Question

A body is suspended from the two light springs separately. The periods of vertical oscillations are T and 3T respectively. The same body is suspended from two springs connected first in parallel and then in series. The periods of oscillation in both cases, respectively, are

• A. \(\frac{2T}{\sqrt{5}},\ \sqrt{5}T\)
• B. \(\frac{3T}{\sqrt{10}},\ \sqrt{10}T\)
• C. \(\frac{4T}{\sqrt{15}},\ \sqrt{15}T\)
• D. \(\sqrt{5}T,\ \frac{2T}{\sqrt{5}}\)

Hint:

Find the spring constants from the periods, then combine them: add in parallel, add reciprocals in series.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
\(T=2\pi\sqrt{\dfrac{m}{k}}\), so \(k\propto\dfrac{1}{T^{2}}\). Take \(k_{1}=k\) for period \(T\), then \(k_{2}=\dfrac{k}{9}\) for period \(3T\).

Step 2: Parallel
\(k_{p}=k+\dfrac{k}{9}=\dfrac{10k}{9}\)
\(T_{p}=T\sqrt{\dfrac{9}{10}}=\dfrac{3T}{\sqrt{10}}\)

Step 3: Series
\(k_{s}=\dfrac{k\cdot k/9}{k+k/9}=\dfrac{k}{10}\)
\(T_{s}=T\sqrt{10}\)

Final Answer:
The periods are \(\frac{3T}{\sqrt{10}}\) in parallel and \(\sqrt{10}\,T\) in series. \[ \boxed{\text{(B) }\dfrac{3T}{\sqrt{10}},\ \sqrt{10}\,T} \]