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Beats in sound waves | Beats class 11

Beats in sound waves | Beats class 11 :- Beats are fluctuations in amplitude(or intensity) produced by superposition of two sound waves of same amplitude and slightly different frequencies travelling in same direction. For example, when two tuning forks with slightly different frequencies are sounded together, then alternate loud and low sounds are heard. These variations in loudness are called beats.

Consider a particular point in space where the two waves of frequencies 16 Hz (red graph) and 18 Hz (blue graph) overlap[Figure (a)]. The displacements of the individual waves are plotted as functions of time. The total length of the time axis represents 1 second. Applying the principle of superposition, we get the resultant wave[Figure (b)].

In above figure(b), at certain times the two waves are in phase(like t = 0.50 sec.); their maxima coincide and their amplitudes add. But because of their slightly different frequencies, the two waves cannot be in phase at all times. Indeed, at certain times (like t = 0.75 sec.) the two waves are exactly out of phase. The two waves then cancel each other, and the total amplitude is zero.

The resultant wave in Figure(b) looks like a single sinusoidal wave with a varying amplitude that goes from a maximum to zero and back. There are two beats contained in a time interval of 1 sec., so beat frequency is 2 Hz. Hence the beat frequency is the difference of frequencies of the two waves.

Mathematical analysis

(Beats in sound waves | Beats class 11)

Let the two waves are :

$\displaystyle {{y}_{1}}=A\sin (2\pi {{f}_{1}}t)$

and

$\displaystyle {{y}_{2}}=A\sin (2\pi {{f}_{2}}t)$

According to Principle of superposition, y = y₁ + y₂

$\displaystyle \Rightarrow y=A[\sin (2\pi {{f}_{1}}t)+\sin (2\pi {{f}_{2}}t)]$

$\displaystyle \Rightarrow y=\left[ 2A\cos \left\{ 2\pi \left( \frac{{{f}_{1}}-{{f}_{2}}}{2} \right)t \right\} \right]\sin \left\{ 2\pi \left( \frac{{{f}_{1}}+{{f}_{2}}}{2} \right)t \right\}..........(1)$

$\displaystyle \Rightarrow y=R\sin (2\pi ft)..........(2)$

where,

$\displaystyle R=2A\cos \left\{ 2\pi \left( \frac{{{f}_{1}}-{{f}_{2}}}{2} \right)t \right\}..........(3)$

and

$\displaystyle f=\left( \frac{{{f}_{1}}+{{f}_{2}}}{2} \right)..........(4)$

Now we know that Intensity ∝ (Amplitude)²

⇒ I ∝ R²

$\displaystyle \Rightarrow I=k4{{A}^{2}}{{\cos }^{2}}\left[ 2\pi \left( \frac{{{f}_{1}}-{{f}_{2}}}{2} \right)t \right]..........(5)$

CONDITION FOR MAXIMA

For maximum intensity,

$\displaystyle {{\cos }^{2}}\left[ 2\pi \left( \frac{{{f}_{1}}-{{f}_{2}}}{2} \right)t \right]=1$

$\displaystyle \Rightarrow \cos \left[ 2\pi \left( \frac{{{f}_{1}}-{{f}_{2}}}{2} \right)t \right]=\pm 1$

$\displaystyle \Rightarrow 2\pi \left( \frac{{{f}_{1}}-{{f}_{2}}}{2} \right)t=n\pi ;\text{ }n=0,1,2,3,...$

$\displaystyle \Rightarrow t=\frac{n}{\left| {{f}_{1}}-{{f}_{2}} \right|};\text{ }n=0,1,2,3,...$

So, MAXIMA are obtained at following instants of time,

$\displaystyle t=0,\text{ }\frac{1}{\left| {{f}_{1}}-{{f}_{2}} \right|}\text{, }\frac{2}{\left| {{f}_{1}}-{{f}_{2}} \right|}\text{, }\frac{3}{\left| {{f}_{1}}-{{f}_{2}} \right|},...$

CONDITION FOR MINIMA

For minimum intensity,

$\displaystyle {{\cos }^{2}}\left[ 2\pi \left( \frac{{{f}_{1}}-{{f}_{2}}}{2} \right)t \right]=0$

$\displaystyle \Rightarrow \cos \left[ 2\pi \left( \frac{{{f}_{1}}-{{f}_{2}}}{2} \right)t \right]=0$

$\displaystyle \Rightarrow 2\pi \left( \frac{{{f}_{1}}-{{f}_{2}}}{2} \right)t=(2n+1)\frac{\pi }{2};\text{ }n=0,1,2,3,...$

$\displaystyle \Rightarrow t=\frac{(2n+1)}{2\left| {{f}_{1}}-{{f}_{2}} \right|};\text{ }n=0,1,2,3,...$

