When two sound waves travel in the same direction in a medium, the displacements of a particle located at 'x' at time 't' is given by :
$$y_1 = 0.05\cos(0.50\pi x - 100\pi t)$$
$$y_2 = 0.05\cos(0.46\pi x - 92\pi t)$$
where $$y_1$$, $$y_2$$ and x are in meters and t in seconds. The speed of sound in the medium is :

The displacements of the particle due to the two sound waves are given by:

$$y_1 = 0.05\cos(0.50\pi x - 100\pi t)$$

$$y_2 = 0.05\cos(0.46\pi x - 92\pi t)$$

where $$y_1$$, $$y_2$$, and $$x$$ are in meters, and $$t$$ is in seconds. We need to find the speed of sound in the medium.

A general wave equation traveling in the positive x-direction is written as $$y = A \cos(kx - \omega t)$$, where $$k$$ is the wave number and $$\omega$$ is the angular frequency. The wave number $$k$$ is related to the wavelength $$\lambda$$ by $$k = \frac{2\pi}{\lambda}$$, and the angular frequency $$\omega$$ is related to the frequency $$f$$ by $$\omega = 2\pi f$$. The speed of sound $$v$$ is given by the formula $$v = f \lambda$$. Substituting the relations, we get:

$$v = f \lambda = \left( \frac{\omega}{2\pi} \right) \times \left( \frac{2\pi}{k} \right) = \frac{\omega}{k}$$

Therefore, the speed can be directly calculated as $$v = \frac{\omega}{k}$$ for each wave. Since both waves are traveling in the same medium, the speed should be identical for both.

For the first wave $$y_1$$, the argument is $$0.50\pi x - 100\pi t$$. Comparing with the general form, we have:

Wave number $$k_1 = 0.50\pi$$ rad/m,

Angular frequency $$\omega_1 = 100\pi$$ rad/s.

Thus, the speed $$v_1 = \frac{\omega_1}{k_1} = \frac{100\pi}{0.50\pi}$$.

Simplifying, the $$\pi$$ cancels out:

$$v_1 = \frac{100}{0.50} = \frac{100}{\frac{1}{2}} = 100 \times 2 = 200 \text{ m/s}$$

For the second wave $$y_2$$, the argument is $$0.46\pi x - 92\pi t$$. Comparing with the general form, we have:

Wave number $$k_2 = 0.46\pi$$ rad/m,

Angular frequency $$\omega_2 = 92\pi$$ rad/s.

Thus, the speed $$v_2 = \frac{\omega_2}{k_2} = \frac{92\pi}{0.46\pi}$$.

Simplifying, the $$\pi$$ cancels out:

$$v_2 = \frac{92}{0.46} = \frac{92}{\frac{46}{100}} = 92 \times \frac{100}{46} = \frac{92 \times 100}{46}$$

Dividing 92 by 46:

$$\frac{92}{46} = 2, \quad \text{so} \quad 2 \times 100 = 200 \text{ m/s}$$

Alternatively, $$92 \div 46 = 2$$, so $$92 = 46 \times 2$$, and thus:

$$\frac{92 \times 100}{46} = \frac{46 \times 2 \times 100}{46} = 2 \times 100 = 200 \text{ m/s}$$

Both waves give the same speed of 200 m/s, which is consistent since they are in the same medium. Therefore, the speed of sound in the medium is 200 m/s.

Now, comparing with the options:

A. 92 m/s

B. 200 m/s

C. 100 m/s

D. 332 m/s

Hence, the correct answer is Option B.