Normal Modes of a Bounded System · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Normal modes of strings and air columns

Science · Waves, Light & Sound · ages 18-20
Name ______________________   Date ____________

  1. An ideal fixed-end string has fundamental frequency 110 hertz. What is its third-harmonic frequency?

    • 330 hertz
    • 220 hertz
    • 440 hertz

  2. An ideal uniform string is 0.80 m long and fixed at both ends. What is its fundamental wavelength?

    • 0.80 m
    • 1.60 m
    • 0.40 m

  3. In the ideal model of a pipe closed at one end, the first overtone has twice the fundamental frequency.

    Circle one:   True   False

  4. An ideal pipe open at both ends is 0.50 m long. Sound speed is 340 m/s. Neglect end corrections. What is its fundamental frequency in hertz?

    Answer: ______________

  5. Why are even harmonics absent from an ideal pipe closed at one end and open at the other?

    • Even harmonics require a different sound speed within the same pipe
    • The air column has zero displacement everywhere inside the pipe
    • They cannot fit a displacement node and antinode at opposite ends

  6. For displacement in a single string normal mode, what distinguishes a node from an antinode?

    • Zero amplitude versus maximum amplitude
    • Maximum amplitude versus zero amplitude
    • Zero frequency versus maximum frequency

  7. A violin and flute sustain the same fundamental frequency. How can their harmonic spectra contribute to different timbres?

    • Equal fundamentals require equal strengths for all their harmonics
    • Equal fundamentals prevent either instrument from having overtones
    • Their harmonics can have different relative strengths

  8. Two ideal pipes have equal lengths and sound speeds. Neglect end corrections. Why is the open-closed pipe's fundamental an octave below the open-open pipe's?

    • Closing an end halves sound speed while leaving wavelength fixed
    • Its fundamental wavelength doubles, so frequency halves
    • Closing an end removes the fundamental and leaves the second harmonic
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Answer key

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Normal modes of strings and air columns W1-mt_650O-YrqJs-s1

  1. 330 hertz · The third harmonic is three times the fundamental: 3 × 110 = 330 Hz.
  2. 1.60 m · The fundamental fits half a wavelength in length L, so λ_1 = 2L = 1.60 m.
  3. False · The first overtone is the next allowed mode above the fundamental. This pipe's modes begin f_1, 3f_1, 5f_1, so its first overtone is 3f_1.
  4. 340 · The fundamental wavelength is 2 × 0.50 = 1.00 m. Frequency is 340 / 1.00 = 340 Hz.
  5. They cannot fit a displacement node and antinode at opposite ends · The closed end requires a displacement node and the open end an antinode. Only odd quarter-wave counts connect them.
  6. Zero amplitude versus maximum amplitude · A displacement node stays at zero. An antinode has the greatest oscillation amplitude, but still passes through zero during each cycle.
  7. Their harmonics can have different relative strengths · Matching fundamentals do not require matching harmonic strengths. Different spectral mixtures can contribute to different timbres.
  8. Its fundamental wavelength doubles, so frequency halves · The fundamental wavelengths are 4L for the open-closed pipe and 2L for the open-open pipe. At equal speed, twice the wavelength gives half the frequency.
Worksheet · LightMySky