If $x= 5sin (\pi t+\frac{\pi}{3})$m represents the motion of a partcile executing simple harmonic motion, the amplitude and time period of motion, respectively, are :

5 m, 2s

The correct answer is option (2) : 5 m, 2s

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

$A$ is the amplitude.

$\omega $ is the angular frequency.

$\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 $\omega $ 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 $T$ :

$\pi =\frac{2\pi}{T}⇒T=\frac{2\pi}{\pi }= 2 $ seconds .

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

Option 2 : 5m , 2s