Question

You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 92.2 kg hop on board for a ride through the woods and the springs (one for each wheel) each compress by 5.97 cm. When you pull the trailer over a tree root in the trail, it oscillates with a period of 1.14 s. Determine the force constant of the springs in N/m.

Answer 1

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 = $\sqrt{ \frac{k}{m} }$

where angular velocity, frequency and period are related

          w = 2π f = 2π / T

           

we substitute

          2π / T = \sqrt{ \frac{k}{m} }

         T² = 4π² $\frac{m}{k}$

          k =  π²  $\frac{m}{T^{2} }$

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