Two harmonic waves moving in the same direction superimpose to form a wave $$x = a \cos(1.5t) \cos(50.5t)$$ where t is in seconds. Find the period with which they beat (close to nearest integer)

We are given that two harmonic waves moving in the same direction superimpose to form:

$$$x = a \cos(1.5t) \cos(50.5t)$$$

where $$t$$ is in seconds.

Step 1: Decompose using the product-to-sum identity

We use the trigonometric identity:

$$$\cos A \cos B = \frac{1}{2}[\cos(A - B) + \cos(A + B)]$$$

Applying this with $$A = 50.5t$$ and $$B = 1.5t$$:

$$$x = a \cos(1.5t) \cos(50.5t) = \frac{a}{2}[\cos(50.5t - 1.5t) + \cos(50.5t + 1.5t)]$$$ $$$x = \frac{a}{2}[\cos(49t) + \cos(52t)]$$$

Step 2: Identify the two component waves

The superposition consists of two harmonic waves with angular frequencies:

$$\omega_1 = 49 \text{ rad/s}$$ and $$\omega_2 = 52 \text{ rad/s}$$

Step 3: Find the beat frequency and beat period

The beat angular frequency is:

$$$\omega_{beat} = |\omega_2 - \omega_1| = |52 - 49| = 3 \text{ rad/s}$$$

The beat period is:

$$$T_{beat} = \frac{2\pi}{\omega_{beat}} = \frac{2\pi}{3}$$$ $$$T_{beat} = \frac{2 \times 3.14}{3} = \frac{6.28}{3} \approx 2.09 \text{ s}$$$

Rounding to the nearest integer: $$T_{beat} \approx 2 \text{ s}$$.

Hence, the correct answer is Option D.