Answer:
k = 1400.4 N / m
Explanation:
When the springs are oscillating a simple harmonic motion is created where the angular velocity is
w² = k / m
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
where angular velocity, frequency and period are related
w = 2π f = 2π / T
we substitute
2π / T = \sqrt{ \frac{k}{m} }
T² = 4π² [tex]\frac{m}{k}[/tex]
k = π² [tex]\frac{m}{T^{2} }[/tex]
in this case the period is T = 1.14s, the combined mass of the children is
m = 92.2 kg and the constant of the two springs is
k = 4π² 92.2 / 1.14²
k = 2800.8 N / m
to find the constant of each spring let's use the equilibrium condition
F₁ + F₂ - W = 0
k x + k x = W
indicate that the compression of the two springs is the same, so we could replace these subtraction by another with an equivalent cosecant
(k + k) x = W
2k x = W
k_eq = 2k
k = k_eq / 2
k = 2800.8 / 2
k = 1400.4 N / m