So, MINIMA are obtained at following instants of time,

$\displaystyle t=\text{ }\frac{1/2}{\left| {{f}_{1}}-{{f}_{2}} \right|}\text{, }\frac{3/2}{\left| {{f}_{1}}-{{f}_{2}} \right|}\text{, }\frac{5/2}{\left| {{f}_{1}}-{{f}_{2}} \right|},...$

Beat Period (T_b) and Beat Frequency (f_b)

(Beats in sound waves | Beats class 11)

Beat Period (T_b) :- The time interval between two successive maxima or minima is called beat time. So

$\displaystyle {{T}_{b}}=\frac{1}{\left| {{f}_{1}}-{{f}_{2}} \right|}$

Beat Frequency (f_b) :- The reciprocal of beat time, gives us the beat frequency. Hence

$\displaystyle {{f}_{b}}=\frac{1}{{{T}_{b}}}=\left| {{f}_{1}}-{{f}_{2}} \right|$

Alternate method to find beat frequency

Suppose two waves of frequencies f₁ and f₂ (<f₁) are meeting at some point in space. The corresponding periods are T₁ and T₂ (>T₁). If the two waves are in phase at t=0, they will again be in phase when the first wave has gone through exactly one more cycle than the second. This will happen at a time t = T_b, the Beat Period (T_b). Let n be the number of cycles of the first wave in time T_b, then the number of cycles of the second wave in the same time is
(n–1). Hence,

$\displaystyle {{T}_{b}}=n{{T}_{1}}=(n-1){{T}_{2}}$

Eliminating n between these two equations, we find

$\displaystyle {{T}_{b}}=\frac{{{T}_{1}}{{T}_{2}}}{{{T}_{2}}-{{T}_{1}}}$

So beat frequency,

$\displaystyle {{f}_{b}}=\frac{1}{{{T}_{b}}}=\frac{1}{\frac{{{T}_{1}}{{T}_{2}}}{{{T}_{2}}-{{T}_{1}}}}=\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}}=\frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}}=\left| {{f}_{1}}-{{f}_{2}} \right|$

Note :-

The resultant wave [equation(2)] is also a harmonic wave with a frequency $\displaystyle f=\frac{{{f}_{1}}+{{f}_{2}}}{2}$ . You can also check the resultant wave in the above figure (b), that in a time interval of 1 sec., there are 17 waves $\displaystyle \left[ \frac{18+16}{2}=17 \right]$ .
The amplitude R is not constant but varies harmonically with a frequency $\displaystyle f=\frac{{{f}_{1}}-{{f}_{2}}}{2}$ . In the resultant wave in the above figure (b) if we trace the amplitude variation, we get a wave of frequency 1 Hz $\displaystyle \left[ \frac{18-16}{2}=1Hz \right]$ , as shown in figure below :

However the number of beats per second[Beat Frequency (f_b)] is twice of this frequency $\displaystyle \left[ \frac{{{f}_{1}}-{{f}_{2}}}{2} \right]$ , i.e., [f₁ – f₂], because human ear perceives the magnitude variations and not its sign(±).

Condition for hearing beats

(Beats in sound waves | Beats class 11)

The sensation of hearing of any sound remains in human brain for 0.1 second. So we can distinguish between two sounds if time interval between them is more than 0.1 seconds or frequency difference between them is less than 10 Hz.( ν = 1/T = 1/0.1 = 10 Hz) So if the frequency difference between two sound waves is more than 10 Hz then it becomes difficult for human ear to distinguish between rise and fall in the intensity of sound. Hence Beats can not be heard.

Difference between standing waves and beats

(Beats in sound waves | Beats class 11)

Standing waves are formed by the superposition of two waves of same amplitude and frequency travelling in opposite directions, but beats are formed by the superposition of two waves of same amplitude but slightly different frequencies travelling in same directions.

Applications of beats

(Beats in sound waves | Beats class 11)

Musicians use the phenomenon of beat in turning various instruments. When an instrument is tuned against a standard frequency and till the beat has not disappeared it is tuned, then the instrument is in a tune with that standard frequency. Beats are also used to determine an unknown frequency.

At the end of this blog post if you want to hear and see the production of beats by superposition of two sound waves of same amplitude and slightly different frequencies travelling in same direction, then see the video given below:-

Beats in sound waves | Beats class 11

Mathematical analysis

(Beats in sound waves | Beats class 11)

Beat Period (Tb) and Beat Frequency (fb)

(Beats in sound waves | Beats class 11)

Condition for hearing beats

(Beats in sound waves | Beats class 11)

Difference between standing waves and beats

(Beats in sound waves | Beats class 11)

Applications of beats

(Beats in sound waves | Beats class 11)

Like this:

Beat Period (T_b) and Beat Frequency (f_b)