Difficulty: Intermediate | Prerequisites: oscillations and SHM (see Part 1 notes), basic thermodynamics (temperature in kelvin), logarithms
Waves and sound build directly on the oscillation material. Once you understand a single oscillator, a wave is what happens when many oscillators are coupled together and energy propagates through the medium. This unit introduces how fast sound travels in a gas, how loud a sound is (intensity and decibels), the Doppler effect, standing waves on strings, beats and double-slit interference. These topics are heavily tested on Midterm 1 and form the foundation for everything that follows in PHYS 116.
Sound is a longitudinal mechanical wave whose speed depends on the gas temperature and molecular mass. Intensity follows an inverse-square law and is measured on a logarithmic decibel scale. The Doppler effect shifts perceived frequency when source or observer moves. Standing waves form at specific resonant frequencies, and beats arise when two close frequencies interfere. Double-slit interference produces bright and dark fringes whose spacing depends on wavelength, slit separation and screen distance.
Wave speed in an ideal gas (v_sound)
The speed at which sound propagates through a gas: v = √(γRT/M), where γ is the adiabatic index, R is the gas constant, T is temperature in kelvin, and M is molar mass. Think of it as: hotter, lighter gases carry sound faster.
Sound intensity (I)
The power per unit area carried by a sound wave, measured in W/m². In simple terms, this means how much sound energy hits each square metre of surface every second.
Sound intensity level (β)
A logarithmic measure of sound intensity relative to a reference: β = 10 log₁₀(I/I₀), measured in decibels (dB). I₀ = 10⁻¹² W/m² is the threshold of human hearing.
Doppler effect
The change in perceived frequency of a wave when the source or observer is moving. Think of it as: the pitch of an ambulance siren rising as it approaches and falling as it moves away.
Standing wave
A wave pattern that forms when two identical waves travel in opposite directions and interfere. The result has fixed nodes (zero displacement) and antinodes (maximum displacement).
Antinode
A point on a standing wave where the amplitude of oscillation is at its maximum. In simple terms, this means the spots that move the most.
Beat frequency
The frequency of the amplitude variation heard when two waves of slightly different frequencies interfere: f_beat = |f₁ − f₂|. Think of it as: the "wah-wah-wah" pulsing you hear when two nearly matched notes play together.
Double-slit interference
The pattern of bright and dark fringes produced when coherent light passes through two narrow slits. The fringe spacing depends on wavelength, slit separation and screen distance.
Linear mass density (μ)
Mass per unit length of a string, measured in kg/m. Determines how fast transverse waves travel along the string: v = √(T/μ), where T is the tension.
The speed of sound in an ideal gas is v = √(γRT/M), where γ is the ratio of heat capacities (adiabatic index), R = 8.314 J/(mol·K), T is the absolute temperature in kelvin, and M is the molar mass in kg/mol.
Temperature must be in kelvin: T(K) = T(°C) + 273.15.
For diatomic nitrogen (N₂): M = 2 × 14.0 g/mol = 28.0 g/mol = 0.0280 kg/mol, γ = 1.40.
Example: at −30.0°C (243.15 K), v = √(1.40 × 8.314 × 243.15 / 0.0280) = √(1.40 × 2021.7 / 0.0280) = √(101,085) ≈ 318 m/s.
Sound intensity level: β = 10 log₁₀(I/I₀), where I₀ = 10⁻¹² W/m².
To find intensity from decibels: I = I₀ × 10^(β/10).
Example: β = 92.0 dB. I = 10⁻¹² × 10^(92/10) = 10⁻¹² × 10^9.2 = 10⁻²·⁸ ≈ 1.58 × 10⁻³ W/m².
When the observer moves towards the source: f' = f × (v_sound + v_observer) / v_sound.
When the source moves towards the observer: f' = f × v_sound / (v_sound − v_source).
The key detail in the exam question: the student notices the wave speed has increased. The speed of sound is affected by the medium, not by source motion. If the observer is moving towards the source, the wave speed relative to the observer increases.
