# Beats

## What are Beats?

We all like to hear the melody 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.

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?

### 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 a beat is and how it looks.

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

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.

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

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

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 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 wave 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.

### Interference and Beats

When two or more waves travelling in a medium meet, the resulting phenomenon is called interference and beats are an excellent example of the phenomenon of interference.

### Waxing and Waning of the Sound

Supposing, the two waves at any given medium having a similar amplitude and slightly varying frequencies travel in the same direction and reach a given point, interfere. The resultant intensity of the sound, the maximum sound is called ‘waxing’ and the minimum sound is termed to be ‘waning’. The simultaneous waxing and waning are defined as beats.

### 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 super position principle, we get:

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

Using the formula: SinA + SinB = 2Cos$\frac{1}{2}$(A−B). Sin$\frac{1}{2}$(A+B)

, 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 = $\frac{1}{(f_1 − f_2)}$, t3 = $\frac{2}{(f_1− f_2)}$....., tn = $\frac{n}{(f_1− f_2)}$

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

tINTERVAL = t2   - t1 = $\frac{1}{(f_1 − f_2)}$ - 0 = $\frac{1}{(f_1 − f_2)}$

We know that frequency =   $\frac{1}{t_\text{INTERVAL} }$ .....(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). $\frac{\pi}{2}$

, where k = 0, 1, 2, 3, 4,...

Now, we get:

π(f1 - f2)t = (2k + 1).$\frac{\pi}{2}$

⇒ t = $\frac{(2k+1)}{2(f_1 − f_2)}$.….(4)

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

t1 = $\frac{1}{2(f_1 − f_2)}$, t2 = $\frac{3}{2(f_1 − f_2)}$, t3 = $\frac{5}{2(f_1 − f_2)}$,..... tn = $\frac{2n+1}{2(f_1 − f_2)}$

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

t2 - t1 = $\frac{3}{2(f_1 − f_2)}$ - $\frac{1}{2(f_1 − f_2)}$ = $\frac{1}{(f_1 − f_2)}$

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|

### The Phenomenon of Superimposition

When a wave of varying amplitude and frequency interferes with another wave, then the resulting interference is called superimposition. This phenomenon is widely used in the making of various sounds as well as in the music industry.

### What is Interference?

When two or more waves travelling in a medium meet, the resulting phenomenon is called interference and beats are an excellent example of the phenomenon of interference.

### The Application of Beats

There are various applications of beats, though the three are discussed in detail below.

• Beats are used in determining the unknown frequency

• Beats are used in determining the existence of poisonous gases in mines.

To identify the presence of poisonous gases in the mine, the following experiment is carried out. Two pipes equal in size are taken, and one is filled with the pure ai. The other pipe is filled with the air in the mine. The pipes are blown together. The pipe that consists of pure air does not produce any sound whereas if some amount of sound is produced by the pipe that consists of mine air, then it indicates the presence of poisonous gases. If the beats are produced, then the mine air is not pure.

• Beats are used in determining the tone of musical instruments.

Two musical instruments are sounded together to notice if two instruments vibrate in a similar tone. If the tones of both the instruments are similar, then no beats will be heard. In such a way, the tones of various musical instruments can be adjusted according to the beats.

• Beats are used to adjust the vibrating length between two bridges in the sonometer experiment.

• Based on the phenomenon of the beats, the doppler ultrasonography and echocardiography works.

• The aeroplane speed can be determined by the Doppler RADAR. This is based on the phenomenon of beats.

### The Salient Features of Sound and Noise

The features of the sound:

• The musical sounds are pleasant as well as comfortable to hear.

• The sound is created by regular and periodic vibrations.

• It has its quality.

The features of the noise.

• The noise is unpleasant to hear and is irritating.

• The noise is created by the irregular vibrations of the source.

• The noise has no quality.

### Sound and Its Applications

The vibration that propagates as an acoustic wave in a transmission medium like gas, liquid or solid is defined as sound.

There are various applications of sound in day-to-day life. Some are listed below:

• With sound, One can communicate and hear other human beings, animals and insects. Sound is our primary mode of communication. The ability to communicate by listening to the sound is a main application of the sound.

• Sound is used in measuring the flying objects and the lightning.

• It is seen that the bats use ultrasound waves to sense.

• The musical sound and rhythmic sound are pleasant to hear and are often used as the mode of entertainment.

• The sonar is used in sound navigation and ranging on the sea by the ships.

• In the medical field, ultrasound waves are used for testing. These ultrasound waves are generally of higher frequency than the normal sound waves.

• As and when needed, sound can be stored and used later on.

### Types of Sound

The sounds are of various kinds. Although they are mainly categorized as infrasound waves and ultrasound waves.

### Infrasound Waves

The frequency of this kind of wave is below 20hz which makes it inaudible for the human ear to hear. The infrasound waves are applied in determining volcanic eruptions and earthquakes. It has various other applications as well like studying the human heart. Plenty of animals use infrasound waves as a mode of communication.

### Ultrasound Waves

The frequency of this kind of wave is above 20hz which makes it inaudible for the human ear to hear. The ultrasound waves are used in the field of medicine for sonograms. Some other applications of ultrasound include imaging, navigation, testing, and sample mixing.

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

 Standing Waves Beats Sound waves travel in opposite directions. Sound waves travel in the same direction. Equal wavelength & amplitude. Varying wavelength and amplitude. f1 & f2 are same. f1 & f2 are different.

2. How do you find the frequency of unknown tuning forks?

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.

3. Write the real-life applications of Beats.

We use beats in the following ways:

• For tuning the musical instruments.

• In designing low-frequency oscillators.

• For ascertaining the existence of dangerous gases in mines.

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?

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

1/1.0−1/1.01

⇒ 16/20 = v x 0.01 / 4 x 1.01

On solving, we get:

v = 332 m/s

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