A block of mass m = 2 kg is attached to an ideal spring of spring constant k = 200 N/m. The block is at rest at its equilibrium position. An impulsive force acts on the block, giving it an initial speed of 2 m/s. Find the amplitude of the resulting oscillations.
Correct answer: C. 0.2
- A. 2
- B. 3
- C. 0.2
- D. 0.02
Explanation
The total energy of the system is conserved. Initially, the block has only kinetic energy, which is converted into potential energy when it reaches the maximum displacement (amplitude).Initial kinetic energy = 1/2 * m * v^2 = 1/2 * 2 * 2^2 = 4 JMaximum potential energy = 1/2 * k * A^2 = 4 JSolving for the amplitude A:A = √(2 * 4 / k) = √(8 / 200) = 0.2 mTherefore, the amplitude of the resulting oscillations is 0.2 m.
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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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