Oscillation Of Two Particle System

Oscillation of a Two-Particle System

Two blocks of masses $m_1 \text { and } m_2$ 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 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 m_r is given by $\frac{1}{m_r}=\frac{1}{m_1}+\frac{1}{m_2}$

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

Or

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

$ \text { then } m_1 A_1=m_2 A_2$

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

$ \text { so ,} \Delta x_{c m}=0 \text { ) }$)

By energy conservation,
$
\begin{aligned}
& \frac{1}{2} k\left(A_1+A_2\right)^2=\frac{1}{2} k x^2 \\
& A_1+A_2=x_0 \text { or, } \quad A_1+\frac{m_1}{m_2} A_1=x_0 \\
& \quad A_1=\frac{m_2 x_0}{m_1+m_2}
\end{aligned}
$

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

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Solved Examples Based on Oscillation of Two Particle System

Example 1: A system shown in Fig can move on a smooth surface. They are initially compressed by 6 cm and then released

1) System performs SHM with $t=\frac{\pi}{10} s$
2) The time period of 2 blocks is in the ratio $1: \sqrt{2}$
3) The system performs SHM with $t=\frac{\pi}{5} s$
4) The system performs SHM with $t=\frac{\pi}{15} s$

Solution:

Oscillation of a two-particle system

$T=2 \pi \sqrt{\frac{\mu}{K}}$

wherein

$\mu=\frac{m_1 m_2}{m_1+m_2} \text { is called reduced mass. }$

K is the spring constant

$\begin{aligned}
& M=\text { reduced mass }=\frac{m_1 m_2}{m_1+m_2}=2 \mathrm{~kg} \\
& T=2 \pi \sqrt{\frac{M}{K}} \Rightarrow 2 \pi \sqrt{\frac{2}{800}}=\frac{\pi}{10} \mathrm{~s}
\end{aligned}$

Example 2: In the block spring system of mass 1 kg and 2 kg, the Force constant of a spring is K = 6 N/m. Spring is Stretched by 12cm and then left. Find out the angular frequency (in rad/s) of oscillation

1) 3

2) 1.3

3) 6

4) 1

Solution:

Oscillation of a two-particle system

$T=2 \pi \sqrt{\frac{\mu}{K}}$

wherein

${ }_{\text {where }} \mu=\frac{m_1 m_2}{m_1+m_2}$ is called reduced mass.
K is the spring constant
$$
\begin{aligned}
& \omega=\sqrt{\frac{K}{\text { reduced mass }}} \Rightarrow \omega=\sqrt{\frac{K}{\mu}} \\
& M=\frac{m_1 m_2}{m_1+m_2}=\frac{2}{3} \mathrm{~kg} \\
& \omega=\sqrt{\frac{6}{\frac{2}{3}}}=3 \mathrm{rad} / \mathrm{s}
\end{aligned}
$$

Hence, the answer is the option (3).

Example 3: A circular spring of natural length $l_0$ is cut and welded with two beads of masses is cut and welded with two beads of masses $m_1$ and $m_2$ each such that the ratio of the original spring is k then find the frequency of oscillation of the heads in a smooth horizontal rigid tube. Assume $m_1=m$ and $m_2=3 m^{\prime}$.

1) $25 \sqrt{\frac{k}{3 m}}$
2) $5 \sqrt{\frac{k}{3 m}}$
3) $5 \sqrt{\frac{k}{m}}$
4) $\frac{5}{3} \sqrt{\frac{k}{m}}$

Solution:

When $m_1$ is displaced x, each spring will be deformed by the same amount. Hence, the springs are connected in parallel. The equivalent spring constant is

$k_{\text {eq }}=k_1+k_2$

If the spring is cut, the force constant of spring $\propto \frac{1}{l}$ $\Rightarrow k_1 l_1=k_2 l_2=k l$
Substituting $l_1=l / 5$ and $l_2=4 / 5$, we have $k_1=5 k$ and $k_2=\frac{5}{4} k$
Then $k_{\text {eq }}=\frac{25}{4} k$
Now we have two partical of masses $m_1$ and $m_2$ and one spring of stiffness $k_{e q}=\frac{25}{4} k$
The reduced mass is $\mu=\frac{m_1 m_2^4}{\left(m_1+m_2\right)}$ where $m_1=m$ and $m_2=3 m$
This gives $\mu=3 / 4 m$
Substituting $\mu=3 / 4 m$ and $k_{\text {eq }}=25 / 4 k$ in the formula
$$
\omega=\sqrt{\frac{k_{\text {eq }}}{\mu}} \Rightarrow \omega=\sqrt{\frac{\frac{25}{\frac{4}{3}} k}{\frac{3}{4} m}}=\sqrt{\frac{25 k}{3 m}}=5 \sqrt{\frac{k}{3 m}}
$$

Hence, the answer is the option (2).

