Body of mass 4 kg attached to a spring is displaced through 0.04 m from its equilibrium position and then released. If the spring constant is 400 N/m, find the time period of the vibration?
Correct answer: D. 0.628 s
- A. 4.567 s
- B. 3.416 s
- C. 2.315 s
- D. 0.628 s
Explanation
The time period of vibration can be calculated using the formula: T = 2π √(m/k) In this case, the mass of the body is 4 kg and the spring constant is 400 N/m. T = 2π √(4 kg / 400 N/m) = 2π √(0.01 s /kg) = 2π 0.1 s 2 T ≈ 0.628 s Therefore, the time period of vibration for the body attached to the spring is approximately 0.628 seconds.
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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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