JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Elasticity & SHM

WhatsApp Email

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. New amplitude of oscillation will be:

(A) $A \frac{M}{M - m}$
(B) $A \frac{M}{M + m}$ ✓ Correct
(C) $A \frac{M - m}{M}$
(D) $A \frac{M + m}{M}$

Answer: (B)

Solution

Initially, the velocity at equilibrium is

v = A ω = A \frac{k}{M}

.

After mass

m

is added, by conservation of momentum:

M v = (M + m) v^{'}

The new amplitude is

A^{'} = \frac{v ^{'}}{ω ^{'}} = \frac{v ^{'}}{\frac{k}{M + m}}

.

Substituting:

v^{'} = \frac{M v}{M + m}

, we get:

A^{'} = \frac{M v}{M + m} \cdot \frac{M + m}{k} = \frac{M}{M + m} \cdot A \frac{k}{M} \cdot \frac{M + m}{k} = A \frac{M}{M + m}

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Elasticity & SHM →

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Elasticity & SHM

Solution

More JEE Main questions on Elasticity & SHM