Q64 / 75 · JEE Main 2026 (28 January Shift 2) · Physics
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 : $k=150 \mathrm{~N} / \mathrm{m}$
- A5
- B27
- C20
- Correct answerD30
Solution
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 $\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{1 \times 0.2}{1.2} = \frac{1}{6}$ kg.
The angular frequency is $\omega = \sqrt{\frac{k}{\mu}} = \sqrt{\frac{150}{1/6}} = \sqrt{900} = 30$ rad/s.
This two-body problem reduces to SHM with reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{1 \times 0.2}{1.2} = \frac{1}{6}$ kg.
The angular frequency is $\omega = \sqrt{\frac{k}{\mu}} = \sqrt{\frac{150}{1/6}} = \sqrt{900} = 30$ rad/s.
Hence, the correct answer is option: 'D'.