JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2018
- Hard
- MCQ
An oscillator of mass \(M\) is at rest in its equilibrium position in a potential \(V\, = \,\frac{1}{2}\,k{(x - X)^2}.\) A particle of mass \(m\) comes from right with speed \(u\) and collides completely inelastically with \(M\) and sticks to it . This process repeats every time the oscillator crosses its equilibrium position .The amplitude of oscillations after \(13\) collisions is: \((M = 10,\, m = 5,\, u = 1,\, k = 1 ).\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac {1}{\sqrt 3}\)
Step-by-step Solution
Detailed explanation
In first collision \(mu\) momentum will be imparted to system, in second collision when momentum of \((\mathrm{M}+\mathrm{m})\) is in opposite direction \(mu\) momentum of particle will make its momentum zero.
On \(13^{\text {th }}\) collision,
\(\mathrm{m} \rightarrow {\mathrm{M}+12} ; \quad \mathrm{M}+13 \mathrm{m} \rightarrow \mathrm{V}\)
\(\mathrm{mu}=(\mathrm{M}+13 \mathrm{m}) \mathrm{v} \Rightarrow \mathrm{v}=\frac{\mathrm{mu}}{\mathrm{M}+13 \mathrm{m}}=\frac{\mathrm{u}}{15}\)
\(v=\omega A \Rightarrow \frac{u}{15}=\sqrt{\frac{K}{M-13 m}} \times A\)
Putting value of \(M, m, u\) and \(K\) we get amplitude
\(A=\frac{1}{15} \sqrt{\frac{75}{1}}=\frac{1}{\sqrt{3}}\)
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