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# Three sound waves of equal amplitudes have frequencies $(v-1), v, (v+1)$. They are supposed to give beats. The number of beats produced per second will be:A.)4B.)3C.)2D.)1

Last updated date: 13th Sep 2024
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Hint: Recall that the number of beats produced per second is also called beat frequency. We know that waves superpose to form beats. But when waves of equal amplitudes superpose, what property of the waves undergo interference? In other words, beats are formed when the waves that superpose “occur” at least once together. This “occurrence” can be obtained by finding the difference between the very same property of the waves that undergo interference.

Formula used:
Beat frequency $=|{{f}_{min}}-{{f}_{max}}|$ or just $={{f}_{max}}-{{f}_{min}}$

Let us first try to use a rather intuitive approach to solving this problem.

The frequency of a sound wave is the number of times the wave passes through a fixed point in a given amount of time. Since frequency of sound waves is usually measured in Hz or $s^{-1}$ it basically describes the number of times the wave passes through a fixed point in 1 second (per second).

Now what are beats? Beats are the vibrations that are produced as a result of superimposing two or more sound waves of similar frequencies. They occur as a result of constructive interference of the waves where their crests or troughs overlap simultaneously.

Now if an interference is to occur between waves 1 and 2, where $v-1$ and $v$ number of waves pass through together in a second, then there is only $|(v-1)-(v)|$ chance for them to overlap. Thus there is only 1 beat produced from waves 1 and 2.

Similarly, if an interference is to occur between waves 2 and 3, where $v$ and $v+1$ number of waves pass through together in a second, then there is only $|(v)-(v+1)|$ chance for them to overlap. Thus, there is only 1 beat produced from waves 2 and 3 as well.

Similarly, if an interference is to occur between waves 1 and 3, where $v-1$ and $v+1$ number of waves pass through together in a second, then there are $|(v-1)-(v+1)|$ chances for them to overlap. Thus, there are 2 beats produced from waves 1 and 3.
We can conclude that the number of beats per second can be quantified as the modulus of the maximum difference in interfering wave frequencies i.e.,

Number of beats per second (beat frequency) $=|{{f}_{min}}-{{f}_{max}}|$ or just $={{f}_{max}}-{{f}_{min}}$

Therefore, the maximum number of beats produced per second will be via the interference of waves 1 and 3 and they will produce 2 beats per second.

So, the correct answer is “Option C”.

Note:
In case of just two waves with similar frequencies the number of beats produced will just be the difference in their frequencies. Remember that beat frequency and wave frequency are different as beat frequency is obtained from the superposition of individual waves while wave frequency is an innate property of the wave produced.