A

A mass $m$ is connected between two springs of spring constants $K_1$ and $K_2$ on a frictionless surface.

If the mass is displaced by a small distance $x$ to the right:

the left spring gets stretched by $x$ and pulls the mass left with force $K_1 x$ ,
the right spring gets compressed by $x$ and also pushes the mass left with force $K_2 x$ .

So both springs provide restoring force in the same direction.

Hence, total restoring force is

F=-K_1x-K_2x=-(K_1+K_2)x

Using Newton's second law,

m\frac{d^2x}{dt^2}=-(K_1+K_2)x

\frac{d^2x}{dt^2}+\frac{K_1+K_2}{m}x=0

This is the standard equation of SHM:

\frac{d^2x}{dt^2}+\omega^2 x=0

Therefore,

\omega=\sqrt{\frac{K_1+K_2}{m}}

We know

T=\frac{2\pi}{\omega}

So,

T=2\pi\sqrt{\frac{m}{K_1+K_2}}

Option A is

2\pi \sqrt{\frac{m}{K_1+K_2}}

which matches exactly.

Thus, the correct answer is A.