Answer:
The amplitude of the oscillation is 2.82 cm
Explanation:
Given;
mass of attached block, m = 4.1 kg
energy of the stretched spring, E = 3.8 J
period of oscillation, T = 0.13 s
First, determine the spring constant, k;

where;
T is the period oscillation
m is mass of the spring
k is the spring constant

Now, determine the amplitude of oscillation, A;

where;
E is the energy of the spring
k is the spring constant
A is the amplitude of the oscillation

Therefore, the amplitude of the oscillation is 2.82 cm