Consider the simple harmonic oscillator involving the mass m = 0.550 kg, which i
ID: 1770843 • Letter: C
Question
Consider the simple harmonic oscillator involving the mass m = 0.550 kg, which is attached to the spring and hung vertically. The oscillator is submerged in a resistive medium. At time t = 0, the mass is displaced from the equilibrium position by 5.50 cm in the downward direction and is given an initial shove, imparting a speed of 0.450 m/s on the mass in the downward direction toward the floor.
Obtain an expression for the displacement of the mass in the form x(t) = A^(t/2) cos(t + ), giving numerical values for A, , , and . Notice that the oscillator is experiencing very light damping and should be treated as such.
Explanation / Answer
given, m = 0.55 kg
yi = -5.5 cm
vi = -0.45 m/s
now, from force balance
my" + cy' + ky = 0
now the solution is of the form
y(t) = Ae^(-gamma*t/2)cos(wt - phi)
dy/dt = Ae^(-gamma*t/2)(-wsin(wt - phi)) + Ae^(-gamma*t/2)(-gamma/2)cos(wt - phi)
at t = 0
y(t) = -5.5/100 = Acos(phi)
v(t) = 0.45 = -A(wsin(phi)) + A(gamma/2)cos(phi)
0.45 = -Awsin(phi) + 0.0275gamma
0.45 = - w*sqroot(1 - (-0.055)^2) + 0.0275*gamma
0.45 = -0.99848635w + 0.0275*gamma
now, gamma = c/2m
and w = sqroot(k/m - gamma^2)
now for very light damping, w = sqroot(k/m)
0.45 = -1.3463*sqroot(k) + 0.0275*gamma
from energy balance
0.5*k*A^2 = 0.5kyi^2 + 0.5mvi^2
kA^2 = 0.003025k + 0.111375
A^2 = 0.003025 + 0.111375/k
hence A = sqroot(0.003025 + 0.111375/k)
where k is the spring constant
phi= arcos(-0.055/sqroot(0.003025 + 0.111375/k))
w = 1.348sqroot(k)
gamma = 16.3636 + 48.94399*sqroot(k)