Difficulty: Intermediate | Prerequisites: SHM and oscillations notes, trigonometry, basic wave concepts
This material builds directly on SHM. Once you understand a single oscillator, waves are what happens when many oscillators are coupled together and energy propagates through the medium. The course moves from single pulses (reflection, superposition) through continuous waves (interference from two sources, nodal and antinodal lines) to sound (intensity, decibels, Doppler effect). These topics make up the bulk of the PHYS 123 midterm.
Waves carry energy, and how that energy spreads depends on geometry (1D strings vs 2D circular ripples). Two sources create interference patterns with predictable nodal and antinodal lines. Sound intensity drops with distance squared, and the Doppler effect shifts frequency depending on relative motion between source and observer.
Superposition
When two or more waves overlap in the same region, the net displacement at any point is the sum of the individual displacements. This is how interference patterns form.
Reflection at a fixed end
A pulse hitting a fixed boundary reflects inverted (flipped). The fixed end must remain stationary, so the reflected pulse has the opposite sign.
Reflection at a free end
A pulse hitting a free boundary reflects upright (not flipped).
Antinodal line
A line in space where waves from two coherent sources always arrive in phase and interfere constructively. The path-length difference is an integer number of wavelengths: Δr = nλ.
Nodal line
A line where waves from two coherent sources always arrive out of phase and interfere destructively. The path-length difference is a half-integer number of wavelengths: Δr = (n + ½)λ.
Doppler effect
The shift in observed frequency when there is relative motion between a wave source and an observer. Approaching makes the frequency higher; receding makes it lower.
Intensity
Power per unit area, measured in W/m². For a point source radiating uniformly, I = P/(4πr²).
Sound intensity level (decibels)
A logarithmic scale for comparing intensities: β = 10 log₁₀(I/I₀), where I₀ = 10⁻¹² W/m² is the threshold of hearing. In simple terms, every increase of 10 dB means the intensity is ten times larger.
Inverse square law
For a source radiating uniformly in all directions, intensity falls as 1/r². Double the distance, quarter the intensity.
A pulse reaching a fixed end reflects inverted.
A pulse reaching a free end reflects upright.
Inversion at a splice depends on the relative densities: a pulse going from a lighter string to a heavier string reflects inverted (the heavier string acts like a more rigid boundary). A pulse going from heavier to lighter reflects upright.
None of the usual changes (increasing μ₂, changing tension, changing pulse width) can reverse an inverted reflection at a splice when the fundamental density ordering stays the same. The only fix is to reverse which string is denser.
To find the shape of a string after a pulse reflects off a fixed wall:
Imagine the pulse continues through the wall as if the wall were not there (the "incident" pulse at its new position).
Create a "virtual" pulse: a mirror image of the incident pulse, flipped both horizontally (reflected about the wall) and vertically (inverted for a fixed end).
Add the two pulses point by point (superposition) to get the actual string shape.
For a circular ripple on water (2D), total energy is conserved but spreads over a growing circumference.
At radius R₁ the wavefront length is 2πR₁. Energy per unit length = E/(2πR₁).
At radius R₂ = 2R₁, the wavefront is twice as long: 2πR₂ = 4πR₁.
Energy per unit length = E/(4πR₁) = half the value at R₁.
This is different from the 3D inverse-square law. In 2D, energy per unit length falls as 1/R, not 1/R².
Two point sources oscillating in phase produce a stable interference pattern.
Path-length difference: Δr = |r₁ - r₂|, where r₁ and r₂ are distances from the point to each source.
Constructive interference (antinodal line): Δr = nλ, where n = 0, 1, 2, ...
Destructive interference (nodal line): Δr = (n + ½)λ, where n = 0, 1, 2, ...
Worked example: Sources at (0, 0) and (0, 5 m). The point (12, 0) has r₁ = 12 m and r₂ = √(144 + 25) = 13 m, so Δr = 1 m.
For an antinodal line: 1 = nλ, so λ = 1/n. Possible wavelengths: 1, ½, ⅓, ¼, ⅕, ... The value ⅔ is not in this list, so it is NOT a possible wavelength for an antinodal line.
For a nodal line: 1 = (n + ½)λ. The largest λ comes from n = 0: λ = 2 m.
The general formula for sound:
f_observed = f_source × (v_sound ± v_observer) / (v_sound ∓ v_source)
Convention: upper signs when approaching, lower signs when receding.
Both approaching: f_obs = f_s × (v + v_obs)/(v - v_s)
Both receding: f_obs = f_s × (v - v_obs)/(v + v_s)
Worked example: An ambulance and car approach at the same speed. f_s = 800 Hz, f_obs = 820 Hz, v_obs = v_s = u.
820 = 800 × (v + u)/(v - u). Solving: (v + u)/(v - u) = 41/40, so v + u = 41v/40 - 41u/40. This gives 81u/40 = v/40, so u = v/81.
When they drive apart: f_obs = 800 × (v - u)/(v + u) = 800 × (80/81)/(82/81) = 800 × 80/82 = 800 × 40/41 ≈ 780.5 Hz.
\Delta r = |r_1 - r_2|Constructive: Δr = nλ. Destructive: Δr = (n + ½)λ.
f_{\text{obs}} = f_s \cdot \frac{v \pm v_{\text{obs}}}{v \mp v_s}I = \frac{P}{4\pi r^2}\beta = 10\,\log_{10}\!\left(\frac{I}{I_0}\right) \quad \text{where } I_0 = 10^{-12}\,\text{W/m}^2P_{\text{absorbed}} = I \cdot A_{\text{surface}}A source emitting power P uniformly in all directions produces intensity I = P/(4πr²) at distance r. Rearranging: r = √(P/(4πI)).
