General physicsLesson 1 of 1 in the course Oscillatory Motion and Waves

Simple Harmonic Motion: Mass–Spring Period and Oscillator Kinematics

5 flashcards · 5 quiz questions · 5 match pairs · 4 fill-in sentences · English

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Summary

Simple harmonic motion (SHM) is a particularly important kind of oscillation: it occurs when the net restoring force follows Hooke’s law. This lesson introduces the simple harmonic oscillator and explains what determines its period and frequency.

Here, xx is displacement from equilibrium, kk is the force constant (a measure of stiffness), and the minus sign indicates that the force acts opposite to the displacement. A system with this kind of restoring force is called a simple harmonic oscillator.

All flashcards

What is a simple harmonic oscillator?
A system that oscillates about equilibrium under a restoring force proportional to its displacement and directed toward equilibrium.
What is amplitude?
Amplitude, denoted by XX, is the greatest displacement of an oscillator from its equilibrium position.
How are period and frequency related?
They are reciprocals: f=1/Tf=1/T and T=1/fT=1/f. The period is the time for one cycle, while frequency is the number of cycles per unit time.
How does spring stiffness affect mass–spring oscillations?
Increasing the force constant kk makes the oscillator stiffer, decreases its period, and increases its frequency.
Does amplitude determine the period of ideal SHM?
No. For an ideal simple harmonic oscillator, the period and frequency are independent of amplitude.

Quiz questions

  1. Which force relationship defines simple harmonic motion for displacement xx from equilibrium?Answer: The restoring force is proportional to xx and directed opposite to it, as in F=−kxF=-kx.
  2. For a mass–spring oscillator, what happens to the period when mass increases while spring stiffness stays fixed?Answer: The period increases, so each oscillation takes longer.
  3. For a fixed mass, how does increasing the spring force constant kk affect oscillation?Answer: The period decreases and the frequency increases.
  4. An ideal oscillator starts at maximum positive displacement. Which expression gives its displacement at time tt?Answer: x(t)=Xcos⁡(2πtT)x(t)=X\cos\left(\frac{2\pi t}{T}\right)
  5. How is acceleration related to displacement xx for an ideal mass–spring oscillator?Answer: a=−(k/m)xa=-(k/m)x, so acceleration points opposite to displacement.

Based on College Physics 2e by Paul Peter Urone, Roger Hinrichs (OpenStax), licensed CC BY 4.0. Adapted into a summary, flashcards and quizzes.

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