Oscillations

JEE Main level, single correct, medium

146 of 194

Question 146 of 194

The frequency of oscillation of a mass mm suspended by a spring is v1v_1. If the length of the spring is cut to half, the same mass oscillates with frequency v2v_2. The value of v2/v1v_2/v_1 is ________.

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Answer: (C) 2\sqrt{2}

The frequency of oscillation of a spring-mass system is given by v=12πkmv = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}, where kk is the spring constant and mm is the mass.

The spring constant kk is inversely proportional to the length of the spring LL, so k∝1Lk \propto \dfrac{1}{L}.

When the length of the spring is cut to half, the new length is L′=L2L' = \dfrac{L}{2}. The new spring constant becomes k′=2kk' = 2k.

The new frequency of oscillation is v2=12πk′m=12π2km=2v1v_2 = \dfrac{1}{2\pi} \sqrt{\dfrac{k'}{m}} = \dfrac{1}{2\pi} \sqrt{\dfrac{2k}{m}} = \sqrt{2} v_1.

Therefore, the ratio v2v1=2\dfrac{v_2}{v_1} = \sqrt{2}.

Answer: 2\sqrt{2}

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