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# Oscillation Of Two Particle System - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• 5 Questions around this concept.

## Solve by difficulty

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

## 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 mr is given by  $\frac{1}{m_{r}}=\frac{1}{m_{1}}+\frac{1}{m_{2}}$

and $T=2 \pi \sqrt{\frac{m_{r}}{k}}$

Or

• 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 $\Delta x_{cm}=0 \\$ )

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

## Study it with Videos

Oscillation of two particle system

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