Hanging Spring.
A 10 kilogram mass suspended from the end of a vertical spring
stretches the spring $\frac{49}{90}$ metres. At time t = 0, the mass is started in motion from the equilibrium position with an initial velocity of $1 $ $m/s$ in the upward direction. At the same time, a constant downward force of $360$ Newtons is applied to the system.
Assume that air resistance is equal to $60 $ times the instantaneous velocity and that the acceleration due to gravity is $g = 9.8 $ m/s^2.
(a) Determine the spring constant.
(b) Show that the equation of motion is
$$ \ddot{x} + 6 \dot{x} + 18x = 36 $$
where $x(t)$ is the displacement of the mass below the equilibrium position at time
$t$ . In your answer include a diagram of all forces acting on the mass.
(c) Find the position of the mass at any time. Would you describe the motion as
overdamped, underdamped or critically damped?
I'm struggling with part (c) and any help would be appreciated.
Solving for $\lambda$ I know that
$$ X_p =A\cdot t\cdot e^{-3t}\cdot\cos(3t)+B\cdot t\cdot e^{-3t}\cdot\sin(3t) $$
since without $t$, using the equation gives a result of zero,
using
$$ \ddot{x} + 6 \dot{x} + 18x = 36, $$
with $X_p$ I get
$$ 6=e^{-3t} (b \cos(3t)-a \sin(3t))$$
but the answer say
$$ x(t) = −2 e^{−3t} \cos (3t) −\frac73 e^{−3t}\sin (3t) + 2 $$
where does this $+2 $ come from, and how is the equation used without multiplying by t if it equals to zero? Sorry if this question is unclear and if any clarification is needed please ask.
Best Answer
Your $X_p$ without the extra $t$ is the solution of the homogeneous equation, i.e. when you set the RHS equal to zero.
To get the complete solution, you need to add a particular solution of the non-homogeneous equation.
In $$ \ddot{x} + 6 \dot{x} + 18x = 36, $$ if you try a constant solution $x=C$, the equation simplifies to
$$0+0+18x=36$$ and the constant is $2$.
Hence the general solution
$$e^{-3t}(A\cos 3t+B\sin 3t)+2.$$
Remains to determine the constants using the initial conditions.