To solve this problem, we need to identify the amplitude and time period of the given simple harmonic motion (SHM) equation.$\\$The equation of motion is given as:$\\ x = 5 \sin\left(\pi t + \frac{\pi}{3}\right) \\$In the standard form of SHM, the equation is:$\\ x = A \sin(\omega t + \phi) \\$where:$\\$• $A$ is the amplitude,$\\$• $\omega$ is the angular frequency, and$\\$• $\phi$ is the phase constant.$\\$From the given equation, we can identify:$\\ A = 5 \\$Therefore, the amplitude of the motion is $5$ m.$\\$Next, we need to find the time period of the motion. The angular frequency $\omega$ is given by:$\\ \omega = \pi \\$The time period $T$ is related to the angular frequency by the formula:$\\ T = \frac{2\pi}{\omega} \\$Substitute the value of $\omega: \\ T = \frac{2\pi}{\pi} \\ T = 2$ s$\\$Thus, the time period of the motion is $2$ s.$\\$Therefore, the amplitude and time period of the motion are $5$ m and $2$ s, respectively.$\\$This corresponds to Option 4.