Question

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 : 

• A. $A \sqrt{\frac{M+m}{M}}$
• B. $A \sqrt{\frac{M}{M+m}}$
• C. $A \sqrt{\frac{M-m}{M}}$
• D. $A \sqrt{\frac{M}{M-m}}$

Hint:

When a mass is added at the equilibrium position, use conservation of linear momentum. If the mass is added at the extreme position, the amplitude remains the same but the time period changes.

Answer 1

The Correct Option is B

Step 1: At the equilibrium position, the velocity of mass $M$ is maximum: $v_{max} = \omega A = \sqrt{\frac{k}{M}} A$.
Step 2: When mass $m$ is placed gently, the momentum is conserved in the horizontal direction: $M v_{max} = (M+m) v'$.
Step 3: $v' = \frac{M}{M+m} v_{max} = \frac{M}{M+m} \sqrt{\frac{k}{M}} A$.
Step 4: For the new system, $v' = \omega' A' = \sqrt{\frac{k}{M+m}} A'$.
Step 5: $\sqrt{\frac{k}{M+m}} A' = \frac{M}{M+m} \sqrt{\frac{k}{M}} A$.
Step 6: $A' = \frac{M}{M+m} \sqrt{\frac{M+m}{M}} A = \sqrt{\frac{M}{M+m}} A$.