When two sound waves travel in the same direction in a medium, the displacement of a particle located at \(\mathrm{x}\) at time \(\mathrm{t}\) is given by $\mathrm{y}_{1}=0.05 \cos (0.50 \mathrm{p} \mathrm{x}-100 \mathrm{pt}) \&$ \(y_{2}=0.05 \cos (0.46 p x-92 p t)\), where \(y_{1}, y_{2}\) and \(x\) are in meter and \(t\) is in seconds. How many times per second does an observer hear the sound of maximum intensity ? (A) 4 (B) 8 (C) 12 (D) 16

Constructive interference is a phenomenon in which two waves combine to form a wave with a greater amplitude (intensity) than either of the individual waves alone. This occurs when the crests (or troughs) of two waves align, causing the resulting wave's amplitude to increase. In this problem, we will find the frequency at which maximum constructive interference occurs.

According to the superposition principle, when two waves travel in the same direction, the displacement of a particle located at any point will be the algebraic sum of the displacements of the particle in each wave. This means that the resulting wave equation can be obtained by adding the two given wave equations: \(y_{r} = y_{1} + y_{2}\) Substitute the given wave equations: \(y_{r} = 0.05 \cos (0.50px - 100pt) + 0.05 \cos (0.46px - 92pt)\)

Since the maximum intensity occurs when the waves are in-phase, their phase difference (\(\Delta \phi = \phi_2 - \phi_1\)) would be a multiple of \(2 \pi\). Analyzing the phase difference: \(\Delta \phi = (0.46px - 92pt) - (0.50px - 100pt)\) \(\Delta \phi = 8pt - 0.04px\) To reach the maximum intensity, the phase difference must be a multiple of \(2 \pi\): \(8pt - 0.04px = 2n\pi\), where \(n\) is an integer. We can ignore the spatial \(x\) term for this problem since we are focused on the frequency in time. The interference pattern will repeat itself in time at each \(x\), it's the temporal frequency at which maximum intensity occurs that interests us. \(8pt = 2n\pi\) Now, we can solve for frequency (\(f\)) by isolating \(t\): \(t = \frac{2n\pi}{8p}\) \(t = \frac{n\pi}{4p}\) Since frequency is the reciprocal of time: \(f = \frac{4p}{n\pi}\) To find the minimum frequency at which maximum intensity occurs, we choose the lowest integer value of \(n\), which is 1 (excluding 0, since a frequency cannot be 0). \(f = \frac{4p}{\pi}\)

Now, substitute the given value for \(p\) and calculate the frequency: \(f = \frac{4(0.50)}{\pi}\) \(f = \frac{2}{\pi}\) Since we are asked for the frequency in terms of how many times per second, multiply by \(2\pi\): \(f = 2 \times \frac{2}{\pi} \times \pi\) \(f = 4\) Thus, the observer hears the sound of maximum intensity 4 times per second (answer: A).