A block is placed on a frictionless horizontal table. The m…

Physics Simple Harmonic Motion (Oscillations) Spring Pendulum Single Correct MCQ

Published on: September 12, 2026

A block is placed on a frictionless horizontal table. The mass of the block is m and springs are attached on either side with force constants $K_{1}$ and $K_{2}$ . If the block is displaced a little and left to oscillate, then the angular frequency of oscillation will be

$\left(\frac{K_1 + K_2}{m}\right)^{1/2}$

$\left[\frac{K_1 K_2}{m(K_1 + K_2)}\right]^{1/2}$

$\left[\frac{K_1 K_2}{(K_1 - K_2) m}\right]^{1/2}$

$\left[\frac{k_1^2 + k_2^2}{(k_1 + k_2) m}\right]^{1/2}$

Text Solution

Verified by Experts

The correct answer is:

In this case springs are in parallel, so $\mathbf{k_{eq}} = \mathbf{k_1} + \mathbf{k_2}$

and $\omega = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{k_1 + k_2}{m}}$

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