- #1
WeiShan Ng
- 36
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Homework Statement
A carriage is mounted on a spring, as shown in the diagram.
The bottom of the spring is fixed to the ground. The carriage (loaded with its passenger) has a mass of 150kg. The carriage can only move vertically. The natural length of the spring is 10m and its spring constant is ##k=600Nm^{-1}##. At equilibrium the spring is compressed by a distance ##s## metres. Air resistance acts on the carriage with a drag coefficient ##300Nsm^{-1}##. Assume that the gravitational constant is ##g=10ms^-2##.
The ride begins with the carriage released from rest 0.5m above ground level.
Let ##y(t)## be the height of the carriage below its equilibrium position at time ##t## seconds after the ride begoins. Take the positive direction to be downwards.
Homework Equations
Show that the equation of motion for this system is
$$150y''+300y'+600y=0$$
The Attempt at a Solution
##\vec{T}## is the spring force, acting upwards, and ##\vec{R}## is the air resistive force, acting downwards, opposite to the moving direction of the carriage.
So the equation will be:
$$m\ddot{y}=mg - \vec{T} + \vec{R}$$
$$m\ddot{y} = mg - k(s+y) + \beta \dot{y}$$
$$m\ddot{y} = -ky + \beta \dot{y}$$
Since ##ks=mg##
And I get $$150\ddot{y}-300\dot{y}+600y = 0$$
which have a negative coefficient for ##\dot{y}##. I've double check with all the directions of the forces, and pretty sure this is the case, what have I done wrong?