a) The spring constant is 12,103 N/m
b) The mass of the trailer 2,678 kg
c) The frequency of oscillation is 0.478 Hz
d) The time taken for 10 oscillations is 20.9 s
Explanation:
a)
When the two children jumps on board of the trailer, the two springs compresses by a certain amount
[tex]\Delta x = 6.17 cm = 0.0617 m[/tex]
Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:
[tex]2mg = k'\Delta x[/tex] (1)
where
m = 76.2 kg is the mass of each children
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]k'[/tex] is the equivalent spring constant of the 2-spring system
For two springs in parallel each with constant k,
[tex]k'=k+k=2k[/tex]
Substituting into (1) and solving for k, we find:
[tex]2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m[/tex]
b)
The period of the oscillating system is given by
[tex]T=2\pi \sqrt{\frac{m}{k'}}[/tex]
where
And for the system in the problem, we know that
T = 2.09 s is the period of oscillation
m is the mass of the trailer
[tex]k'=2k=2(12,103)=24,206 N/m[/tex] is the equivalent spring constant of the system
Solving the equation for m, we find the mass of the trailer:
[tex]m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg[/tex]
c)
The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:
[tex]f=\frac{1}{T}[/tex]
where
f is the frequency
T is the period
In this problem, we have
T = 2.09 s is the period
Therefore, the frequency of oscillation is
[tex]f=\frac{1}{2.09}=0.478 Hz[/tex]
d)
The period of the system is
T = 2.09 s
And this time is the time it takes for the trailer to complete one oscillation.
In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.
Therefore, the time needed for 10 oscillations is:
[tex]t=10T=10(2.09)=20.9 s[/tex]
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