A particle performs simple harmonic motion of amplitude 0.02m and frequency 2.5 Hz what is the maximum speed?
Correct answer: B. 0.314m/s
- A. 0.008m/s
- B. 0.314m/s
- C. 0.125m/s
- D. 0.05m/s
Explanation
To determine the maximum speed of a particle in simple harmonic motion with an amplitude of 0.02 m and frequency of 2.5 Hz, we can use the equation vmax = ωA, where ω is the angular frequency and A is the amplitude. 0.314 m/s: This option is correct. By substituting the values, vmax = (2π × 2.5 Hz) × 0.02 m = 0.314 m/s. Therefore, the correct option is "0.314 m/s" as it represents the maximum speed of the particle in simple harmonic motion with the given parameters.
This question appeared on the ETEA MDCAT 2009 paper, which you can sit online with every answer explained.
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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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