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

Let the equations of the three sound waves be: $y_1 = A \sin 2\pi(n-1)t$ $y_2 = A \sin 2\pi nt$ $y_3 = A \sin 2\pi(n+1)t$ According to the principle of superposition, the resultant displacement is: $y = y_1 + y_2 + y_3$ $y = A [\sin 2\pi(n-1)t + \sin 2\pi(n+1)t] + A \sin 2\pi nt$ Using the trigonometric identity $\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$, we get: $y = A [2 \sin 2\pi nt \cos 2\pi t] + A \sin 2\pi nt$ $y = A(1 + 2 \cos 2\pi t) \sin 2\pi nt$ The resultant amplitude is $A_{res} = A(1 + 2 \cos 2\pi t)$. The intensity $I$ of the sound is proportional to the square of the amplitude: $I \propto A^2(1 + 2 \cos 2\pi t)^2$ To find the maxima (beats), we differentiate intensity with respect to time and equate to zero: $\frac{dI}{dt} = 0 \implies 2(1 + 2 \cos 2\pi t)(-2 \sin 2\pi t)(2\pi) = 0$ This gives $\sin 2\pi t = 0$ or $\cos 2\pi t = -1/2$. The condition $\cos 2\pi t = -1/2$ corresponds to $I = 0$ (minima). The condition $\sin 2\pi t = 0$ corresponds to maxima. This occurs at $t = 0, 0.5, 1, 1.5, \dots$ seconds. At $t = 0, 1, 2 \dots$, $\cos 2\pi t = 1 \implies I \propto 9A^2$ (Primary maxima). At $t = 0.5, 1.5 \dots$, $\cos 2\pi t = -1 \implies I \propto A^2$ (Secondary maxima). Since there are two maxima (one primary and one secondary) in a time interval of $1$ second, the number of beats heard per second is $2$. Alternatively, the number of beats is simply given by the maximum difference in frequencies present in the mixture: $(n+1) - (n-1) = 2$.