A student is performing a lab experiment on simple harmonic motion. He has two different springs (with force constants k1 and k2) and two different blocks (of masses m1 and m2). If k1 =2k2, and m1 = 2m2, which of the following combinations would give the student the spring-block simple harmonic oscillator with the shortest period?
Correct answer: B. The spring with force constant k1 and the block of mass m2
- A. The spring with force constant k1 and the block of mass m1
- B. The spring with force constant k1 and the block of mass m2
- C. The spring with force constant k2 and the block of mass m1
- D. The spring with force constant k2 and the block of mass m2
- E. All the combinations above would give the same period
Explanation
The period of the spring-block simple harmonic oscillator is given by the equation T =2π√m/k so, to make T as small as possible, we want m to be as small as possible and k to be as large as possible. Since m2 is the smaller mass and k1 is the larger spring constant, this combination will give the oscillator the shortest period.
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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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