Two loudspeakers: does a 6 dB dip really disprove interference?
Two loudspeakers S1 and S2 are 1.50 m apart and are driven in phase by the same signal generator at 680 Hz. The speed of sound is 340 m/s. A listener stands at point P, 4.00 m from S1 and 4.75 m from S2. The sound level at P is 52 dB. At a nearby point Q, where the path difference is 1.00 m, the level is 58 dB. The student claims that P must be a point of perfect silence because the path difference is 1.5 wavelengths, and uses the 6 dB difference to argue that the experiment is invalid. The signal generator frequency can drift by ±5 Hz. The room has hard walls, so reflections are possible. Decide whether the two-source interference model explains the observations, and what the 6 dB difference implies.
1. Calculate the wavelength and the path difference at P in wavelengths. Is P expected to be a minimum or a maximum?
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The wavelength is
λ = v/f = 340/680 = 0.500 m.
The path difference at P is
Δr = 4.75 − 4.00 = 0.75 m = 1.5λ.
For two in-phase sources, destructive interference occurs when Δr = (n + ½)λ. With n = 1, Δr = 1.5λ, so P is expected to be a minimum. At Q, Δr = 1.00 m = 2.0λ, which satisfies Δr = nλ and is expected to be a maximum. The simple path-difference rule therefore predicts a minimum at P and a maximum at Q, in qualitative agreement with the observed 52 dB and 58 dB levels.
2. Explain why P is not silent. Use the inverse-square-like amplitude variation and the observed 6 dB difference to judge whether the simple model is adequate.
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The two waves do not arrive with equal amplitudes because the distances differ. For spherical spreading, amplitude is proportional to 1/r, so at P
A2/A1 = 4.00/4.75 = 0.842.
At destructive interference the resultant amplitude is A1 − A2 = 0.158A1, not zero. Compared with a nearby constructive point where the amplitudes are roughly equal and add to about 1.84A1, the intensity ratio is approximately
(0.158/1.84)² ≈ 0.0074,
which is about −21 dB. The observed dip is only 6 dB, far shallower than this ideal prediction. Reflections from hard walls and background sound can fill in the minimum and change the local amplitudes, so the simple two-source model explains the existence and approximate position of the minimum but not its depth. The 6 dB dip does not invalidate the interference interpretation; it shows that the idealised equal-amplitude, reflection-free model is incomplete.
3. The generator can drift by ±5 Hz. Assess whether this drift undermines using interference minima to determine the speed of sound.
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At 685 Hz, λ = 340/685 = 0.4964 m, so the path difference is 0.75/0.4964 = 1.511λ. The phase difference is 2π(1.511), which is 0.022π rad away from exact destructive interference. At 675 Hz, λ = 0.5037 m and the path difference is 1.489λ, again 0.022π rad away. These deviations are small: near a minimum the intensity varies as the square of the small phase deviation, so the minimum stays deep and its position barely moves.
A 5 Hz drift is 5/680 ≈ 0.7%. If the speed is found from v = fλ using the generator frequency and a wavelength measured from minima spacing, this gives about 0.7% uncertainty in v, which is acceptable for a school determination but not negligible. To reduce the problem, measure several minima and use the average spacing, and monitor or measure the frequency directly. The drift does not destroy the interference pattern; it mainly limits the precision of a speed measurement.