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Q13.8: NCERT Physics - Part 2 OSCILLATIONS 13.8 Solution: A spring balance has a scale that reads from 0 to 50 kg. The…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently equate the 20 cm length of the balance’s scale with the spring’s total extension for the 50 kg load and compute $k=Mg/\ell$ directly. The scale marks force, not displacement, so unless the problem explicitly states the spring stretches the full 20 cm at 50 kg, using the scale length as the extension leads to an incorrect spring constant and weight.

A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

Step-by-Step Verified Solution

**Answer:** Weight = 220.5 N **Step-by-Step Explanation:** Maximum mass M=50M = 50 kg, maximum displacement l=0.2l = 0.2 m. Spring constant:
k=Mgl=50×9.80.2=2450 N/mk = \frac{Mg}{l} = \frac{50 \times 9.8}{0.2} = 2450 \text{ N/m}
Time period T=0.6T = 0.6 s. Using T=2πm/kT = 2\pi\sqrt{m/k}:
m=kT24π2=2450×(0.6)24π2=22.5 kgm = \frac{kT^2}{4\pi^2} = \frac{2450 \times (0.6)^2}{4\pi^2} = 22.5 \text{ kg}
Weight W=mg=22.5×9.8=220.5W = mg = 22.5 \times 9.8 = 220.5 N [citation:13].

All Questions in this Chapter

Q13.1 Q13.2 Q13.3 Q13.4 Q13.5 Q13.6 Q13.7 Q13.8 Q13.9 Q13.10 Q13.11 Q13.12 Q13.13 Q13.14 Q13.15 Q13.16 Q13.17 Q13.18

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