Question

A mass $m$ is suspended from a spring of spring constant $K$ and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is

• A. $2\pi \sqrt{\frac{m}{K}}$
• B. $\pi \sqrt{\frac{m}{K}}+\pi \sqrt{\frac{m}{\frac{K}{2}}}$
• C. $\pi \sqrt{\frac{m}{\frac{3K}{2}}}$
• D. $\pi \sqrt{\frac{m}{K}}+\pi \sqrt{\frac{m}{2K}}$

Answer 1

The correct option is D $\pi \sqrt{\frac{m}{K}}+\pi \sqrt{\frac{m}{2K}}$

The mass $m$ completes first half of oscillation (upward) under the effect of upper spring alone.
Since, $T=2\pi \sqrt{\frac{m}{K}}$, time taken for one half of oscillation $t_1=\frac{T}{2}=\pi \sqrt{\frac{m}{K}}$
When the mass goes downwards, it is under the combined action of both the springs.
Both springs are in parallel combination.
So, $K_{\text{eq}}=K_1+K_2=2K$
Time taken to complete second half of oscillation
$t_2=\frac{T}{2}=\frac{2\pi \sqrt{\frac{m}{2K}}}{2}=\pi \sqrt{\frac{m}{2K}}$
Therefore, total time for 1 complete oscillation
$=t_1+t_2=\pi \sqrt{\frac{m}{K}}+\pi \sqrt{\frac{m}{2K}}$