Beats: frequency of resulting wave vs. beats frequency

  • #1
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The beats frequency heard from the interference of two sound waves with frequencies ##f_1## and ##f_2## is $$\nu=|f_1-f_2|$$

Nevertheless the frequency of the resulting wave is not ##\nu## but the mean value of the two frequencies
$$f_{resulting}=\frac{f_1+f_2}{2}$$

As far as I understood ##\nu## is the frequency at which the maxima of intensity are heard, while ##f_{resulting}## is indeed the frequency of the resulting wave.

In particular I can notice that ##f_{resulting} > \nu## always.

The distincion between the two is clear in theory, but in practice I still have doubts.

Suppose to have an instrument that can measure sound waves iff the frequency is, say, bigger than ##f_{min}## and lower than ##f_{max}##.

Now suppose that two waves interefere in a way such that ##f_{resulting} > f_{min}## but ##\nu <f_{min}##, or, in a way such that ##\nu < f_{max}## but ##f_{resulting} > f_{max}##.

What does the instrument measure in these cases?

Which of the two frequencies "determine" the upper of lower limits for the frequency that can be measured by such instrument?
 

Answers and Replies

  • #2
sophiecentaur
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The beats frequency heard from the interference of two sound waves with frequencies ##f_1## and ##f_2## is $$\nu=|f_1-f_2|$$

Nevertheless the frequency of the resulting wave is not ##\nu## but the mean value of the two frequencies
$$f_{resulting}=\frac{f_1+f_2}{2}$$

As far as I understood ##\nu## is the frequency at which the maxima of intensity are heard, while ##f_{resulting}## is indeed the frequency of the resulting wave.

In particular I can notice that ##f_{resulting} > \nu## always.

The distincion between the two is clear in theory, but in practice I still have doubts.
This link says it all, I think.
The reason for hearing a beat of twice the frequency that you would expect is that you hear the two peaks in loudness and your ear ignores the phase inversion of the sum at frequency (f1+f2)/2
The ear is an imperfect instrument for measuring what you present it with and that accounts for the apparent paradox, I think.
The instrument that you describe would be like an AM receiver, I think and, to measure the beat, it would need a wide enough bandwidth to include the two input tones. But, as with an AM receiver, it only needs to admit those tones and needs no response at the difference frequency - to take an extreme case, a 1MHz AM receiver doesn't need any response to audio signals in its antenna input in order to receive audio signals carried on the carrier.
 
  • #3
Svein
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[itex]\sin(x)+\sin(y)=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2}) [/itex]
Let [itex]x=\omega_{1}t [/itex] and [itex]y=\omega_{2}t [/itex].
 
  • #4
sophiecentaur
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[itex]\sin(x)+\sin(y)=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2}) [/itex]
Let [itex]x=\omega_{1}t [/itex] and [itex]y=\omega_{2}t [/itex].
Did you want to add some more? :smile:
 

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