In the equation for simple harmonic motion (SHM), $$x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m},$$ the general form $$x = A \sin (\omega t + \phi)$$ can be used to identify the parameters of SHM, where:

• A is the amplitude.
• ω (omega) is the angular frequency.
• φ (phi) is the phase constant.
• t is the time.

Comparing the given equation with the standard form:

• The amplitude A is 5 m, as that is the coefficient of sine in the equation.
• The angular frequency ω is $\pi$ rad/s.

The angular frequency $\omega$ is related to the time period $T$ of the motion through the formula:

$$\omega = \frac{2\pi}{T}$$.

Given that $\omega = \pi$, we can substitute and solve for $T$:

$$\begin{aligned}

\pi = \frac{2\pi}{T} & \implies T = \frac{2\pi}{\pi} = 2 \text{ seconds}.

\end{aligned}$$

Hence, the amplitude of the motion is 5 m and the time period is 2 s. Thus, the correct answer is:

Option B: 5 m, 2 s