So the observer (student) is moving towards the stationary source: f' = f(v + v_obs)/v. With f' = 1.07f: 1.07 = (v + v_obs)/v, so v_obs = 0.07v = 0.07 × 343 ≈ 24.0 m/s.
Wave speed on a string: v = √(T/μ), where T is tension and μ is linear mass density.
For a string fixed at both ends, the resonant wavelengths are λₙ = 2L/n, where n = 1, 2, 3, ...
The resonant frequencies are fₙ = nv/(2L). The number of antinodes equals n.
Example: L = 0.750 m, T = 40.0 N, μ = 0.00711 kg/m. v = √(40.0/0.00711) = √(5627) ≈ 75.0 m/s.
At f = 150.0 Hz: n = 2Lf/v = 2 × 0.750 × 150.0 / 75.0 = 3. So 3 antinodes.
When two frequencies f₁ and f₂ are close together, beats are heard at f_beat = |f₁ − f₂|.
Beat frequency tells you how far apart the two frequencies are, but not which is higher.
Example: a violin string produces 4 beats per second with a 440 Hz tuning fork. The violin is "too tightly stretched," meaning its tension is too high, which raises its frequency. So the string frequency is 440 + 4 = 444 Hz.
The position of bright fringes on a distant screen: y_m = mλL/d, where m is the order, λ is the wavelength, L is the distance to the screen, and d is the slit spacing.
The fringe spacing (distance between adjacent bright fringes) is Δy = λL/d.
If the fringe spacing changes by some factor, check: Δy_new/Δy_old = (λ_new/λ_old)(L_new/L_old)(d_old/d_new).
The second pattern in the exam appears to have wider fringe spacing (roughly 1.5× the original). You need to find which option gives a factor of 1.5 for the ratio λL/d.
Option B: (675/450)(4L/L)(d/(4d)) = 1.5 × 4 × 0.25 = 1.5. This works.
Option C: (450/450)(2L/L)(d/(3d)) = 1 × 2 × (1/3) = 0.667. This gives narrower fringes.
Check all options to determine if more than one works.
v_{\text{sound}} = \sqrt{\frac{\gamma R T}{M}}\beta = 10 \log_{10}\!\left(\frac{I}{I_0}\right), \quad I_0 = 10^{-12}\;\text{W/m}^2I = I_0 \times 10^{\,\beta/10}f' = f\,\frac{v \pm v_{\text{obs}}}{v \mp v_{\text{src}}}v_{\text{string}} = \sqrt{\frac{T}{\mu}}f_n = \frac{n\,v}{2L}, \quad n = 1, 2, 3, \ldotsf_{\text{beat}} = |f_1 - f_2|\Delta y = \frac{\lambda L}{d}The speed of sound in air determines how quickly you hear thunder after seeing lightning, and it changes noticeably in cold weather. Concert sound engineers use the decibel scale daily to manage speaker levels and protect hearing. The Doppler effect is the principle behind police radar guns and medical ultrasound blood-flow measurements. Standing waves on strings are how every stringed instrument produces its tone, and beats are how musicians tune by ear.
Students often think the speed of sound depends on frequency or amplitude. It does not: in an ideal gas, v depends only on temperature and the molecular properties of the gas.
A common error with decibels is dividing by 10 instead of using the power-of-10 relationship. Remember: β = 10 log₁₀(I/I₀), so I = I₀ × 10^(β/10), not I₀ × β/10.
With the Doppler effect, students frequently mix up which formula to use. The observer moving changes the numerator; the source moving changes the denominator. A useful mnemonic: the observer is "on top."
Students sometimes think beat frequency tells you which of two sources has the higher pitch. It does not: beats give you |f₁ − f₂|, not the sign. You need additional information (such as "the string is too tight") to determine which frequency is higher.
In double-slit problems, students sometimes forget that increasing slit spacing d makes the fringes closer together, not further apart. Wider slits, narrower pattern.
⚠️ Wave speed in a gas calculations are common. Always convert temperature to kelvin and molar mass to kg/mol before plugging in.
