Oscillation Of Two Particle System MCQ - Practice Questions & Answers

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A trolley of mass 8kg as shown in figure, is connected to two identical springs, each of spring constant 400N/m. If the trolley is displaced from its equilibrium position by 9cm and released, the maximum speed of the trolley

0.9m/s

0.4m/s

0.3m/s

0.5m/s

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Concepts Covered - 1

Oscillation of two particle system

Two blocks of masses $m_{1}$ and $m_{2}$ are connected with a spring of natural length l and spring constant k. The system is lying on a
frictionless horizontal surface. Initially, the spring is compressed by a distance $x_{0}$ as shown in below Figure.

If we release these blocks from the compressed position, then they will oscillate and will perform SHM about their equilibrium position.

The time period of the blocks-

In this case, the reduced mass m_ris given by $\frac{1}{m_{r}}=\frac{1}{m_{1}}+\frac{1}{m_{2}}$

and $T=2 \pi \sqrt{\frac{m_{r}}{k}}$

The amplitude of the blocks- Let the amplitude of the blocks as A₁ and A₂

$then \ \ m_{1} A_{1}=m_{2} A_{2}$

(As net external force is zero and initially the centre of mass was at rest

so $\Delta x_{cm}=0 \\$ )

$\\ \text{By energy conservation,} \\ \frac{1}{2} k\left(A_{1}+A_{2}\right)^{2}=\frac{1}{2} k x^2 \\ A_{1}+A_{2}=x_{0} \quad or, \quad A_{1}+\frac{m_{1}}{m_{2}} A_{1}=x_{0} \\ \begin{gathered} A_{1}=\frac{m_{2} x_{0}}{m_{1}+m_{2}} \\ \text { Similarly, } A_{2}=\frac{m_{1} x_{0}}{m_{1}+m_{2}} \end{gathered}$