Two audio speakers in an acoustics laboratory emit identical sound waves in phase with a w… — Armed Services Vocational Aptitude Battery (ASVAB)

Two audio speakers in an acoustics laboratory emit identical sound waves in phase with a wavelength of $0.5\text{ m}$ . At a specific observation point, the distance to the first speaker is $4.0\text{ m}$ and the distance to the second speaker is $4.75\text{ m}$ . Which of the following describes the type of interference that occurs at this location and its effect on the detected sound?

A
Constructive interference, producing a louder sound because the path difference equals an integer multiple of the wavelength.
Destructive interference, producing a softer sound because the two waves arrive exactly half a wavelength out of phase.Answer
C
Constructive interference, producing a higher pitch because the frequencies of the two sound sources combine additively.
D
Destructive interference, producing a lower pitch because the speed of sound decreases over unequal path lengths.

Answer

Destructive interference occurs, producing a softer sound because the path difference is an odd multiple of half-wavelengths (

1.5\lambda

), causing crests to meet troughs.

The difference in distance from the two coherent speakers to the observer is

4.75\text{ m} - 4.0\text{ m} = 0.75\text{ m}

. Dividing by the wavelength (

0.5\text{ m}

) yields

1.5\lambda

, or three half-wavelengths. When two in-phase coherent waves travel distances differing by an odd number of half-wavelengths, they arrive exactly out of phase (

180^\circ

). The crests of one wave coincide with the troughs of the other, resulting in destructive interference and a reduction in sound intensity (loudness).

Step-by-Step Solution

Calculate the path length difference (

\Delta d

) between the two sound sources and the observation point.

\Delta d = |4.75\text{ m} - 4.0\text{ m}| = 0.75\text{ m}

Interference patterns depend on the relative distance traveled by waves from coherent sources.

Determine the path difference in terms of the wave's wavelength (

\lambda = 0.5\text{ m}

\frac{\Delta d}{\lambda} = \frac{0.75\text{ m}}{0.5\text{ m}} = 1.5 = \frac{3}{2}\lambda

Comparing the path difference to the wavelength reveals the phase relationship upon arrival.

Apply the physical criteria for wave interference and evaluate the effect on the acoustic signal.

Because

\Delta d = \left(m + \frac{1}{2}\right)\lambda

with

m = 1

, the waves arrive

180^\circ

(

\pi\text{ radians}

) out of phase, producing destructive interference that minimizes sound amplitude (volume).

Odd half-integer multiples of wavelength cause wave crests from one source to superimpose onto wave troughs from the other, reducing overall loudness without changing wave frequency.

Key Concept

Wave Superposition and Path Difference Interference

Estimated Time:1m 15s

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