Beats phenomenon — how two waves of slightly different frequency create beats

easy CBSE JEE-MAIN NEET NCERT Class 11 3 min read
Tags Waves

Question

Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. How many beats per second will be heard? Explain the phenomenon of beats mathematically.

(NCERT Class 11, Chapter 15)


Solution — Step by Step

When two sound waves of slightly different frequencies f1f_1 and f2f_2 superpose, the resultant intensity fluctuates — rising and falling periodically. Each cycle of rising and falling is called a beat. The number of beats per second equals the difference in frequencies.

Let the two waves be: y1=Asin(2πf1t)y_1 = A\sin(2\pi f_1 t) and y2=Asin(2πf2t)y_2 = A\sin(2\pi f_2 t).

By superposition: y=y1+y2y = y_1 + y_2.

Using sinC+sinD=2cos(CD2)sin(C+D2)\sin C + \sin D = 2\cos\left(\frac{C-D}{2}\right)\sin\left(\frac{C+D}{2}\right):

y=2Acos(2πf1f22t)sin(2πf1+f22t)y = 2A\cos\left(2\pi \cdot \frac{f_1 - f_2}{2} \cdot t\right) \sin\left(2\pi \cdot \frac{f_1 + f_2}{2} \cdot t\right)

The resultant is a wave of average frequency f1+f22\frac{f_1 + f_2}{2} whose amplitude varies at frequency f1f22\frac{|f_1 - f_2|}{2}.

The amplitude term 2Acos(π(f1f2)t)2A\cos\left(\pi(f_1 - f_2)t\right) completes one full cycle in time 1f1f2\frac{1}{|f_1 - f_2|} seconds.

But the intensity (proportional to amplitude squared) has maxima twice per cycle of the amplitude variation. So:

Beat frequency=f1f2\text{Beat frequency} = |f_1 - f_2|

For our problem: fbeat=260256=4 beats per secondf_{beat} = |260 - 256| = \mathbf{4 \text{ beats per second}}.


Why This Works

Beats arise from the constructive and destructive interference of the two waves. When the waves are in phase (crests align), the sound is loud. When they are out of phase (crest meets trough), the sound is soft. This alternation happens f1f2|f_1 - f_2| times per second.

The human ear can perceive beats when the frequency difference is up to about 7 Hz. Beyond that, the fluctuations are too rapid to hear distinctly.


Alternative Method — Practical Application in Tuning

Musicians use beats to tune instruments. They play two strings together — if beats are heard, the strings are slightly out of tune. They adjust tension until the beats disappear (frequencies match).

A classic NEET question: “A tuning fork of unknown frequency produces 5 beats/s with a fork of 512 Hz. When the unknown fork is loaded with wax, the beat frequency becomes 3 beats/s. Find the unknown frequency.” Loading with wax decreases the frequency. Since beats decreased from 5 to 3, the unknown was above 512 Hz. Answer: 517 Hz (now it’s closer to 512 after loading).


Common Mistake

Students sometimes write beat frequency = f1+f2f_1 + f_2 instead of f1f2|f_1 - f_2|. Beats arise from the difference, not the sum. Also, beat frequency is always positive — use the absolute value. A quick check: if both forks have the same frequency, there should be zero beats (perfect harmony), and f1f2=0|f_1 - f_2| = 0 confirms this.

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