Question

two small speakers a and b are driven in step at 745 hz by the same audio oscillator. these speakers both start out 4.50 m from the listener, but speaker a is slowly moved away (see figure 1).
part a
at what distance d will the sound from the speakers first produce destructive interference at the location of the listener?
express your answer in meters.
d = m
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part b
if a keeps moving, at what distance d will the speakers next produce destructive interference at the listener?
express your answer in meters.
d = m
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Explanation:

Step1: Calculate the wavelength

The speed of sound \(v = 344\space m/s\), and the frequency \(f=745\space Hz\). Using the formula \(v = f\lambda\), we can find the wavelength \(\lambda=\frac{v}{f}\).

$$ \lambda=\frac{344}{745}\approx0.462\space m $$

Step2: Use the condition for destructive interference

For destructive interference, the path - difference \(\Delta r=(n +\frac{1}{2})\lambda\). Initially, the path - difference \(\Delta r = d - 4.50\) (where \(d\) is the distance of speaker A from the listener and the distance of speaker B from the listener is \(4.50\space m\)). For the first destructive interference \(n = 0\), so \(\Delta r=\frac{\lambda}{2}\).

$$ d-4.50=\frac{\lambda}{2} $$

Substitute \(\lambda = 0.462\space m\) into the above equation:

$$ d=4.50+\frac{0.462}{2}=4.50 + 0.231=4.731\space m $$

Step3: For the next destructive interference

For the next destructive interference \(n = 1\), so \(\Delta r=(1+\frac{1}{2})\lambda=\frac{3\lambda}{2}\)

$$ d-4.50=\frac{3\lambda}{2} $$

Substitute \(\lambda = 0.462\space m\) into the equation:

$$ d=4.50+\frac{3\times0.462}{2}=4.50+0.693 = 5.193\space m $$

Answer:

Part A: \(d = 4.73\space m\)
Part B: \(d = 5.19\space m\)