JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Elasticity & SHM

A block with mass

M

is connected by a massless spring with stiffness constant

k

to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude

A

about an equilibrium position

x_{0}

. Consider two cases: (i) when the block is at

x_{0}

; and (ii) when the block is at

x = x_{0} + A

. In both the cases, a particle with mass

m (< M)

is softly placed on the block after which they stick to each other. Which of the following statement(s) is(are) true about the motion after the mass

m

is placed on the mass

M

(A)The amplitude of oscillation in the first case changes by a factor of $\frac{M}{m + M}$ , whereas in the second case it remains unchanged✓ Correct
(B)The final time period of oscillation in both the cases is same✓ Correct
(C)The total energy decreases in both the cases
(D)The instantaneous speed at $x_{0}$ of the combined masses decreases in both the cases✓ Correct

Answer: (A), (B), (D)

Solution

Case (i): Mass added at mean position (maximum velocity).

By momentum conservation:

M v_{0} = (M + m) v^{'}

, so

v^{'} = \frac{M}{M + m} v_{0}

New amplitude:

A^{'} = \frac{v ^{'}}{ω ^{'}} = \frac{M v _{0}}{M + m} \frac{M + m}{k} = A \frac{M}{M + m}

✓ (A)

Energy:

E^{'} = \frac{1}{2} (M + m) v^{'2} = \frac{M}{M + m} E_{0} < E_{0}

✓ (C)

Case (ii): Mass added at extreme position (zero velocity).

Amplitude remains

A

(no velocity change). Energy remains same. ✗ (C) for this case

Both cases:

T^{'} = 2 π \frac{M + m}{k}

(same) ✓ (B)

Speed at

x_{0}

decreases in case (i) by momentum conservation, and in case (ii) because

v = A ω^{'} < A ω_{0}

✓ (D)

Note: Source says (C) is correct, but energy remains constant in case (ii).

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · One or more correct options