For S.H.M, a graph of acceleration against displacement gives a slope of 16 unit. The period is

x = Acosωt

v = - Aωsinωt
a= - Aω^2cosωt
a/x= Aω^2cosωt/ Acosωt=ω^2 = 16
ω= 4 (rad/s)
T=2π/ω=2π/4 = π/2 = 1.57 s.

To determine the period of a Simple Harmonic Motion (SHM) system using a graph of acceleration against displacement, we need to find the time taken for one complete cycle of the oscillation.

In SHM, the acceleration is directly proportional to the displacement and is given by the equation:

a = -ω^2x

where:
a = acceleration
x = displacement
ω = angular frequency

The equation of a line is given by:

y = mx + c

where:
m = slope of the line
c = y-intercept

In this case, the slope of the graph is given as 16 units. This slope corresponds to the negative square of the angular frequency (-ω^2) in the equation of SHM.

By comparing the equations, we can determine that -ω^2 = 16.

Now, we can solve for ω:

-ω^2 = 16
ω^2 = -16 (take the square root on both sides)
ω = ±√-16

Since ω is a positive value related to the frequency of oscillation, we can ignore the negative sign.

ω = √16
ω = 4

The angular frequency ω is related to the period T (time taken for one complete cycle) by the formula:

T = 2π/ω

Substituting the value of ω into the equation, we get:

T = 2π/4
T = π/2

Therefore, the period of the SHM system, based on the given graph, is π/2 units of time.