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JEE Advanced 2016Paper 2
Exam
JEE Advanced
Year
2016
Paper
Paper 2
Subject
Physics
Chapter
Oscillations

JEE Advanced 2016 · Paper 2 · Physics · Oscillations

A block with mass M is connected by a massless spring with...

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 x0{x}_{0}. Consider two cases : (i) when the block is at x0;{x}_{0} \text{;} and (ii) when the block is at x=x0+A.x = {x}_{0} + \text{A} \text{.} In both the cases, a particle with mass m (<M)\left(< \text{M}\right) 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?

Options

  1. AThe amplitude of oscillation in the first case changes by a factor of \sqrt{\dfrac{\text{M}}{\text{m} + {\text{M}}^{&prime;}}} whereas in the second case it remains unchanged
  2. BThe final time period of oscillation in both the cases is same
  3. CThe total energy decreases in both the cases
  4. DThe instantaneous speed at x0{x}_{0} of the combined masses decreases in both the cases.

Correct answer

A. The amplitude of oscillation in the first case changes by a factor of \sqrt{\dfrac{\text{M}}{\text{m} + {\text{M}}^{&prime;}}} whereas in the second case it remains unchanged

Step-by-step solution

{\text{T}}_{\text{i}} = 2 &pi; \sqrt{\dfrac{\text{M}}{\text{K}}} , {\text{T}}_{\text{f}} = 2 &pi; \sqrt{\dfrac{\text{M} + \text{m}}{\text{K}}}
Case (i):
\text{M} \left(\text{A} &omega;\right) = \left(\text{M} + \text{m}\right) \text{V}
&there4; Velocity decreases at equilibrium position.
By energy conservation
Af=AiMM+m{\text{A}}_{f} = {\text{A}}_{\text{i}} \sqrt{\dfrac{\text{M}}{\text{M} + \text{m}}}
Case (ii):
No energy loss, amplitude remains same
At equilibrium \left({x}_{0}\right) velocity = \text{A} &omega;
In both cases &omega; decrease so velocity decreases in both cases
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