Is the Total Force in Damped Harmonic Motion Always Opposite to Velocity?

[fcoulomb]

Homework Statement

Reading chapter 4 of Morin's "Introduction to classical mechanics" I came across to the explanation of the damped harmonic motion.

The mass m is subject to a drag force proportional to its velocity, $F_f = -bv$.
He says that the total force of the mass is $F= -b \dot{x} -kx$ and considering that $F= m\ddot{x}$ we get this differential equation $$ \ddot{x} +\frac{b}{m} \dot{x} + \frac{k}{m} x=0$$

But the total force should be $F= -b \dot{x} +kx$, shouldn't it? These two forces have opposite direction!

 

[drvrm]

But the total force should be F=−b˙x+kxF=−bx˙+kxF= -b \dot{x} +kx, shouldn't it? These two forces have opposite direction!

if x is extension or compression in the spring the force experienced is always opposite to the
direction of change in x so spring force must be -k.x

 

[Orodruin]

Also, $x$ is not the same as $\dot x$. Depending on their relative sign, the drag and restoring force may or may not be in the same direction.

 

[drvrm]

Also, xxx is not the same as ˙xx˙\dot x. Depending on their relative sign, the drag and restoring force may or may not be in the same direction.

however the velocity /motion of the body will always be opposed by the drag so it should be -damping coefficient times the velocity.
take two cases...

1. .x is increasing ..extension..so -k.x and velocity is in the same direction so again the drag will be opposite

2. take compression x is reducing so so restoring force will oppose compression and velocity is in the direction of compression so drag will be outward.

Am i right...

 

[Orodruin]

1. .x is increasing ..extension..so -k.x and velocity is in the same direction so again the drag will be opposite

This depends on what you mean with "x is increasing". If you assume positive $x$, yes. It would be more accurate to say that the forces act in the same direction whenever you move away from the equilibrium position and in opposite directions whenever you move towards the equilibrium position.