A travelling harmonic wave is represented by the equation $$y(x, t) = 10^{-3} \sin(50t + 2x)$$, where x and y are in meter and t is in seconds. Which of the following is a correct statement about the wave?

We are given the displacement of the wave as $$y(x,t)=10^{-3}\,\sin(50t+2x)$$ where $$x$$ and $$y$$ are in metres and $$t$$ is in seconds.

The standard mathematical form for a one-dimensional harmonic wave of angular frequency $$\omega$$, wave number $$k$$ and amplitude $$A$$ is

$$y(x,t)=A\,\sin(\omega t-kx) \qquad\text{(wave travelling towards +x)},$$

or

$$y(x,t)=A\,\sin(\omega t+kx) \qquad\text{(wave travelling towards -x)}.$$

We now compare the argument of the sine function in the given expression with these standard forms. The given phase is

$$\phi = 50t + 2x.$$

Because the sign in front of the $$x$$-term is positive, the expression matches $$\sin(\omega t + kx)$$. Therefore the wave is propagating along the negative $$x$$-axis.

Next, we identify the angular frequency $$\omega$$ and the wave number $$k$$ directly from the coefficients:

Comparing $$\phi = \omega t + kx$$ with $$\phi = 50t + 2x$$, we have

$$\omega = 50 \text{ s}^{-1},\qquad k = 2 \text{ m}^{-1}.$$

The speed $$v$$ of a harmonic wave is obtained from the relation

$$v = \frac{\omega}{k}.$$

Substituting the identified values, we get

$$v = \frac{50}{2} = 25 \text{ m s}^{-1}.$$

So, the wave travels with a speed of $$25 \text{ m s}^{-1}$$, and its direction is towards the negative $$x$$-axis.

Hence, the correct answer is Option C.