Question

Two point source three meter apart are driven by the same oscillator at a frequency of 800 hz. The sources emit energy in the form of spherical waves. An observer initially hears a maximum at a distance 10 m along the perpendicular bisector of the line joining the two sources. What distance (in m) must observer move along a line parallel to the line joining the two sources before reaching a minimum intensity? (Assume the speed of sound is 343 m/s.)

Answer 1

In order to answer this question, the equation for the velocity of a wave is needed, along with the equation for double slit interference.

The velocity v of a wave of frequency f and wavelength is:

The equation for the distance y from the central point on a distant screen that is a distance D from a pair of wave sources (of wavelength ) that are a distance d apart is approximately:

with m being the order of the minimum, which is integer numbers starting with 0.

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Step 1) Since there is a pair of point sources for the waves, it is functionally identical to the waves being emitted from a pair of closely-spaced slits. That is why the equation for double-slit interference is used.

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Step 2) In order to use the equation, the wavelength of the waves is needed. To do that, use the wave equation, and rearrange it briefly to solve for the wavelength :

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Step 3) Evaluate the wavelength by using 343 m/s for the velocity v, and 800 hz (800 s^-1) for the frequency f:

So the wavelgnth of the waves is 0.42875 m.

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Step 4) Now that the wavelength is known, use that in the equation for double-slit interference for , and use 10 m for the distance D that the observer is from the wave-sources, 3 m for the wave-source separation d, and 0 for the order m, since the first minimum from the central point is needed:

So the observer needs to move about 0.715 m along the line that is parallel with the oscillators in order to encounter a point where the sound is at a minimum.