Answer: 3. 2 Add the three waves of equal amplitude and see how the intensity repeats. y=A[cos2π(n-1)t+cos2π nt+cos2π(n+1)t]=Acos2π nt (1+2cos2π t) . The resultant amplitude is |1+2cos2π t| , so intensity ∝(1+2cos2π t)² . Over one second: At t=0 the value is 3²=9 , a big maximum. It falls to 0 at t=1/3 , rises to a small maximum of 1 at t=1/2 , falls to 0 at t=2/3 , and returns to 9 at t=1 . So there are 2 intensity maxima per second, i.e. 2 beats per second. Option (3).

Topic: Beats· Oscillations and Waves

Three sound waves of equal amplitudes have frequencies (n-1), n, (n+1). They superimpose to give beats. The number of beats produced per second will be

1. 4

2. 3

3. 2

4. 1

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Answer: 3. 2

Add the three waves of equal amplitude and see how the intensity repeats.

\(y=A[\cos2\pi(n-1)t+\cos2\pi nt+\cos2\pi(n+1)t]=A\cos2\pi nt\,(1+2\cos2\pi t)\).

The resultant amplitude is \(|1+2\cos2\pi t|\), so intensity \(\propto(1+2\cos2\pi t)^2\). Over one second:

At \(t=0\) the value is \(3^2=9\), a big maximum.
It falls to 0 at \(t=\tfrac13\), rises to a small maximum of \(1\) at \(t=\tfrac12\), falls to 0 at \(t=\tfrac23\), and returns to 9 at \(t=1\).

So there are 2 intensity maxima per second, i.e. 2 beats per second. Option (3).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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Three sound waves of equal amplitudes have frequencies (n-1), n, (n+1). They superimpose to give beats. The number of beats produced per second will be

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