Question

You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 62.2 kg hop on board for a ride through the woods and the springs (one for each wheel) each compress by 5.57 cm. When you pull the trailer over a tree root in the trail, it oscillates with a period of 1.24 s. Determine the following.

(a) force constant of the springs

(b) mass of the trailer

(c) frequency of the oscillation

(d) time it takes for the trailer to bounce up and down 10 times

I figured out that the answer for part A is 5471.81, but I can't find part b.

Answer 1

Part A. (Already Solved)

Spring Force = Fs = k_eff*dx

k_eff = Fs/dx = 62.2*9.81/0.0557

k_eff = 10943.6 N/m

Since both spring are attached in parallel, So

k_eff = k + k = 2*k

k = spring constant of each spring = 10943.6/2 = 5471.8 N/m

Part B.

Angular frequency in SHM is given by:

w = sqrt (k_eff/M)

M = k_eff/w^2

Given that period of oscillation of trailer = 1.24 sec

w = 2*pi/T = 2*pi/1.24 = 5.07 rad/sec

So,

M = 10943.6/5.07^2

M = 425.74 kg

M = mass of trailer + children = 425.74

Mass of trailer = 425.74 - 62.2

Mass of trailer = 363.54 kg

Part C.

frequency = 1/T = 1/1.24

frequency = 0.81 Hz

Part D.

time taken by trailer to bounce up and down 10 times = t = 10*T

t = 10*1.24

t = 12.4 sec

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