Question

All the springs in fig (a), (b) and (c) are identical, each one having force constant K. Mass m is attached to each system. If \(T_a\), \(T_b\) and \(T_c\) are the periodic time of oscillations of the three systems in fig (a), (b) and (c) respectively, then

• A. \(T_a = \sqrt{2}\,T_b\)
• B. \(T_b = 2T_a\)
• C. \(T_a = \frac{T_c}{\sqrt{2}}\)
• D. \(T_b = 2T_c\)

Hint:

Find the effective spring constant for each arrangement.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For a mass on a spring, \(T = 2\pi\sqrt{\frac{m}{k_{eff}}}\). The figure shows (a) one spring, (b) two springs one below the other (series) and (c) two springs side by side (parallel).

Step 2: Key Formula or Approach:
Series: \(\frac{1}{k_{eff}} = \frac1K + \frac1K\), so \(k_b = \frac K2\). Parallel: \(k_c = K + K = 2K\). Single: \(k_a = K\).

Step 3: Detailed Explanation:
\(T_a = 2\pi\sqrt{\frac mK}\).
\(T_b = 2\pi\sqrt{\frac{2m}{K}} = \sqrt2\,T_a\).
\(T_c = 2\pi\sqrt{\frac{m}{2K}} = \frac{T_a}{\sqrt2}\).
Now \(\frac{T_b}{T_c} = \frac{\sqrt2\,T_a}{T_a/\sqrt2} = 2\), so \(T_b = 2T_c\).
Option A would need \(T_a = \sqrt2\,T_b\), but \(T_b\) is larger. Option B would need \(T_b = 2T_a\), but it is \(\sqrt2\,T_a\). Option C would need \(T_a = \frac{T_c}{\sqrt2}\), but \(T_a = \sqrt2\,T_c\).

Final Answer:
\(T_b = 2T_c\), option (D). \[ \boxed{T_b = 2T_c} \]