Oscillation Of Two Particle System MCQ - Practice Questions with Answers

Two blocks of masses $m_1$ and $m_2$ are connected with a spring of natural length $I$ and spring constant $k$. The system is lying on a
frictionless horizontal surface. Initially, spring is compressed by a distance $x_0$ as shown in the 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 $\mathrm{m}_{\mathrm{r}}$ is given by $\frac{1}{m_r}=\frac{1}{m_1}+\frac{1}{m_2}$

  $$
  \text { and } T=2 \pi \sqrt{\frac{m_r}{k}}
  $$

  
  Or
  - The amplitude of the blocks- Let the amplitude of the blocks as $A_1$ and $A_2$ then $m_1 A_1=m_2 A_2$
  (As net external force is zero and initially the center of mass was at rest so $\Delta x_{c m}=0$ )

  By energy conservation, $\frac{1}{2} k\left(A_1+A_2\right)^2=\frac{1}{2} k x \frac{1}{0}$ or, $A_1+A_2=x_0 \quad$ or, $\quad A_1+\frac{m_1}{m_2} a A_1=x_0$

  $$
  A_1=\frac{m_2 x_0}{m_1+m_2}
  $$

  or,
  Similarly, $A_2=\frac{m_1 x_0}{m_1+m_2}$