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: Understanding the Question:
We have three different configurations of two identical springs with a mass \(m\). We need to compare their time periods of oscillation.
Step 2: Key Formula or Approach:
1) Time period of spring-mass system: \(T = 2\pi \sqrt{\frac{m}{K_{\text{eff}}}}\).
2) For springs in parallel: \(K_{\text{eff}} = K_{1} + K_{2} = 2K\).
3) For springs in series: \(\frac{1}{K_{\text{eff}}} = \frac{1}{K_{1}} + \frac{1}{K_{2}} \implies K_{\text{eff}} = \frac{K}{2}\).
Step 3: Detailed Explanation:
- System (a): Two springs in parallel.
\(K_{a} = 2K \implies T_{a} = 2\pi \sqrt{\frac{m}{2K}}\).
- System (b): Two springs in series.
\(K_{b} = \frac{K}{2} \implies T_{b} = 2\pi \sqrt{\frac{m}{K/2}} = 2\pi \sqrt{\frac{2m}{K}}\).
- System (c): One spring above the mass and one below. This is equivalent to a parallel combination because both springs exert force in the same direction when displaced.
\(K_{c} = 2K \implies T_{c} = 2\pi \sqrt{\frac{m}{2K}}\).
Comparing the results:
\[ T_{b} = 2\pi \sqrt{\frac{2m}{K}} = 2 \times \left( 2\pi \sqrt{\frac{m}{2K}} \right) = 2 T_{a} \]
Also, since \(T_{a} = T_{c}\), we have \(T_{b} = 2 T_{c}\).
Looking at the options, both (C) and (D) express this correct relationship.
Step 4: Final Answer:
The time period \(T_{b} = 2 T_{a}\).