Two springs fixed at one end are stretched by 5 cm and 10 cm, respectively, when masses 0.5 kg and 1 kg are suspended at their lower ends. When displaced slightly from their mean positions and released, they will oscillate with time periods in the ratio
Correct answer: A. 1:√2
- A. 1:√2
- B. 1:2
- C. 2:1
- D. √2:1
Explanation
This is the correct answer. The time period of a spring-mass system is given by T = 2π√(m/k), where T is the time period, m is the mass, and k is the spring constant. Since the spring constant is the same for both springs (as the same force produces the same extension), the ratio of the time periods will only depend on the ratio of the masses: T1/T2 = √(m1/m2). Substituting the given masses, we get T1/T2 = √(0.5 kg / 1 kg) = √(1/2) = 1/√2.
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About Simple Harmonic Motion
Simple harmonic motion is oscillation in which acceleration is directly proportional to displacement and directed toward the equilibrium position. Work includes displacement, velocity, acceleration, phase, period, frequency, amplitude and energy, with applications to springs and simple pendulums. The restoring force and the conditions for SHM distinguish it from general periodic motion.
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