Test Index

Oscillations Part 2

Section: Physics

Question : 8 of 81

Marks: +1, -0

Solution:

Concept:

For a two-body spring system on a frictionless surface, the angular frequency depends on the reduced mass.

Formula:

\omega = \sqrt{\frac{k}{\mu}} \quad , \quad \mu = \frac{m_1 m_2}{m_1 + m_2}

Solution:

Given

m_1 = 1 \text{ kg}

m_2 = 200 \text{ g} = 0.2 \text{ kg}

, and

k = 150 \text{ N/m}

(from the figure).

Calculate the reduced mass:

\mu = \frac{1 \times 0.2}{1 + 0.2} = \frac{0.2}{1.2} = \frac{1}{6} \text{ kg}

Now find the angular frequency:

\omega = \sqrt{\frac{150}{1/6}} = \sqrt{150 \times 6} = \sqrt{900} = 30 \text{ rad/s}

Answer:

30 rad/s, corresponding to option C.

[28 Jan 2026 Shift 2]

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