Notes
Waves
- Waves carry energy, not matter. Mechanical waves need a medium; electromagnetic waves do not.
- Transverse: medium moves perpendicular (crests, troughs). Longitudinal: medium moves parallel (compressions, rarefactions).
- $v = f\lambda$ and $\displaystyle T = \frac{1}{f}$. Speed depends on the medium; amplitude relates to energy.
Sound
- Longitudinal pressure wave; about $343 \text{ m/s}$ in room-temperature air, faster in liquids and solids.
- Pitch follows frequency; loudness follows amplitude (decibels, $+10 \text{ dB}$ is 10 times the intensity).
- Echo distance $\displaystyle = \frac{vt}{2}$.
Doppler effect
- Source and observer approaching: higher observed frequency. Moving apart: lower.
Interference and standing waves
- Superposition: displacements add. Constructive (crest on crest) and destructive (crest on trough).
- Beat frequency $= |f_1 - f_2|$.
- String fixed at both ends: $\displaystyle \lambda_n = \frac{2L}{n}$, $f_n = nf_1$. Resonance happens at a natural frequency.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle \lambda = \frac{v}{f} = \frac{3.0 \times 10^8}{1.0 \times 10^8} = 3.0 \text{ m}$.
Light arrives almost instantly; the sound travels $343 \times 3 = 1029 \text{ m}$ one way.
Frequency sets pitch; amplitude sets loudness.
A receding source gives a lower observed frequency.
$\displaystyle \lambda = \frac{v}{f}$: larger $v$ with the same $f$ gives a longer wavelength.
The $n$th harmonic on a string has $n$ antinodes (loops).
Free response
Write your answer first, then open the scoring guide.
Free-response question
A guitar string is 0.65 m long and fixed at both ends. Waves travel along it at $390 \text{ m/s}$. The string's sound travels through air at $343 \text{ m/s}$.
(a) What are the wavelength and frequency of the string's fundamental?
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$\lambda_1 = 2L = 2(0.65) = 1.3 \text{ m}$. $\displaystyle f_1 = \frac{v}{\lambda_1} = \frac{390}{1.3} = 300 \text{ Hz}$.
Check off the points you earned:
(b) What is the frequency of the second harmonic, and how many nodes does that pattern have (including the ends)?
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$f_2 = 2f_1 = 600 \text{ Hz}$. The pattern has 2 loops, so 3 nodes: one at each end and one in the middle.
Check off the points you earned:
(c) What is the wavelength in air of the sound from the fundamental? Explain why it differs from the wavelength on the string.
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The frequency in air is still $300 \text{ Hz}$, so $\displaystyle \lambda = \frac{343}{300} \approx 1.14 \text{ m}$. It is different because the wave speed in air ($343 \text{ m/s}$) differs from the speed on the string, while the frequency stays the same.
Check off the points you earned: