Interference diffraction and sound

Sound is a longitudinal mechanical wave: the air pressure oscillates along the direction of propagation, and the wave travels with the speed $v$ set by the medium and the wavelength $\lambda = v/f$.

When two coherent sources (same frequency, constant phase difference) radiate into the same region, the waves superpose, and the amplitude at each point depends on the path difference $\Delta r = r_2 - r_1$. Constructive interference (louder) occurs when

$$\Delta r = n\lambda, \qquad n = 0, \pm1, \pm2, \dots$$

and destructive interference (quieter) occurs when

$$\Delta r = \left(n+\tfrac12\right)\lambda.$$

For two sources separated by a distance $d$, the maxima lie in directions $\theta$ measured from the centre line, where

$$d\sin\theta = n\lambda.$$

Diffraction is the bending and spreading of a wave at an edge or an opening; it is strong when the obstacle or aperture size $a$ is comparable to or smaller than the wavelength ($a \lesssim \lambda$). For a doorway with $a \approx 0.8$ m the condition $a \lesssim \lambda$ holds only when $\lambda \gtrsim 0.8$ m, that is $f \lesssim v/a \approx 400$ Hz: only low and mid audible frequencies diffract strongly at the opening, whereas for $\lambda \ll a$ (high frequencies) the doorway transmits a nearly straight, only slightly spreading beam. Single-slit minima satisfy $a\sin\theta = n\lambda$, so when $\lambda > a$ there is no minimum at all, and the sound spreads behind the aperture almost uniformly.

Two loudspeakers are driven in phase at $f = 680$ Hz in air where $v = 340$ m/s. A listener stands where the distances to the speakers are $r_1 = 4.00$ m and $r_2 = 5.25$ m. Find the wavelength and decide whether the listener is at a maximum (loud) or a minimum (quiet).

Two in-phase loudspeakers stand on a stage a distance $d = 2.0$ m apart and radiate a tone of $f = 343$ Hz ($v = 343$ m/s). A listener walks along a far arc whose points are all much farther from the sources than $d$. (a) At what angle from the centre line is the first off-axis maximum? (b) How many maxima in total can be found on that arc?

A sound source in a room plays a tone that reaches a listener outside through an open doorway of width $a = 0.80$ m. Take $v = 340$ m/s. (a) For $f = 200$ Hz, is a diffraction minimum possible? Estimate how the loudness changes as the listener moves to the side. (b) Compare with $f = 4000$ Hz and explain why bass notes are heard around a corner but high notes are not.

Self-study plan (safe, at home or in class).

Safety. Keep the volume moderate, never hold loudspeakers against your ears, and do not use headphones for this work; if your ears ring or feel uncomfortable, stop and rest in silence.

Option A — simulation (if you want to avoid noise). Open the PhET "Wave Interference" simulation in two-source mode. Set a wavelength equivalent to $f = 2000$ Hz at $v = 343$ m/s ($\lambda \approx 0.17$ m) and a source separation $d = 3\lambda$. Mark the lines of maxima and minima on the screen, then change $d$ and $\lambda$ and record how the number and spacing of the fringes change. Then switch to one-slit mode and vary the slit width $a$ from $0.2\lambda$ to $5\lambda$, noting at which ratio the spreading behind the slit becomes noticeably narrow.

Option B — measurement with a smartphone. Place two small loudspeakers (or two phones) a distance $d = 1.00$ m apart, playing the same tone $f = 2000$ Hz. At a distance $D = 2.00$ m draw a line parallel to the line of sources and record a sound-level app reading every 0.05 m. Prediction: the spacing between neighbouring minima is

$$\Delta x \approx \frac{\lambda D}{d} = \frac{0.17 \cdot 2.00}{1.00} \approx 0.34 \text{ m}.$$

Plot "sound level versus position", locate the minima, and compare the measured $\Delta x$ with the prediction. Repeat at $f = 200$ Hz and explain why the pattern becomes coarser and why a low tone at an open door is heard from almost any angle.

To show how the path difference of two coherent sources creates hyperbolic loud and quiet zones, and how the number and spacing of the maxima depend on the ratio $d/\lambda$.

Diagram (labelled sketch). On a horizontal line sit two points $S_1$ and $S_2$ separated by $d$; straight segments $r_1$ and $r_2$ run from them to the listener point $P$, and the angle $\theta$ is measured from the centre line (the perpendicular bisector of $S_1S_2$). Solid curves through points with $\Delta r = n\lambda$ are labelled "maxima (loud)", and dashed curves with $\Delta r = \left(n+\tfrac12\right)\lambda$ are labelled "minima (quiet)".

Inset. A plane wave strikes an aperture of width $a$; behind it the wavefronts bend into the geometrical shadow; a caption reads: $a \lesssim \lambda$ — strong spreading, $a \gg \lambda$ — nearly straight propagation.

In headphones, a microphone samples outside noise at $200\ \text{Hz}$; with speed of sound in air $343\ \text{m/s}$, wavelength is $\lambda=v/f=343/200\approx1.7\ \text{m}$. Electronics sends a wave of the same frequency but opposite phase to the speaker. At the ear, these two sound waves add, and destructive interference reduces loudness by about $20\ \text{dB}$.

This uses superposition of two sound waves and the destructive-interference condition: at phase difference $\Delta\phi=(2n+1)\pi$ the waves cancel. In active noise cancellation the phase shift is made electronically, not only by path difference; a path difference $\Delta r=(n+1/2)\lambda$ gives the same result when waves arrive from two sources.

Two speakers stand $2.0\ \text{m}$ apart and emit sound at $686\ \text{Hz}$. With $v=343\ \text{m/s}$, the wavelength is $\lambda=v/f=343/686=0.50\ \text{m}$. At point A the distances to the speakers are $6.00\ \text{m}$ and $6.50\ \text{m}$, so the path difference is $\Delta r=0.50\ \text{m}=\lambda$, and loudness is maximum. At point B the distances are $6.00\ \text{m}$ and $6.25\ \text{m}$, so $\Delta r=0.25\ \text{m}=\lambda/2$, and the sound is almost cancelled.

This is a direct application of two-source interference conditions: maxima occur at $\Delta r=n\lambda$, minima at $\Delta r=(n+1/2)\lambda$. The scenario tests how path difference measured in wavelengths controls loudness at different points in the hall.

A band plays on an open stage. A bass note at $85\ \text{Hz}$ has wavelength $\lambda=v/f=343/85\approx4.0\ \text{m}$, while a cymbal at $8.0\ \text{kHz}$ has $\lambda=343/8000\approx0.043\ \text{m}$. A gate or doorway is about $1.0\ \text{m}$ wide. The bass bends around the building corner and is heard around the corner, while high frequencies hardly bend around the obstacle.

Diffraction appears here: sound bends noticeably around an obstacle or through an opening when the wavelength is comparable to the obstacle or opening size. For the bass, $\lambda\approx4.0\ \text{m}$ with a $1.0\ \text{m}$ opening gives strong diffraction; for the cymbal, $\lambda\approx0.043\ \text{m}\ll1.0\ \text{m}$ gives weak diffraction.