Worked example: A helicopter emits 12.56 W. Intensity at the ground is 10⁻⁴ W/m².
4π r² × 10⁻⁴ = 12.56, so r² = 12.56 / (4π × 10⁻⁴) = 12.56 / (1.256 × 10⁻³) = 10,000. Therefore r = 100 m.
β = 10 log₁₀(I/I₀) = 10 log₁₀(10⁻⁴ / 10⁻¹²) = 10 log₁₀(10⁸) = 80 dB.
If an eardrum absorbs 10⁻⁸ J in one second facing the source, the power absorbed is 10⁻⁸ W.
P_ear = I × A_ear, so A_ear = P_ear / I = 10⁻⁸ / 10⁻⁴ = 10⁻⁴ m² = 1 cm².
The Doppler effect is why an ambulance siren sounds higher-pitched as it approaches and lower-pitched as it drives away. Noise-cancelling headphones use destructive interference, generating a wave that is out of phase with ambient noise. Sonar and ultrasound imaging rely on pulse reflection and the travel time of reflected pulses to measure distances and map surfaces.
Students often think that energy per unit length in a circular wave falls as 1/r². It does not. The wavefront is a circle (1D curve), not a sphere. Energy per unit length falls as 1/r. The 1/r² law applies to 3D spherical waves (energy per unit area).
Students assume the Doppler formula is symmetric: that approaching and receding give equal and opposite frequency shifts. They do not. The formula has v_s in the denominator and v_obs in the numerator, so the shifts are not symmetric.
Students mix up nodal and antinodal conditions. Constructive (antinodal) requires whole-number multiples of λ. Destructive (nodal) requires half-integer multiples.
Students forget that a reflected pulse at a fixed end is inverted. They draw it upright and get superposition wrong.
⚠️ Two-source interference problems test your ability to compute path-length differences and check divisibility. Get comfortable quickly with Δr = nλ and Δr = (n + ½)λ.
⚠️ Doppler problems often ask you to find the speed first from one scenario, then compute the frequency in a second scenario. Do not try to shortcut the algebra.
⚠️ Intensity and decibel calculations are straightforward but easy to trip on if you forget the 4π in the spherical area formula or misremember I₀.
⚠️ Pulse reflection and superposition (the virtual-pulse method) is a staple of the tutorial section.
True or false: a pulse reflecting off a fixed end comes back upright. (False. It comes back inverted.)
Fill in the blank: the energy per unit length of a circular ripple is proportional to ______. (1/R)
True or false: two in-phase sources produce constructive interference when the path-length difference is (n + ½)λ. (False. That is the condition for destructive interference.)
Fill in the blank: for a point source radiating uniformly, the intensity at distance r is ______. (P / (4πr²))
True or false: the Doppler shift is the same magnitude whether source and observer approach or recede at the same speed. (False. The shifts are asymmetric.)
Q: A reflected pulse at a string splice is inverted. Which change makes the reflected pulse non-inverted: increasing μ₂, increasing tension, decreasing tension, or decreasing pulse width?
A: None of the above. The inversion depends on the relative mass densities of the two strings, not on tension or pulse width. You would need to swap which string is denser.
Q: A circular ripple doubles its radius. If energy is conserved, what happens to the energy per unit length along the wavefront?
A: It halves. The circumference doubles (2πR → 2π(2R)), so the same total energy is spread over twice the length.
Q: Two in-phase sources at (0,0) and (0,5). The point (12,0) is on an antinodal line. Is λ = ⅔ m possible?
A: r₁ = 12 m, r₂ = 13 m, Δr = 1 m. For an antinodal line, 1 = nλ, so λ = 1/n. The value ⅔ is not of the form 1/n for any integer n, so it is not possible.
Q: Same geometry, but the point is on a nodal line. What is the largest possible wavelength?
A: 1 = (n + ½)λ, so λ = 1/(n + ½). For the largest λ, use n = 0: λ = 1/(0.5) = 2 m.
Q: An ambulance (800 Hz) and a car approach at speed u and the driver hears 820 Hz. They then drive apart at the same speed. What does the driver hear?
A: From the approach: (v + u)/(v - u) = 820/800 = 41/40. When driving apart: f = 800 × (v - u)/(v + u) = 800 × 40/41 ≈ 780.5 Hz.
Q: A helicopter at altitude d emits 12.56 W. Intensity at the ground is 10⁻⁴ W/m². What is d, and what is the intensity level in dB?
A: 4πd² × 10⁻⁴ = 12.56. d² = 10⁴, d = 100 m. β = 10 log₁₀(10⁻⁴/10⁻¹²) = 10 × 8 = 80 dB.
Q: At 10⁻⁴ W/m², an eardrum absorbs 10⁻⁸ J per second. What is its area?
A: Power = 10⁻⁸ W. Area = Power / Intensity = 10⁻⁸ / 10⁻⁴ = 10⁻⁴ m² = 1 cm².
Interference connects back to SHM through the principle of superposition: you add displacements, just as you did when combining incident and reflected pulses. The Doppler effect reappears in optics (redshift of light from receding galaxies). Intensity and the inverse square law are foundational for any later work in electromagnetism or radiation.
Wave reflection, fixed end, free end, pulse inversion, superposition, virtual pulse, constructive interference, destructive interference, antinodal line, nodal line, path-length difference, two-source interference, Doppler effect, Doppler shift, approaching frequency, receding frequency, sound intensity, decibels, dB, inverse square law, intensity level, threshold of hearing, eardrum area, power and intensity, circular wave energy, PHYS 123, UW physics