Beats

"Beats," a concept often explored in physics, particularly in acoustics, refers to the periodic variation in sound intensity caused by the interference of two waves of slightly different frequencies. Imagine listening to two musical instruments that are almost, but not perfectly, in tune. As they play together, you might notice a fluctuating rhythm—a pulsing or throbbing sound. This phenomenon is the beat, an audible representation of the constructive and destructive interference of sound waves. In real life, beats can be likened to the ebb and flow of conversations or the harmony and discord in relationships, where differences in perspectives or emotions can lead to moments of resonance or tension. Just as beats can be a tool for tuning instruments, understanding and navigating these fluctuations in our interactions can lead to a more balanced and harmonious life. In this article, we will discuss the concept of Beats with solved examples.

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Beats

When any two sound waves of slightly different frequencies, travel along the same direction in a medium and superimpose on each other then the intensity of the resultant sound at a particular position rises and falls regularly with time. This phenomenon of regular variation in the intensity of sound with time at a particular position is called beats. If we struck two tuning forks of slightly different frequencies, one hears a sound of periodically varying amplitude. This phenomenon is called beating. Beat frequency equals the difference in frequency between the two sources which we will see below.

Let us consider two sound waves travelling through a medium having equal amplitude with slightly different frequencies $f_1$ and $f_2$. We use equations similar to equation $y=A\sin (kx-\omega t)$ to represent the wave functions for these two waves at a point such that $kx=\pi /2$

$\begin{array}{rl}& y_1=A\sin (\frac{\pi }{2}-{\omega }_{2}t)=A\cos \left(2\pi f_{1}t\right)\\ & y_2=A\sin (\frac{\pi }{2}-{\omega }_{2}t)=A\cos \left(2\pi f_{2}t\right)\end{array}$

By using the superposition principle

$y=y_1+y_2=A(\cos 2\pi f_1+\cos 2\pi f_{2}t)$

We can also write the above equation by using trigonometric identity as

$y=[2A\cos 2\pi \left(\frac{f_1-f_2}{2}\right)t]\cos 2\pi \left(\frac{f_1+f_2}{2}\right)t$

The graph is like this

Graphs of the individual waves and the resultant wave are shown in the figure. We can see that the resultant wave has an effective frequency equal to the average frequency $\frac{f_1+f_2}{2}$. From the figure, we can see that this wave is multiplied by the envelope whose equation is given as

$y_{\text{envelope}}=2A\cos 2\pi \left(\frac{f_1-f_2}{2}\right)t$

A maximum in the amplitude of the resultant sound wave is detected whenever
$\cos 2\pi \left(\frac{f_1-f_2}{2}\right)t=\pm 1$

Hence, there are two maxima in each period of the envelope wave. Because the amplitude varies with frequency as $\frac{(f_1-f_2)}{2}$ the beat frequency is two times this value and given by

$f_{\text{beat}}=|f_1-f_2|$

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Solved Examples Based on Beats

Example 1: A source of sound S of frequency 500 Hz situated between a stationary observer O and a wall, moves towards the wall with a speed of 2 m/s. If the velocity of sound is 332 m/s, then the number of beats per second heard by the observer is (approximately)

1) 8 Hz

2) 6 Hz

3) 4 Hz

4) 2 Hz

Solution:

$\begin{array}{rl}{n}_1=n\left(\frac{v}{v+v_s}\right)& =500\left(\frac{332}{332+2}\right)\\ & =500\left(\frac{332}{334}\right)=497\mathrm{Hz}\end{array}$

For sound reflected from the wall-
$\begin{array}{rl}& n_2=n\left(\frac{v}{v-v_5}\right)\\ & \begin{array}{rl}{n}_2=500\left(\frac{3.32}{332-2}\right)& =503\mathrm{Hz}\\ =n_2-h_1& =503-457\\ \text{Beat frequency}& =6\mathrm{HZ}\end{array}\end{array}$

Hence, the answer is the option (2).

Example 2: A tuning fork A of unknown frequency produces 5 beats /with a fork of known frequency 340Hz. When fork A is filled, the beat frequency decreases to 2 beats /s. What is the frequency of fork A?

