Question

A circular spring of natural length $l$ is cut and welded with two beads of masses $m_1$ and $m_2$ such that the ratio of the lengths of the springs between the beads is $4:1$. If the stiffness of the original spring is $k$, then find the angular frequency of oscillation of the beads in a smooth horizontal rigid tube.
[Assume $m_1=m$ and $m_2=3m$].

• A. $\omega =\sqrt{\frac{25k}{3m}}$
• B. $\omega =\sqrt{\frac{5k}{3m}}$
• C. $\omega =\sqrt{\frac{25k}{4m}}$
• D. $\omega =\sqrt{\frac{5k}{4m}}$

Answer 1

The correct option is A $\omega =\sqrt{\frac{25k}{3m}}$

When $m_1$ is displaced relative to $m_2$ by a distance $x$, each spring will be deformed by the same amount. Hence, the springs are connected in parallel. The equivalent spring constant is
$k_{\text{eq}}=k_1+k_2$


We know that, the relation between force constant of the spring and length of the spring is given by $k=\frac{YA}{l}$
$\Rightarrow k\propto \frac{1}{l}\Rightarrow k_1{l}_1=k_2{l}_2=kl$
From the data given in the question, $l_1=\frac{l}{5}$ and $l_2=\frac{4l}{5}$.
So, $k_1=5k$ and $k_2=\frac{5}{4}k$
Then, $k_{\text{eq}}=\frac{25}{4}k$
Now, we have two particles of masses $m_1$ and $m_2$ and one spring of stiffness
$k_{\text{eq}}=\frac{25}{4}k$.
To reduce a two-body problem into a one-body problem, we use the concept of reduced mass.
Reduced mass $\left(\mu \right)=\frac{m_1{m}_2}{m_1+m_2}$
where $m_1=m$ and $m_2=3m$
This gives $\mu =\frac{3}{4}m$
Substituting $\mu =\frac{3}{4}m$ and $k_{\text{eq}}=\frac{25}{4}k$ in the formula
$\omega =\sqrt{\frac{k_{\text{eq}}}{\mu }}$
we get, $\omega =\sqrt{\frac{\frac{25}{4}k}{\frac{3}{4}m}}=\sqrt{\frac{25k}{3m}}$
Thus, option (a) is the correct answer.