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JEE Main 2026 (28 Jan, Shift 2) · Spring Systems

Spring Systems question from JEE Main 2026 (28 Jan, Shift 2)

Q22JEE Main 2026 · 28 JAN · SHIFT 2Difficult

Spring Systems

As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses 11 kg and 0.20.2 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:

Figure

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Answer: (c)

  1. Two free masses joined by a spring oscillate with ω=kμ\omega = \sqrt{\frac{k}{\mu}}, μ=m1m2m1+m2\mu = \frac{m_1 m_2}{m_1+m_2}.
  2. μ=1×0.21.2=16 kg\mu = \frac{1 \times 0.2}{1.2} = \frac{1}{6}\text{ kg}, k=150 N/mk = 150\text{ N/m} (from figure).
  3. ω=150×6=900=30 rad/s\omega = \sqrt{150 \times 6} = \sqrt{900} = 30\text{ rad/s}
  4. So the answer is (c) 30.
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