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Physics · Simple Harmonic Motion

JEE Main 2026 — 8 April, Evening Shift — Question 65

The frequency of oscillation of a mass m suspended by spring is v1v_1. If the length of the spring is cut to half, the same mass oscillates with frequency v2v_2. The value of v2v1 \frac{v_2}{v_1} is _______\_\_\_\_\_\_\_.

  1. Option A:

    1

  2. Option B:

    2

  3. Option C:

    2 \sqrt{2}

    Correct
  4. Option D:

    3 \sqrt{3}

Answer: C

Step-by-step solution

& \text{For\; a\; spring,\; } kl = \text{constant} \\ \\ & k_1 l = k_2 \frac{l}{2} \Rightarrow k_2 = 2k_1 \\ \\ & f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \\ \\ & \Rightarrow \frac{f_1}{f_2} = \sqrt{\frac{k_1}{k_2}} \\ \\ & \Rightarrow \frac{f_1}{f_2} = \sqrt{\frac{1}{2}} \Rightarrow \frac{f_2}{f_1} = \sqrt{2} \end{aligned}$$

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
8 April, Evening Shift
Subject
Physics
Chapter
Simple Harmonic Motion
Topic
Linear SHM and Spring-Pulley-Block Systems
The frequency of oscillation of a mass m suspended by spring is v 1 .… | JEE Main 2026 PYQ with Solution · DhiX AI