17.4 Normal modes of a standing sound wave  (Page 2/9)

The phase difference at each point is due to the different path lengths traveled by each wave. When the difference in the path lengths is an integer multiple of a wavelength,

the waves are in phase and there is constructive interference. When the difference in path lengths is an odd multiple of a half wavelength,

the waves are $180\text{°}(\pi \text{rad})$ out of phase and the result is destructive interference. These points can be located with a sound-level intensity meter.

Interference of sound waves

Two speakers are separated by 5.00 m and are being driven by a signal generator at an unknown frequency. A student with a sound-level meter walks out 6.00 m and down 2.00 m, and finds the first minimum intensity, as shown below. What is the frequency supplied by the signal generator? Assume the wave speed of sound is $v=343.00\text{m/s}\text{.}$

Strategy

The wave velocity is equal to $v=\frac{\lambda }{T}=\lambda f.$ The frequency is then $f=\frac{v}{\lambda }.$ A minimum intensity indicates destructive interference and the first such point occurs where there is path difference of $\text{Δ}r=\lambda \text{/}2,$ which can be found from the geometry.

Solution

1. Find the path length to the minimum point from each speaker.
   
   
   $r_1=\sqrt{{\left(6.00\text{m}\right)}^2+{\left(2.00\text{m}\right)}^2}=6.32\text{m,}{r}_2=\sqrt{{\left(6.00\text{m}\right)}^2+{\left(3.00\text{m}\right)}^2}=6.71\text{m}$
2. Use the difference in the path length to find the wavelength.
   
   $\text{Δ}r=\left|r_2-r_1\right|=\left|6.71\text{m}-6.32\text{m}\right|=0.39\text{m}$
   
   
   $\lambda =2\text{Δ}r=2\left(0.39\text{m}\right)=0.78\text{m}$
3. Find the frequency.
   
   $f=\frac{v}{\lambda }=\frac{343.00\text{m/s}}{0.78\text{m}}=439.74\text{Hz}$

Significance

If point P were a point of maximum intensity, then the path length would be an integer multiple of the wavelength.

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The concept of a phase shift due to a difference in path length is very important. You will use this concept again in Interference and Photons and Matter Waves , where we discuss how Thomas Young used this method in his famous double-slit experiment to provide evidence that light has wavelike properties.

Noise reduction through destructive interference

[link] shows a clever use of sound interference to cancel noise. Larger-scale applications of active noise reduction by destructive interference have been proposed for entire passenger compartments in commercial aircraft. To obtain destructive interference, a fast electronic analysis is performed, and a second sound is introduced $180\text{°}$ out of phase with the original sound, with its maxima and minima exactly reversed from the incoming noise. Sound waves in fluids are pressure waves and are consistent with Pascal’s principle; that is, pressures from two different sources add and subtract like simple numbers. Therefore, positive and negative gauge pressures add to a much smaller pressure, producing a lower-intensity sound. Although completely destructive interference is possible only under the simplest conditions, it is possible to reduce noise levels by 30 dB or more using this technique.