⚠️ Decibel-to-intensity conversions appear frequently. Drill the formula I = I₀ × 10^(β/10) until it is automatic.
⚠️ Doppler problems require you to identify who is moving and apply the correct form of the equation. Read the problem carefully for clues about whether the wave speed changes (observer moving) or the wavelength changes (source moving).
⚠️ Standing wave questions on strings nearly always ask for the harmonic number n or the number of antinodes. Find v first, then use n = 2Lf/v.
⚠️ Beats problems will often give you physical context ("too tight," "too loose") to determine which frequency is higher. Do not ignore this information.
⚠️ Double-slit questions test whether you understand how λ, L and d each affect the fringe spacing. Be prepared to work out ratios.
True or false: Sound travels faster in cold air than in warm air. (False, speed increases with temperature)
Fill in the blank: Every increase of 10 dB corresponds to a factor of ______ in intensity. (10)
True or false: When you walk towards a stationary siren, the pitch you hear is higher than the emitted pitch. (True)
Fill in the blank: A string fixed at both ends vibrating in its 3rd harmonic has ______ antinodes. (3)
True or false: Increasing the slit spacing in a double-slit experiment makes the fringes wider apart. (False, increasing d makes fringes closer together)
Q: Air at −30.0°C is pure N₂ (atomic mass 14.0 g/mol, γ = 1.40). What is the speed of sound?
A: T = 243.15 K, M = 0.0280 kg/mol. v = √(1.40 × 8.314 × 243.15 / 0.0280) ≈ 318 m/s. Answer C.
Q: The sound intensity level at a concert is 92.0 dB. What is the sound intensity?
A: I = 10⁻¹² × 10^(92/10) = 10⁻¹² × 10^9.2 = 10⁻².⁸ ≈ 1.58 × 10⁻³ W/m². Answer E.
Q: A fire engine emits frequency f. A student hears f' = 1.07f and notices the wave speed has increased. What is the speed of the moving party?
A: The wave speed increasing means the observer is moving (wave speed relative to observer = v + v_obs). So f' = f(v + v_obs)/v. 1.07 = 1 + v_obs/343. v_obs = 0.07 × 343 ≈ 24.0 m/s. Answer D.
Q: A 75.0 cm string (μ = 0.00711 kg/m, tension 40.0 N) oscillates at 150.0 Hz. How many antinodes?
A: v = √(40.0/0.00711) ≈ 75.0 m/s. n = 2Lf/v = 2(0.750)(150)/75.0 = 3. Answer C.
Q: A violin string and a 440 Hz tuning fork produce 4 beats per second. The string is too tightly stretched. What is the string's frequency?
A: Too tight means the frequency is too high. f_string = 440 + 4 = 444 Hz. Answer B.
Q: A 450 nm laser through a double slit (spacing d, screen distance L) produces a pattern. A second pattern has wider fringes. Which change could produce this?
A: The fringe spacing ratio is (λ_new/λ_old)(L_new/L_old)(d_old/d_new). Option B: (675/450)(4/1)(1/4) = 1.5. This gives fringes 1.5× wider. Check all options to see if more than one works. Answer B (or E if multiple work).
Wave speed in gases connects to thermodynamics and the kinetic theory of gases. If you go on to study fluid dynamics or atmospheric science, this same √(γRT/M) relationship will reappear.
The decibel scale and logarithmic measurement come up again in electrical engineering (signal gain), seismology (the Richter scale) and chemistry (pH).
Double-slit interference is one of the foundational experiments in quantum mechanics. When you reach modern physics, you will see the same fringe pattern used to demonstrate wave-particle duality with electrons.
speed of sound, ideal gas, adiabatic index, gamma, molar mass, kelvin, sound intensity, decibels, dB, threshold of hearing, inverse square law, Doppler effect, Doppler shift, observer moving, source moving, standing waves, harmonics, resonance, nodes, antinodes, fundamental frequency, overtones, linear mass density, string tension, beat frequency, tuning fork, interference, double slit, Young's experiment, fringe spacing, wavelength, slit separation, PHYS 116 midterm 1, University of Washington general physics