Two-speaker interference: a 'quiet point' that is not a minimum
A student sets up two identical loudspeakers on a bench, separated by 1.10 m, facing a flat wall 3.00 m away. Both speakers are driven in phase by the same signal generator at 1.20 kHz. The speed of sound in the room is 343 m/s. A microphone slides along the wall on a line parallel to the speakers, with its position x measured from the central axis. At x = 0 the signal is strong. The student finds a pronounced quiet point at x = 0.95 m and claims it is the first-order minimum. The room has hard walls and the speakers are not perfectly matched in amplitude. The student also notices that the quiet point is not completely silent and that the positions of the minima drift slightly when the temperature changes by 2 °C.
1. Calculate the wavelength and the path difference at x = 0.95 m. Determine whether this point is a first-order minimum, and if not, calculate the expected position of the first minimum using the small-angle approximation.
Read the worked answer
Wavelength: λ = v/f = 343/1200 = 0.286 m.
With speakers at x = ±0.55 m and the wall 3.00 m away, the path difference at microphone position x is Δ = sqrt((x + 0.55)² + 3.00²) − sqrt((x − 0.55)² + 3.00²).
At x = 0.95 m: Δ = sqrt(1.50² + 9.00) − sqrt(0.40² + 9.00) = sqrt(11.25) − sqrt(9.16) = 3.354 − 3.027 = 0.327 m.
A first-order minimum requires Δ = λ/2 = 0.143 m. The measured point is not a first-order minimum; it is not even close. The small-angle approximation gives Δ ≈ xd/L, so the first minimum is at x ≈ (λ/2)L/d = (0.143 × 3.00)/1.10 = 0.390 m. The second minimum would be at x ≈ (3λ/2)L/d = (0.429 × 3.00)/1.10 = 1.17 m. Thus 0.95 m lies between the first and second minima and cannot be explained as a simple two-source minimum.
2. The student reverses the connections to one speaker. Predict what happens at x = 0 and at the expected first minimum position, and explain why the quiet point at 0.95 m is not completely silent even before reversing.
Read the worked answer
Reversing one speaker adds a phase difference of π. At x = 0 the path difference is zero, so the π phase shift makes the two waves destructive: the centre becomes a minimum (ideally a quiet point). At the position where the path difference is λ/2, the extra π phase shift makes the total phase difference 2π, so the waves arrive in phase and the point becomes a maximum.
The quiet point is not completely silent because destructive interference only gives zero amplitude if the two waves have equal amplitude and are perfectly coherent. The speakers are at different distances from the microphone, so the inverse-square fall-off makes their amplitudes unequal; the speakers may also have different efficiencies. Hard-wall reflections add extra waves with different path lengths, and the room is not anechoic. The finite size of the speakers also means the simple point-source model is imperfect. All of these reduce the depth of the minimum.
3. The temperature rises by 2 °C. Estimate the fractional change in wavelength and discuss whether this explains the drift of the minima. Use v ∝ sqrt(T) with T in kelvin.
Read the worked answer
At about 20 °C, T = 293 K. A 2 K rise gives ΔT/T = 2/293 = 0.0068. Since v ∝ T^0.5, Δv/v ≈ ½(ΔT/T) = 0.0034, or 0.34%. The wavelength changes by the same fraction because f is fixed.
The fringe positions scale with λ, so the first minimum at 0.390 m would shift by about 0.0034 × 0.390 = 1.3 mm. A drift of a few millimetres could therefore be explained by the temperature change, but a drift of centimetres could not. The student should measure the actual drift before attributing it to temperature. Other causes include air currents, speaker drift, and reflections changing with microphone position.