JEE PYQ: Oscillations Question 9
Question 9 - 2021 (24 Feb 2021 Shift 1)
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, another mass $m$ is gently fixed upon it. The new amplitude of oscillation will be:
(1) $A\sqrt{\frac{M}{M+m}}$
(2) $A\sqrt{\frac{M+m}{M-m}}$
(3) $A\sqrt{\frac{M-m}{M}}$
(4) $A\sqrt{\frac{M+m}{M}}$
Show Answer
Answer: (1)
Solution
At equilibrium, momentum is conserved: $MA\omega = (M+m)v’$. New amplitude $A_f = A\sqrt{\frac{M}{M+m}}$.
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