Sound Waves and Beats
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Direct answer
Sound is a longitudinal mechanical wave; its speed in a gas follows v = sqrt(gamma P/rho) = sqrt(gamma R T/M), about 331 m/s in air at 0 °C and rising roughly 0.61 m/s per degree. In pipes, an open end reflects without inversion so both odd and even harmonics sound (f = v/2L, 2v/2L, ...), while a closed end inverts and permits only odd harmonics (f = v/4L, 3v/4L, ...). Superposing two tones of nearby frequencies f(1) and f(2) produces beats at |f(1) − f(2)| per second, the number of intensity maxima the ear counts each second, and each waxing-fork question resolves by tracking which frequency moved.
What you must remember
- Speed formulae: v = sqrt(gamma P/rho) — Newton assumed isothermal (gamma = 1) and underpredicted by about 15%; Laplace's adiabatic correction (gamma = 1.4 for air) fixed it to 332 m/s.
- Temperature and humidity: v grows about 0.61 m/s per °C; humid air is lighter, so sound travels faster on a damp day — both effects enter through rho.
- Intensity and loudness: I = P/(4 pi r^2) from a point source; sound level beta = 10 log10(I/I(0)) decibels with I(0) = 10^-12 W/m^2; +10 dB means ×10 intensity.
- Open pipe: all harmonics, f(n) = n v/(2L); closed pipe: odd harmonics only, f(n) = (2n − 1) v/(4L); end correction 0.6 r per open end sharpens both.
- Beats: beat frequency = |f(1) − f(2)|; wax added to a tuning fork lowers its frequency, filing raises it — the fork-unknown question turns on this single fact.
- Loudness and pitch: pitch tracks frequency, loudness tracks intensity; the ear judges ratios, hence the logarithmic decibel scale.
- String versus pipe: v(string) = sqrt(T/mu) is speed set by tension; harmonics of a string are all multiples, same pattern as the open pipe.
- Pattern note: Main asks pipe harmonic counting and beat-number numericals; Advanced adds end corrections, resonance-tube experiments and Doppler layers on moving mediums.
One fork, one tube, one decision
A tuning fork of 512 Hz sounds with an unknown fork and produces 4 beats per second; loading the unknown with wax increases the beat count to 6. The unknown is either 516 Hz or 508 Hz, and the wax decides. Wax adds mass, so the unknown's frequency falls. If it were 516, a small dip would move it toward 512 and the beats would shrink toward zero; instead the count rose to 6, so the unknown must have been sitting below: 508 Hz, and waxing carried it to 506. The method is the lesson: assume a side, predict the beat motion, and match it to observation — the numbers always discipline the guess.
Resonance-tube logic mirrors this honesty. A tube closed at one end resonates with a 512 Hz fork at lengths 16.2 cm and 49.8 cm; the difference, 33.6 cm, is half the wavelength, so lambda = 67.2 cm and v = f lambda = 512 × 0.672 = 344 m/s.
Where students slip
The harmonic census of a closed pipe misleads more candidates than any other acoustics item: it supports only odd multiples, so its "second overtone" is the fifth harmonic (5v/4L), not the third, and its octave is missing entirely. Second, in beat problems, the unknown has two candidate frequencies (f ± beats) and only a loading or filing operation chooses between them; guessing without the operation is exactly the trap. Third, intensity falls as the square of distance from a point source — inverse square, not inverse — while the speed of sound is essentially independent of frequency: loud low-frequency boat horns are not "slow".
Frequently asked questions
Why did Laplace correct Newton's speed-of-sound formula?
Compressions and rarefactions are too rapid for heat exchange, so the process is adiabatic; replacing P by gamma P raised the prediction to the observed 332 m/s.
Which harmonics does a pipe closed at one end support?
Only odd multiples of the fundamental: f = v/4L, 3v/4L, 5v/4L ..., because the closed end's inversion forbids even harmonics.
How are beats produced and what is their frequency?
Interference of two waves of nearly equal frequency makes intensity wax and wane |f(1) − f(2)| times per second; the ear hears this as beats.
What happens when wax is applied to a tuning fork?
The prongs gain effective mass, frequency falls; if beats increase after waxing, the fork was the lower of the pair — the standard decision rule.
How is the speed of sound measured with a resonance tube?
Two successive resonance lengths differ by half a wavelength, so v = 2 f (L(2) − L(1)), with end correction cancelling automatically in the difference.