If x = 5sin(πt + π/3)m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion respectively, are :

We are given the displacement of a particle as

$x\; =\; 5 sin(\pi t\; +\; \pi /3)���m$

The standard form for simple harmonic motion is

$x\; =\; A sin(\omega t\; +\; \phi )$

Comparing the two expressions, we can read off the parameters directly.

Amplitude (A): The coefficient in front of the sine function represents the maximum displacement from the equilibrium position. Here it is 5, and the unit attached to the whole expression is metres, so

$A\; =\; 5 m$

Angular frequency (ω): The factor multiplying the time variable t inside the sine argument is the angular frequency. From the given equation, ω = π rad s⁻¹.

Period (T): The period is the time taken for one complete cycle and is related to ω by

$T\; =\; \backslash frac\{2\pi \}\{\omega \}$

Substituting ω = π:

$T\; =\; \backslash frac\{2\pi \}\{\pi \}\; =\; 2 s$

Therefore, the amplitude is 5 m and the period is 2 seconds, which matches the provided correct option.