Spread the love

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, as discussed below :

🎵 Finding the Unknown Frequency of a Tuning Fork Using Beats

When two tuning forks of slightly different frequencies are sounded together, the sound alternately becomes loud and soft at regular intervals. This periodic variation in intensity is called a beat. Suppose :

Frequency of known tuning fork =
Frequency of unknown tuning fork =

Strike both tuning forks simultaneously and count the number of beats per second. If the number of beats per second is , then unknown frequency can be :

$\displaystyle {{f}_{x}}=f+n$

$\displaystyle {{f}_{x}}=f-n$

To determine whether the unknown frequency is higher or lower than the known one :

Load the unknown tuning fork slightly with wax.
Adding wax decreases its frequency.
Observe what happens to the beat frequency.

If beats increase → unknown frequency was originally lower (f_x = f – n).
If beats decrease → unknown frequency was originally higher (f_x = f + n).

For example if a 256 Hz tuning fork is sounded with an unknown fork and 4 beats per second are heard, then :

f_x = 256 ± 4

f_x = 260 Hz or f_x = 252 Hz

After loading the unknown fork :

If beats reduce → correct frequency (f_x) = 260 Hz
If beats increase → correct frequency (f_x) = 252 Hz

Example 1.

A tuning fork having f = 300 Hz produces 5 beats/s with another tuning fork. If impurity (wax) is added on the arm of known tuning fork, the number of beats decreases then calculate the frequency of unknown tuning fork.

Solution :

f_x = (300 ± 5) Hz

f_x = 305 Hz or f_x = 295 Hz

Wax added to known fork
→ Frequency of known fork decreases

Beats decrease
→ Frequencies come closer

If unknown were 305 Hz :
300 ↓ becomes 299
→ Difference becomes 6
→ Beats increase ❌

If unknown were 295 Hz :
300 ↓ becomes 299
→ Difference becomes 4
→ Beats decrease ✅

Hence

⇒ f_x = 295 Hz

Example 2.

To tuning forks A and B produce 4 beats per second when sounded simultaneously. The frequency of A is known to be to 256 Hz. When B is loaded with a little wax, 4 beats per second are again produced. Find the frequency of B before and after loading.

Solution :

Frequency of fork A :

f_A= 256 Hz

⇒ f_B = (256 ± 4) Hz

⇒ f_B = 252 Hz Or f_B = 252 Hz

After loading B with wax, again 4 beats/sec are heard. Therefore,

⇒ f_B = 252 Hz Or f_B = 252 Hz

But by loading a fork its frequency can only decrease. Hence, before loading fork B, its frequency can only be 260 Hz, so that after loading it decreases to 252 Hz.

🔬 Curio Capsules : Finding Unknown Frequency by Filing

“Filing a tuning fork” means carefully rubbing a file on the prongs (tines) of a tuning fork to remove a small amount of metal in order to change its frequency. The frequency of a tuning fork is approximately :

$\displaystyle f\propto \sqrt{\frac{k}{m}}$

Where :

= stiffness
$m$ = effective mass of vibrating prongs

If mass decreases → increases. That’s why filing raises the pitch.

Procedure

(1). Sound standard fork f and unknown fork together. If beat frequency = n, then

f_x = f ± n

(2). File the unknown fork slightly (frequency increases) and Observe beats again :

If beats decrease → originally f_x

beats increase → originally f_x

⚠️ Filing must be done carefully because :

It permanently increases frequency.
Excess filing can damage the tuning fork.

Example 3.

A tuning fork of frequency 158 Hz produces 3 beats per second with another tuning fork. When the prongs of the unknown tuning fork are filed, the beat frequency becomes 7 per second. Find the frequency of the unknown tuning fork.

Solution :

Known frequency :

f= 158 Hz

f_x = (158 ± 3) Hz

f_x = 161 Hz or f_x = 155 Hz

After filing the prongs of unknown tuning fork, beat frequency becomes 7 beats/sec.

f_x = (158 ± 7) Hz

f_x = 165 Hz or f_x = 151 Hz

Now, filing a tuning fork increases its frequency. let us check both initial possibilities :

If initial f_x = 155 Hz
After filing, frequency increases → it can become 165 Hz
(Difference from 158 becomes 7 ✔)
If initial f_x = 161 Hz
After filing, frequency increases → it can again become 165 Hz
(Difference from 158 becomes 7 ✔)

In both cases, the new frequency (165 Hz) can be obtained by filing. Hence, we cannot decide uniquely. So both are possible.

f_x = 155 Hz and f_x = 161 Hz

Example 4.

The wavelengths of two notes in air are (90/175) m and (90/173) m respectively. Each of these notes produces 4 beats per second with a third note of fixed frequency. Calculate the velocity of sound in air.

Solution :

Given :

$\displaystyle {{\lambda }_{1}}=\frac{90}{175}m$ and $\displaystyle {{\lambda }_{2}}=\frac{90}{173}m$

Frequencies :

$\displaystyle {{f}_{1}}=\frac{\upsilon }{{{\lambda }_{1}}}=\frac{\upsilon }{{}^{90}\!\!\diagup\!\!{}_{175}\;}=\frac{175\upsilon }{90}$

$\displaystyle {{f}_{2}}=\frac{\upsilon }{{{\lambda }_{2}}}=\frac{\upsilon }{{}^{90}\!\!\diagup\!\!{}_{173}\;}=\frac{173\upsilon }{90}$

Each produces 4 beats/sec with a third note of frequency Hence,

$\displaystyle {{f}_{1}}-f=4$ and $\displaystyle f-{{f}_{2}}=4$

$\displaystyle \Rightarrow {{f}_{1}}-{{f}_{2}}=8$

$\displaystyle \Rightarrow \frac{175\upsilon }{90}-\frac{173\upsilon }{90}=8$

$\displaystyle \Rightarrow \upsilon =360m/s$

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)