To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.
By Hook's law we know that force is defined as,
[tex]F= kx[/tex]
Where,
k = spring constant
x = Displacement change
PART A) For the case of the spring constant we can use the above equation and clear k so that
[tex]k= \frac{F}{x}[/tex]
[tex]k = \frac{mg}{x}[/tex]
[tex]k= \frac{77.2*9.8}{0.0637}[/tex]
[tex]k = 11876.92N/m[/tex]
Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m
PART B) In the case of speed we can obtain it through the period, which is given by
[tex]T = \frac{2\pi}{\omega}[/tex]
Re-arrange to find \omega,
[tex]\omega = \frac{2\pi}{T}[/tex]
[tex]\omega = \frac{2\pi}{2.14}[/tex]
[tex]\omega = 2.93rad/s[/tex]
Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to
[tex]\omega^2 = \frac{k}{m}[/tex]
[tex]m = \frac{k}{\omega^2}[/tex]
[tex]m = \frac{ 11876.92}{2.93}[/tex]
[tex]m = 4093.55Kg[/tex]
Therefore the mass of the trailer is 4093.55Kg
PART C) The frequency by definition is inversely to the period therefore
[tex]f = \frac{1}{T}[/tex]
[tex]f = \frac{1}{2.14}[/tex]
[tex]f = 0.4672 Hz[/tex]
Therefore the frequency of the oscillation is 0.4672 Hz
PART D) The time it takes to make the route 10 times would be 10 times the period, that is
[tex]t_T = 10*T[/tex]
[tex]t_T = 10 *2.14s[/tex]
[tex]t_T = 21.4s[/tex]
Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s
