Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.00 m.A person is standing 12.0 m away from the wall,equidistant from the loudspeakers.When the person moves 1.00 m parallel to the wall,she experiences destructive interference for the first time.What is the frequency of the sound? The speed of sound in air is 343 m/s.
A) 211 Hz
B) 256 Hz
C) 422 Hz
D) 512 Hz
E) 674 Hz

Question

Quizplus · Accepted Answer

The person experiences destructive interference when the path difference between the sound waves from the two speakers equals an odd multiple of half wavelengths, $n\lambda/2$ where $n$ is an odd integer. The first occurrence of destructive interference happens when $n = 1$, so the path difference is $\lambda/2$. Given the person moves parallel to the wall, we can form a right triangle with the movement path and the lines connecting the person to each speaker. The path difference is the difference in lengths of these two lines. Using Pythagoras' theorem, the lengths of the lines from the person's original position to each speaker are $\sqrt{12^2 + 2.5^2}$ meters, and from the new position, they are $\sqrt{12^2 + 1.5^2}$ meters and $\sqrt{12^2 + 3.5^2}$ meters. The path difference is thus $\sqrt{12^2 + 3.5^2} - \sqrt{12^2 + 1.5^2}$.Calculating the path difference:$$ \Delta L = \sqrt{12^2 + 3.5^2} - \sqrt{12^2 + 1.5^2} \approx \sqrt{144 + 12.25} - \sqrt{144 + 2.25} \approx \sqrt{156.25} - \sqrt{146.25} \approx 12.5 - 12.1 = 0.4 \, \text{m} $$Since this path difference equals $\lambda/2$, we have $\lambda = 0.8 \, \text{m}$.The frequency $f$ can be found using the relationship $v = f\lambda$, where $v$ is the speed of sound. Rearranging for $f$, we get $f = v/\lambda$.Substituting the given values:$$ f = 343 \, \text{m/s} / 0.8 \, \text{m} = 428.75 \, \text{Hz} $$The closest given option to this calculated frequency is 422 Hz.

Two In-Phase Loudspeakers That Emit Sound with the Same Frequency