Two speakers face each other at a distance of 1 m and are driven by a common audio oscillator.A first minimum in sound intensity is found 16.1 cm from the midpoint.If the velocity of sound is 330 m/s,find the frequency of the oscillator.
A)256 Hz
B)1024 Hz
C)512 Hz
D)341 Hz
E)683 Hz

Question

Quizplus · Accepted Answer

The first minimum occurs when the distance from one speaker to the minimum point and back to the other speaker is half of a wavelength. Let's call this distance x. 
From the given information, we can set up the following equation: 
2x + 0.161 m = nλ 
where n is an integer representing the number of wavelengths between the speakers. 
Since the distance between the speakers is 1 m, we know that n must be odd for the first minimum (otherwise we would get a maximum instead). So we can set n = 1, 3, 5, etc. 
Let's solve for λ: 
2x + 0.161 m = λ 
Substituting in the velocity of sound, we get: 
330 m/s = f λ 
where f is the frequency we want to find. 
Substituting in λ from above, we get: 
f = 330/(2x + 0.161) 
Now we can solve for f using the distance x from one speaker to the minimum point: 
x = 0.161/2 + 1/2 = 0.8315 m 
f = 330/(2(0.8315) + 0.161) 
f ≈ 512 Hz

Two Speakers Face Each Other at a Distance of 1