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Physics · Simple Harmonic Motion

JEE Main 2026 — 28 January, Evening Shift — Question 42

As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses 1 kg and 0.2 kg with a separation more than spring natural length and are released. Assuming the horizontal surface to be frictionless, the angular frequency (in SI unit) of the system is :

Question figure
  1. Option A:

    30

    Correct
  2. Option B:

    27

  3. Option C:

    20

  4. Option D:

    5

Answer: A

Step-by-step solution

μ=m1 m2 m1+m2=1×0.21.2\mu=\frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}=\frac{1 \times 0.2}{1.2} μ=16\mu=\frac{1}{6} ω=kμ=1501/6=30\omega=\sqrt{\frac{\mathrm{k}}{\mu}}=\sqrt{\frac{150}{1 / 6}}=30

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Evening Shift
Subject
Physics
Chapter
Simple Harmonic Motion
Topic
Linear SHM and Spring-Pulley-Block Systems

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