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Superposition: Question 4

Syllabus 8.3

Multiple choice AS 1 mark

Two loudspeakers, S1S_1 and S2S_2, are connected to the same signal generator so that they act as coherent sources, emitting sound of wavelength 0.80 m0.80\text{ m} in phase with each other. At a point PP, the path from S1S_1 is 2.8 m2.8\text{ m} longer than the path from S2S_2.

What is the phase difference (reduced to a value between 0°0° and 360°360°) between the two waves arriving at PP, and what type of interference occurs there?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Express the path difference in wavelengths

The path difference is Δ=2.8 m\Delta = 2.8\text{ m} and the wavelength is λ=0.80 m\lambda = 0.80\text{ m}. The number of whole wavelengths contained in Δ\Delta is: Δλ=2.80.80=3.5\frac{\Delta}{\lambda} = \frac{2.8}{0.80} = 3.5

Check (recompute independently): 0.80×3.5=0.80×3+0.80×0.5=2.4+0.4=2.80.80 \times 3.5 = 0.80\times3 + 0.80\times0.5 = 2.4+0.4=2.8, confirms 2.8/0.80=3.52.8/0.80=3.5 exactly.

Step 2: Convert to a phase difference

A path difference of one whole wavelength corresponds to a phase difference of 360°360°. So: phase difference=3.5×360°=1260°\text{phase difference} = 3.5 \times 360° = 1260°

Reducing this modulo 360°360° (since phase difference only matters up to a whole number of cycles): 1260°−3×360°=1260°−1080°=180°1260° - 3\times360° = 1260°-1080°=180°.

Check (using only the fractional part): 3.53.5 wavelengths =3= 3 whole wavelengths (a phase shift of 3×360°=1080°3\times360°=1080°, which has no physical effect) plus a remaining 0.50.5 wavelength, i.e. 0.5×360°=180°0.5\times360°=180°, the same result.

Step 3: Identify the type of interference

A path difference that is a whole number of wavelengths (nλn\lambda) gives constructive interference (phase difference 0°0° or a multiple of 360°360°). A path difference that is an odd number of half-wavelengths ((n+12)λ(n+\tfrac12)\lambda) gives destructive interference (phase difference 180°180°).

Here Δ=3.5λ=(3+12)λ\Delta = 3.5\lambda = (3+\tfrac12)\lambda, an odd multiple of half a wavelength, and the phase difference is 180°180°, so the waves arrive out of phase, and destructive interference occurs at PP.

Why the other options are wrong

  • A (0°0°; constructive): would require the path difference to be a whole number of wavelengths (e.g. 3.0λ3.0\lambda or 4.0λ4.0\lambda), not 3.5λ3.5\lambda.
  • B (90°90°; partial): 90°90° would come from a path difference of 0.250.25 of a wavelength (plus whole wavelengths); the actual path difference corresponds to exactly half a wavelength, not a quarter.
  • D (360°360°; constructive): 360°360° is the same as 0°0° (a full cycle), which again requires a whole-wavelength path difference, not 3.5λ3.5\lambda.

Final answer

  • Phase difference =180°= \boxed{180°}; destructive interference occurs at PP, option C.