What are Beats?

We all like to hear the melody sound of the piano and tuning fork. When you play the piano, the sound you hear by striking black and white keys of your keyboard is the interference of two sound waves. These superimposed waves (having slight variation in their frequencies) have an alternating loudness and softness; this fluctuation is a beat.

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The concept of beat becomes hard to understand because when we can’t see waves, how would we determine a beat’s frequency or its phase difference.

In this article, you will get the best explanation of this concept that will remain in your mind forever. So, let’s get started.


What are Beats in Physics?

Now, let’s say, your friend invited you to his concert, and you hear the pleasant sound of a tuning fork from the loudspeaker. If you’re a science buddy, you might wonder what is a beat and how does it look.

The sound coming out of a speaker S1 looks like this:

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Now, what happens here is, the sound coming out of it strikes with the air molecules. Let’s focus on a single air molecule (A1). A1 on getting hit with sound oscillates about its mean position. 

Now, we’ll plot the graph to demonstrate how the displacement of the air varies with the passing time.

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Now, if we take another speaker S2 and superimpose it on S1, then the graph so formed will be:                           

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Now, on overlapping or interfering these two waves, we can see the frequency difference in the following graph:

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When the two waves overlap, we hear a loud sound, and when they do not overlap, we hear a soft sound. So, this variation in the loudness and softness is a beat. Similarly, when more number of sound waves hit the air molecules, we hear the beats. 

Note: For a beat formation, we must travel in the same direction, and lie in the same phase.   


What is Beat Frequency?

Let’s see from fig.m, the frequency of a pink-colored wave is f1, and that of a green-colored is f2. So, the frequency of the beat is the difference between these two, which is:

                               fBEATS  = |f1 - f2|

 Important Notes: 

  • The reason for keeping a modulus sign is, sometimes f1 is lesser than f2, then we get a negative value, which comes out positive after crossing the modulus sign.

  • For hearing distinctive beats, the frequency of the beats must be less than 3, its because, after this, we can’t hear the beats.

 Now, let us derive the frequency of a beat:


Beat Frequency Formula Derivation

We know that the number of beats in a second is the beat frequency. Now, at time t, the equation of displacements of wave 1 and 2 is given as:

For wave 1:

                             y1  = rsinω1t = rsin2πf1t…..(1)

For wave 2:

                            y2  = rsinω2t = rsin2πf2t….(2)

Now, applying superposition principle, we get:

                  ynet =  y1 + y2 = rt(sin2πf1t + sin2πf2t)


Using the formula: SinA + SinB = 2Cos\[(\frac{A - B}{2})\] . Sin\[(\frac{A + B}{2})\], we get:

                  ynet = 2rCosπ(f1 - f2)t Sinπ(f1 + f2)t

Here, amplitude A = 2rCosπ(f1 - f2)t, is the resultant amplitude of a wave (it varies with time). 

We get our equation as:

                            ynet = ASinπ(f1 + f2)t…..(3)

Now, let us apply the cases for finding the frequency of maxima and minima.


Frequency of Maxima

Amplitude A will be maximum when Cosπ(f1 - f2)t = max = ±1 = Coskπ 

Where, (f1 - f2)t = k

On putting k = 0, 1, 2, 3,...., we get the times when the intensity of the resultant wave will be maximum:

                    t1 = 0, t2 = \[\frac{1}{(f1-f2)}\], t3 = \[\frac{2}{(f1-f2)}\] ....., tn = \[\frac{n}{(f1-f2)}\]

Now, the time interval between two successive maxima of sound will be:

                      tINTERVAL = t2   - t1 = \[\frac{1}{(f1-f2)}\] - 0 = \[\frac{1}{(f1-f2)}\]

We know that frequency = 1/tINTERVAL = (f1 - f2)...(a)


Frequency of Minima

The amplitude A of the resultant wave will be minimum when Cosπ(f1 - f2)t = 0.

∴ Cosπ(f1 - f2)t = Cos (2k + 1) \[\frac{π}{2}\], where k = 0, 1, 2, 3, 4,...

Now, we get:

π(f1 - f2)t = (2k + 1)\[\frac{π}{2}\]

⇒ t = \[\frac{(2k + 1)}{2(f1-f2)}\]….(4)

From eq (4), the times at which the intensity of the sound will be minimum:

 t1 = \[\frac{1}{2(f1-f2)}\], t2 = \[\frac{3}{2(f1-f2)}\] , t2 = \[\frac{5}{2(f1-f2)}\],..... tn = \[\frac{(2n+1)}{2(f1-f2)}\]


Now, finding out the time interval between two successive minima of sound, we have:

 t2 - t1 =  \[\frac{3}{2(f1-f2)}\] - \[\frac{1}{2(f1-f2)}\] = \[\frac{1}{(f1-f2)}\]

So, the frequency of minima = (f1 - f2)....(b)

Comparing values of t in maxima and maxima intensity shows that the maximum and minimum intensity of sound occur alternately.

Comparing eq (a) & (b), we get the frequency of beats as:   

fBEATS  = |f1 - f2|

FAQs (Frequently Asked Questions)

Question 1: Write the primary differences between Standing waves & Beats.

Answer: Two key differences are:

Standing Waves

Beats

Sound waves travel in the opposite direction.

Sound waves travel in the same direction.

Equal wavelength & amplitude.

Varying wavelength & amplitude.

f1 & f2 are same.

f1 & f2 are different.

Question 2: How do you find the frequency of unknown tuning Fork?

Answer: We can use the phenomenon of beats to determine unknown frequencies.

Let’s say, a tuning fork A has a known frequency of f1, and it produces m beats per second with another fork B, which has an unknown frequency of f2. We can find f2 by using the following formula:

f2  = f1 ±m

Here, the positive or negative sign indicates that m can be positive or negative.

Question 3: Write the real-life applications of Beats.

Answer: We use beats in the following ways:

  1. For tuning the musical instruments.

  2. In designing low-frequency oscillators.

  3. For ascertaining the existence of dangerous gases in mines.

Question 4: The two closed organ pipes of length 100 cm and 101 cm, respectively, give 16 beats in 20 seconds when each is sounded in its basic node, then the velocity of sound is?

Solution: Given l1 = 100 cm = 1.0 m

 l2 = 101 cm = 1.01 m

Number of beats, m = f1 - f2 

16/20 = v/(4/1) - v/(4/2)

= v/4 [1/1.0 - 1/1.01]

⇒ 16/20 = v x 0.01 / 4 x 1.01

On solving, we get:   

v = 332 m/s