Example 4: In the reported figure, two bodies A and B masses 200 g and 800 g are attached to the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will ________$\mathrm{rad} / \mathrm{s}_{\text {when }} \mathrm{k}=20 \mathrm{~N} / \mathrm{m}$.

1) 10

2) 12

3) 14

4) 16

Solution:

(A block system is replaced by one block with equivalent mass $\mu_{}$)
$$
\begin{aligned}
\mu & =\frac{m_A m_B}{m_A+m_B}=\frac{200 \times 800}{200+800} \\
\mu & =160 \mathrm{~g}
\end{aligned}
$$
for spring in series combination
$$
\begin{aligned}
& \frac{1}{k_{e q}}=\frac{1}{k_1}+\frac{1}{k_2}=\frac{1}{k}+\frac{1}{4 k} \\
& \frac{1}{k_{e q}}=\frac{5}{4 k} \\
& k_{e q}=\frac{4 k}{5}=\frac{4 \times 20}{5} \\
& =16 \\
& T=2 \pi \sqrt{\frac{\mu}{k_{\text {eq }}}}=2 \pi \sqrt{\frac{160 \times 10^{-3}}{16}} \\
& T=2 \pi \times 10^{-1} s \\
& \omega=\frac{2 \pi}{T}=\frac{1}{10^{-1}}=10\left(\frac{\mathrm{rad}}{\mathrm{s}}\right) \\
&
\end{aligned}
$$

Hence, the answer is the option (1).

Example 5: A system is shown in the figure. The time period for small oscillations of the two blocks will be.

1) $2 \pi \sqrt{\frac{3 m}{k}}$
2) $2 \pi \sqrt{\frac{3 m}{2 k}}$
3) $2 \pi \sqrt{\frac{3 m}{4 k}}$
4) $2 \pi \sqrt{\frac{3 m}{8 k}}$

Solution:

Time period of oscillation for a spring-mass system

$$
T=2 \pi \sqrt{\frac{m}{K}}
$$
wherein
$\mathrm{m}=$ mass of block
$\mathrm{K}=$ spring constant
The series combination of spring

wherein

$$
\frac{1}{K_{e q}}=\frac{1}{K_1}+\frac{1}{K_2}
$$
$K_1$ and $K_2$ are spring constants of spring $1 \& 2$ respectively.
Here, both springs are in a series
$$
\therefore \quad K_{e q}=\frac{K(2 K)}{K+2 K}=\frac{2 K}{3}
$$

Time period

$$
\begin{aligned}
& T=2 \pi \sqrt{\frac{\mu}{K_{e q}}} \quad \mu=\frac{m_1 m_2}{m_1+m_2} \\
& \text { Here } \mu=\frac{m}{2} \therefore T=2 \pi \sqrt{\frac{m}{2} \times \frac{3}{2 K}}=2 \pi \sqrt{\frac{3 m}{4 K}}
\end{aligned}
$$

Method II
$$
\therefore m x_1=m x_2 \Rightarrow x_1=x_2
$$
force equation for the first block
$$
\frac{2 k}{3}\left(x_1+x_2\right)=-m \frac{d^2 x_1}{d t^2}
$$

$\begin{aligned} & \text { Put } x_1=x_2 \Rightarrow \frac{d^2 x_1}{d t^2}+\frac{4 k}{3 m} \times x_1=0 \Rightarrow \omega^2=\frac{4 k}{3 m} \\ & \therefore T=2 \pi \sqrt{\frac{3 m}{4 k}}\end{aligned}$

Hence, the answer is the option (3).

Summary

The oscillation of a two-particle system involves two masses connected by a spring, exhibiting Simple Harmonic Motion (SHM) when displaced from their equilibrium position. Key parameters such as reduced mass, spring constant, and amplitude determine the system's time period and frequency of oscillation. Understanding this motion is crucial for solving problems related to mechanical vibrations and coupled systems, as demonstrated through various examples.