Oscillations

JEE Main level, single correct, medium

8 of 194

Question 8 of 194

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 : k=150 N/mk=150 \mathrm{~N} / \mathrm{m}

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Answer: (D) 30

Two masses (1 kg and 0.2 kg) connected by a spring (k = 150 N/m) on a frictionless surface undergo oscillation about their center of mass.
This two-body problem reduces to SHM with reduced mass μ=m1m2m1+m2=1×0.21.2=16\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{1 \times 0.2}{1.2} = \frac{1}{6} kg.
The angular frequency is ω=kμ=1501/6=900=30\omega = \sqrt{\frac{k}{\mu}} = \sqrt{\frac{150}{1/6}} = \sqrt{900} = 30 rad/s.

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