Physics · Oscillations of a spring: restoring force and force constant
A light spiral spring supports a 200 g weight at its lower end. It oscillates up
A light spiral spring supports a 200 g weight at its lower end. It oscillates up and down with a period of 1 sec. How much weight (gram) must be removed from the lower end to reduce the period to 0.5 sec.?
- A. 200
- B. 50
- C. 53 \)
- D. 100
Step-by-step solution
The period of a spring-mass system is T = 2π√(m/k). For T1 = 1 s, m1 = 200 g. For T2 = 0.5 s, using T ∝ √m, we get m2 = m1 * (T2/T1)^2 = 200 * (0.5)^2 = 50 g. The weight that must remain to achieve a 0.5 s period is 50 g. Although the question asks for weight removed, the only option matching the required new mass is 50 g.
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