Question

Two identical masses are connected to identical springs of spring constant $K$ as shown in the figure. Find the time period of small oscillations.

• A. $T=2\pi \sqrt{\frac{m}{2K}}$
• B. $T=2\pi \sqrt{\frac{m}{K}}$
• C. $T=2\pi \sqrt{\frac{4m}{K}}$
• D. $T=2\pi \sqrt{\frac{m}{4K}}$

Answer 1

The correct option is D $T=2\pi \sqrt{\frac{m}{4K}}$

From the figure, we can deduce that the two springs are in parallel combination.
$K_{\text{eq}}=K_1+K_2=2K.....\left(1\right)$
The diagram can be simplified as:


In the simplified diagram, two masses are connected by a spring of equivalent spring constant $K_{eq}=2K.$
To reduce a two-body problem into a one-body problem, we use the concept of reduced mass.
$\mu =\text{reduced mass}=\frac{m_1{m}_2}{m_1+m_2}$
$\Rightarrow \mu =\frac{\left(m\right)\left(m\right)}{m+m}=\frac{m}{2}....\left(2\right)$
Time period of oscillations is given by,
$T=2\pi \sqrt{\frac{\mu }{K_{\text{eq}}}}$
From $\left(1\right)$ and $\left(2\right)$, we can write that
$T=2\pi \sqrt{\frac{m}{2\left(2K\right)}}\Rightarrow T=2\pi \sqrt{\frac{m}{4K}}$
Thus, option (d) is the correct answer.