Question

Two springs having force constants $k$ each are arranged in parallel and in series. A mass $M$ is attached to two arrangements separately. If time period in first case is $T_{1}$ and in second case is $T_{2}$, then ratio $\frac{T_{1}}{T_{2}}$ is :

• A. 1.5
• B. 3.2
• C. 0.5
• D. 2.1

Answer 1

Answer: (C)

In first set up, the springs are joined in parallel and in second, the springs are joined in series.
When springs are connected in parallel, the effective spring constant is

$k_{1}=k +k=2 k$
Hence, time period
$T_{1}=2 \pi \sqrt{\frac{M}{k_{1}}}=2 \pi \sqrt{\frac{M}{2 k}}$ ...(i)
When springs are connected in series, the effective spring constant is

$\frac{1}{k_{2}}=\frac{1}{k}+\frac{1}{k}=\frac{2}{k}$
$\Rightarrow k_{2} =\frac{k}{2}$
Therefore, time period,
$T_{2}=2 \pi \sqrt{\frac{M}{k_{2}}}$
$=2 \pi \sqrt{\frac{M}{k / 2}}$
$T_{2} =2 \pi \sqrt{\frac{2 M}{k}}$ ...(ii)
Dividing Eq. (i) by Eq. (ii), we have
$\frac{T_{1}}{T_{2}}=\sqrt{\frac{M / 2 k}{2 M / k}}=\sqrt{\frac{1}{4}}=\frac{1}{2}=0.5$