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Lecture 1: Periodic Phenomena

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Lecture Topics

A tuning fork sits next to a rectangular tank of water with a clear bottle in it. A red sphere on a stick rests on top of the bottle.
  • Periodic Phenomena (Oscillations, Waves)
  • Simple Harmonic Oscillators
  • Complex Notation
  • Differential Equations
  • Physical Pendulum

Learning Objectives

By the end of this lecture, you should:

  • recognize examples of periodic motion in everyday life.
  • see some complicated oscillatory motion and production of sound through oscillatory motion.
  • know that sound is pressure oscillations in air of frequency 20 Hz to 20 kHz.
  • explain simple harmonic oscillation (SHO) through its basic equation, relation to circular motion, and complex exponential form.
  • relate phase, angular velocity, frequency, and period.
  • explain the dynamics of springs qualitatively and quantitatively.
  • understand the relation of complex numbers to circular motion.
  • quantitatively explain the dynamics of the pendulum in the small (angle) amplitude approximation.
  • understand the concept of wave polarization.

Lecture Activities

Check Yourself

  • If the speed of sound in air is 1236 km/hour, what are the respective wavelengths of the highest (20 kHz) and lowest (20 Hz) frequency sound waves a human can hear?

› View/hide answer

    0.0172 m; 17.2 m

 

  • For BOTH the cases of a pendulum and a mass on a spring, choose respectively the unit calculations that show that the period formula gives a result with the units of time.

› View/hide answer

\[\frac{l}{g} \sf{\mbox{has units}} \frac{[m]}{[m/{{s}^{2}}]}=\frac{[1]}{[1/{{s}^{2}}]}=[{{s}^{2}}]\]

\[\frac{m}{k} \sf{\mbox{has units}} \frac{[kg]}{[N/m]}=\frac{[kg]}{[kg\cdot m/{{s}^{2}}/m]}=[{{s}^{2}}]\]

The only further operation is a square root which gives [s].

 

  • Again consider the pendulum and the mass on a spring, and describe how the angular frequency formula matches your experience and intuition.

› View/hide answer

    \[\omega =\sqrt{\frac{g}{l}}\]

    makes sense since we would think the higher restoring force from a higher “g” would make a pendulum go faster, while a longer pendulum is know to vibrate more slowly

    \[\omega =\sqrt{\frac{k}{m}}\]

    makes sense since a stiffer spring vibrates with higher frequency, while a larger mass being pushed by a spring moves ponderously.

 

  • For North American AC electricity with a frequency of 60 Hz, what is the angular frequency, including units?

› View/hide answer

    377 rad/s

 

  • If a pendulum swings back and forth by an angle of 5º about its vertical rest point, it is well described by simple harmnonic oscillation in the small angle approximation. After one full swing, by what angle has the phase describing this oscillator advanced?

› View/hide answer

    360º or equivalently, 2π radians

 

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