Alight platform is suspended from the ceiling by a spring. a student with a mass of 90 kg climbs onto the platform. when it stops bouncing and reaches its new equilibrium position (x=0), the student notices that the spring has stretched 0.82 m. the student's friend pulls the platform down 0.32 m further and then releases it at t=0. what is the amplitude of the motion of the student on the platform?

Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or


Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

 
0.32 m (single amplitude), or
0.64 m (double amplitude)

The best answer is (a) which is universal law is in effect at all times. it is constantly at play in everything we do and governs our lives. it is also important to note that universal law applies to not only everything on earth, but everything in the universe as well, which is another reason why it is in effect at all times.