Create a trigonometric function (cosine) to model the situation:

A mass is supported by a spring so that it is at rest 0.5 m above a tabletop. The mass is pulled down 0.4 m and released at time t = 0. It then moves up through its rest position, stops and drops again, creating a periodic up and down motion that can be modelled using a trigonometric function. It takes 1.2 seconds to return to the lowest position each time.

General form of a periodic function:
y = Acos(2π(t - a)/T) + b

A = amplitude
a = horizontal shift
b = vertical shift
T = period

From the problem:
b = 0.5 m (since the mass is suspended 0.5 m above the table)
A = 0.4 m
T = 1.2 s

"The mass is pulled down 0.4 m and released at time t = 0." Therefore, it is 0.4 m below the resting point, which is 0.5 m above the table. Therefore, at t = 0, the mass is 0.1 m above the table.

y = 0.4cos(2π(t - a)/1.2) + 0.5

Find a using the point (0, 0.1):

0.1 = 0.4cos(2π(0 - a)/1.2) + 0.5
0.4cos(-2πa/1.2) = -0.4
cos(-πa/0.6) = -1
-πa/0.6 = π
a = -0.6

Therefore, your model is
y = 0.4cos(2π(t - 0.6)/1.2) + 0.5

Where y is distance above the table in meters
t is time in seconds

simple harmonic motion: x=x'cos(ωt+Φ)

x' = amplitude = .4m
ω = angular frequency = 2π/T = 2π/(1.2) = 5.236
Φ = phase constant = 0 in this case

x=.4cos(5.236t)