In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be:

JEE Main 2021 24 Feb Physics Shift 1 Question 18

a. \(A\sqrt{\frac{M}{M+m}}\)

b. \(A\sqrt{\frac{M}{M-m}}\)

c. \(A\sqrt{\frac{M-m}{M}}\)

d. \(A\sqrt{\frac{M+m}{M}}\)

Answer: (a)

JEE Main 2021 24 Feb Physics Shift 1 Question 18 solution

We know that ωf = \(\sqrt{\frac{k}{M-m}}\) and ωi = \(\sqrt{\frac{k}{M}}\) Ai = A

Also, momentum is conserved just before and just after the block of mass (m) is placed because there is no impulsive force. So

MAiωi = (M + m) v’

JEE Main 2021 24 Feb Physics Shift 1 Question 18 solutin

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