Displacement of a wave is expressed as $$x(t) = 5\cos\left(628t + \frac{\pi}{2}\right)$$ m. The wavelength of the wave when its velocity is 300 m/s is :

The given equation is $$x(t)=5\cos\left(628\,t+\frac{\pi}{2}\right)$$.

This is of standard form $$x(t)=A\cos\left(\omega t+\phi\right)$$, so the angular frequency is
$$\omega = 628\ \text{rad s}^{-1}$$.

Frequency $$f$$ is related to angular frequency by
$$\omega = 2\pi f$$ $$-(1)$$.

Substituting $$\omega = 628$$ rad s$$^{-1}$$ in $$(1)$$,
$$f = \frac{\omega}{2\pi} = \frac{628}{2\pi}\ \text{Hz}$$.

Using $$\pi \approx 3.14$$,
$$2\pi \approx 6.28$$, hence
$$f \approx \frac{628}{6.28} = 100\ \text{Hz}$$.

The relation between wave speed $$v$$, frequency $$f$$, and wavelength $$\lambda$$ is
$$v = f\lambda$$ $$-(2)$$.

Given velocity $$v = 300\ \text{m s}^{-1}$$, substituting $$f = 100\ \text{Hz}$$ in $$(2)$$:
$$\lambda = \frac{v}{f} = \frac{300}{100}\ \text{m} = 3\ \text{m}$$.

Therefore, the wavelength of the wave is $$3\ \text{m}$$.

Option B.