Question

All the springs in fig. (a), (b) and (c) are identical, each having force constant \(K\). Mass \(m\) is attached to each system. If \(T_{a}\), \(T_{b}\) and \(T_{c}\) are the time periods of oscillations in fig. (a), (b) and (c) respectively, then:

• A. \(T_{a} = \sqrt{2} T_{b}\)
• B. \(T_{a} = \frac{T_{c}}{\sqrt{2}}\)
• C. \(T_{b} = 2 T_{a}\)
• D. \(T_{b} = 2 T_{c}\)

Hint:

Springs side-by-side (parallel) increase stiffness; springs end-to-end (series) decrease it.

Answer 1

The Correct Option is A

Step 1: Analyze Case (a)
Single spring: \(K_{eq} = K\). Time period \(T_a = 2\pi\sqrt{\frac{m}{K}}\).

Step 2: Analyze Case (b)
Springs in parallel: \(K_{eq} = K + K = 2K\). Time period \(T_b = 2\pi\sqrt{\frac{m}{2K}}\).

Step 3: Ratio
\(\frac{T_a}{T_b} = \sqrt{\frac{2K}{K}} = \sqrt{2} \implies T_a = \sqrt{2} T_b\).
Final Answer: (A)