Two loudspeakers are placed side by side and driven by the same source at 500 Hz.A listener is positioned in front of the two speakers and on the line separating them,thus creating a constructive interference at the listener's ear.If one of the speakers is gradually pushed toward the listener,how far must it be moved to repeat the condition of constructive interference at the listener's ear? (The speed of sound = 340 m/s. )
A)1.02 m
B)0.68 m
C)0.34 m
D)0.17 m

Question

Quizplus · Accepted Answer

When the two speakers are driven simultaneously, a sound wave with a frequency of 500 Hz is produced. The distance between the two speakers is not provided, but it is mentioned that the listener is positioned in front of the two speakers and on the line separating them. Therefore, the listener hears the sound waves from the two speakers with the same amplitude and phase, resulting in constructive interference.

To repeat the condition of constructive interference, one of the speakers needs to be moved a distance such that the difference in the path length of the sound waves from the two speakers to the listener is equal to an integral multiple of the wavelength of the sound wave.

Let's assume that the distance between the listener and one of the speakers is x. Then, the path length difference between the two speakers is given by (x+d) - x = d, where d is the distance by which the speaker is moved. For constructive interference, d must be equal to an integral multiple of the wavelength.

The wavelength of a wave with frequency 500 Hz can be calculated using the formula λ = v/f, where v is the speed of sound and f is the frequency. Substituting the given values, we get λ = 340/500 = 0.68 m.

Therefore, d = nλ, where n is an integer. To find the minimum distance by which the speaker must be moved, we take n = 1. Hence, d = λ = 0.68 m.

Therefore, the correct choice is B) 0.68 m.

Two Loudspeakers Are Placed Side by Side and Driven by the Same