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Q13.13: NCERT Physics - Part 2 OSCILLATIONS 13.13 Solution: Figure 13.21(a) shows a spring of force constant k clamped…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume that two masses attached to a single spring create an effective spring constant of $2k$ for each mass, leading to the wrong period $2\pi\sqrt{m/2k}$. The correct analysis uses the reduced mass $\mu = m/2$ with the original spring constant $k$, yielding the correct period $T = 2\pi\sqrt{m/2k}$ only when derived via relative motion or reduced mass formalism, not by arbitrarily doubling $k$.

Figure 13.21(a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 13.21 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 13.21(b) is stretched by the same force F. (a) What is the maximum extension of the spring in the two cases? (b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?

Step-by-Step Verified Solution

**Answer:** (a) Case (a): x=F/kx = F/k. Case (b): x=F/kx = F/k. (b) Case (a): T=2πm/kT = 2\pi\sqrt{m/k}. Case (b): T=2πm/(2k)T = 2\pi\sqrt{m/(2k)}. **Step-by-Step Explanation:** (a) In case (a), the spring is clamped at one end and force F is applied at the other. Maximum extension x=F/kx = F/k. In case (b), both ends are pulled by force F. The tension in the spring is F, so the extension is also x=F/kx = F/k [citation:14]. (b) Case (a): One mass m attached to spring of constant k. T=2πm/kT = 2\pi\sqrt{m/k}. Case (b): Two masses m attached to the same spring. The reduced mass is μ=m/2\mu = m/2. The effective spring constant for each mass is 2k2k. Hence, T=2πm/(2k)T = 2\pi\sqrt{m/(2k)} [citation:14].

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