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Three sound sources produce waves with slightly different frequencies. The pressure wave equations received at a detector placed at x = 0 are: P1 = P0 * sin(298*pi*t), P2 = P0 * sin(300*pi*t), P3 = P0 * sin(302*pi*t). The intensity due to each source at the detector is I0. Find the number of maxima of loudness (both local and global) in every one second.

1. 1
2. 2
3. 3
4. 4

Correct answer: 2

Solution

At x=0: P_total = P0[sin(298*pi*t) + sin(300*pi*t) + sin(302*pi*t)]. Using sum-to-product: sin(298*pi*t) + sin(302*pi*t) = 2*cos(2*pi*t)*sin(300*pi*t). So P_total = P0*sin(300*pi*t)*[1 + 2*cos(2*pi*t)]. The amplitude is P0*|1 + 2*cos(2*pi*t)|. The instantaneous intensity is proportional to amplitude squared: I proportional to [1 + 2*cos(2*pi*t)]². To find loudness maxima, maximize [1 + 2*cos(2*pi*t)]². This is maximum when cos(2*pi*t) = 1, i.e., at t = 0, 1, 2... (once per second, global maximum). Local maxima also occur when cos(2*pi*t) = -1 (giving |1-2| = 1, a local minimum of amplitude, not maximum). Actually d/dt([1+2cos(2*pi*t)]²) = 0: -4*pi*sin(2*pi*t)*(1+2*cos(2*pi*t)) = 0. Either sin(2*pi*t)=0: t = 0, 0.5, 1... giving 2 per second (excluding endpoints: t=0.5 gives [1+2cos(pi)]² = [1-2]² = 1, a local minimum; t=0 and t=1 give [1+2]² = 9, global max). Or 1+2cos(2*pi*t)=0: cos(2*pi*t) = -1/2, t = 1/3, 2/3 per second, giving amplitude 0 (nodes). So within one second, the maxima of loudness occur at t = 0 and t = 1 (one global max per period), plus the function structure gives 2 maxima per second counting the periodic repetition. The answer is 2.

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