1) 338 Hz

2) 345 Hz

3) 335 Hz

4) 342 Hz

Solution:

Initially beat frequency $=5\mathrm{Hz}$
so, ${\rho }_{\mathrm{A}}=340\pm 5=345\mathrm{Hz}$, or 335 Hz
so, new value of frequency of $\mathrm{A}>{\rho }_{\mathrm{A}}$
Now, beat frequency $=2\mathrm{Hz}$
$\Rightarrow$ new ${\rho }_{\mathrm{A}}=340\pm 2=342\mathrm{Hz}$, or 338 Hz
hence, original frequency of A is ${\rho }_{\mathrm{A}}=335\mathrm{Hz}$

Hence, the answer is the option (3).

Example 3: The velocity of sound in a gas, in which two wavelengths 4.08 m and 4.16 m produce 40 beats in 12 s will be :

1) $282.8{\mathrm{ms}}^{-1}$
2) $175.5{\mathrm{ms}}^{-1}$
3) $353.6{\mathrm{ms}}^{-1}$
4) $707.2{\mathrm{ms}}^{-1}$

Solution:

$\begin{array}{rl}& \text{beat frequeny}\left({\mathrm{f}}_{\mathrm{b}}\right)=|{\mathrm{f}}_1-{\mathrm{f}}_2|\\ & \frac{40}{12}=\frac{\mathrm{v}}{4.08}-\frac{\mathrm{v}}{4.16}\\ & \mathrm{v}=707.2\mathrm{m}/\mathrm{s}\end{array}$

Hence, the answer is the option (4).

Example 4: Two identical strings are stretched at tension. A turning fork is used to set them in vibration. A vibrates in its fundamental mode and in its second harmonic mode then -

$\begin{array}{rl}& \text{1)}{\mathrm{T}}_{\mathrm{A}}=2{\mathrm{T}}_{\mathrm{B}}\\ & \text{2)}{\mathrm{T}}_{\mathrm{A}}=4{\mathrm{T}}_{\mathrm{B}}\\ & \text{3)}2{\mathrm{T}}_{\mathrm{A}}={\mathrm{T}}_{\mathrm{B}}\\ & \text{4)}4{\mathrm{T}}_{\mathrm{A}}=4{\mathrm{T}}_{\mathrm{B}}\end{array}$

Solution:

$\begin{array}{rl}& \text{Use.}\frac{{\mathrm{V}}_{\mathrm{A}}}{2\mathrm{l}}=2\left(\frac{{\mathrm{V}}_{\mathrm{B}}}{2\mathrm{l}}\right)\\ & {\mathrm{V}}_{\mathrm{A}}=2{\mathrm{V}}_{\mathrm{B}}\text{but}\mathrm{v}\alpha \sqrt{\mathrm{T}}[\because \mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mathrm{v}}}]\\ & \therefore {\mathrm{T}}_{\mathrm{A}}=4{\mathrm{T}}_{\mathrm{B}}\end{array}$

Hence, the answer is the option (2).

Example 5: The frequency of tuning forks A and B are respectively 3% more and 2% less than the frequency of tuning fork C. When A and B are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork A (in Hz) is

1) 98

2) 100

3) 103

4) 105

Solution:

Let n be the frequency of fork C, then

$\begin{array}{rl}& {\mathrm{n}}_{\mathrm{A}}=\mathrm{n}+\frac{3\mathrm{n}}{100}=\frac{103\mathrm{n}}{100}\\ & \text{and}{\mathrm{n}}_{\mathrm{B}}+\mathrm{n}-\frac{2\mathrm{n}}{100}=\frac{98\mathrm{n}}{100}\\ & n_A-n_B=5\\ & \frac{5n}{100}=5\\ & n=100\mathrm{Hz}.\\ & n_A=\frac{103\times 100}{100}=103\mathrm{Hz}.\end{array}$

Hence, the answer is the option (3).

Summary

Beats occur when two sound waves of nearly identical frequencies interact. Interference repeats over time according to the regular fluctuation of the two waves, which provides alternate constructive and destructive interference of the resulting amplitude, hence changing the loudness of the sound. This repetition rate for loudness variation is called the beat frequency, and it is equal in value to the absolute difference of the two frequencies. One especially useful application of beats is in tuning musical instruments, where it is much easier to slightly match pitches with similar frequencies by adjusting the instrument until the beat frequency is reduced to zero.