If $x=5sin(\mathrm{\pi <!--\; \pi \; -->}t+\frac{\mathrm{\pi <!--\; \pi \; -->}}{3})m$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are

The given equation representing the motion of the particle executing simple harmonic motion (SHM) is:
$x=5\sin (\pi t+\frac{\pi }{3})m$

The standard equation of a particle executing simple harmonic motion along the x-axis is given by:
$x=A\sin (\omega t+\varphi )$
where:
- A is the amplitude of the motion,
- ω is the angular frequency,
- t is the time, and
- ϕ is the initial phase.

By comparing the given equation with the standard equation, we can determine the values of amplitude (A) and angular frequency (ω):
1. The amplitude is:
$A=5\text{m}$
2. The angular frequency is:
$\omega =\pi {\text{rad s}}^{-1}$

The time period (T) of the simple harmonic motion is related to the angular frequency (ω) by the formula:
$T=\frac{2\pi }{\omega }$

Substituting the value of ω into the formula, we get:
$T=\frac{2\pi }{\pi }=2\text{s}$

Thus, the amplitude of the motion is 5 m and the time period of the motion is 2 s.

Therefore, the correct option is 5 m, 2 s.