Equation for Damping in an Ideal Mass-Spring System

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In summary, in a mass-spring system with a mass m, spring constant k, and damping coefficient b, the restoring force is given by ##\vec F_r=-kx\hat{x}## and the damping force is given by ##\vec F_d=-bv\hat{v}##. When the oscillating object moves towards the amplitude position from the equilibrium position, the direction of velocity and position are the same, resulting in the net force being written as ##F=m\ddot x=-kx-bv##. However, when the object moves back from the amplitude position to the equilibrium position, the direction of velocity changes, causing the direction of the damping force to change as well. This results in the direction
  • #1
Auror
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In an ideal mass–spring system with mass m, spring constant k, and damping coefficient b,
restoring force,##\vec F_r=-kx\hat{x}##
and damping force,##\vec F_d=-bv\hat{v}##
When the oscillating object goes towards amplitude position from equilibrium position,direction of velocity and position(##\displaystyle \vec x##) is same. For that , net force,##\vec F=\vec F_r+\vec F_d## can be written as(just putting values)- ##F=m\ddot x=-kx-bv...(1)##.
But when the oscillating object moves back from that amplitude position to equilibrium position,direction of velocity changes,but direction of position doesn't. So direction of damping force changes. Hence direction of restoring force and direction of damping force are opposite. So net force should be, ##F=m\ddot x=-kx+bv...(2)##
While solving differential equation for damping- we used (1) only which gave the differential equation,
##m\ddot x+kx+b\dot x=0##
But what about the equation 2 which we didn't deduce but is valid simultaneously? Why doesn't change of direction of damping force count here?
 
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  • #2
When the velocity changes direction, the sign of the damping term automatically switches. You don't need to change the form of the equation manually, the same form is needed in all cases, and the sign takes care of itself. So equation (2) is wrong, there is no need to change the form of equation (1), or the differential equation it results in, when the velocity changes sign. Put another way, when you switched your notation from vector notation to the one-dimensional form, you dropped the unit vectors, which is fine, but v is no longer a magnitude-- it now has a sign too. That's an important difference between the v-hat unit vector and the x-hat unit vector-- the latter always points in the same direction, but not the former. So when you write x times x-hat, the x has a sign, but when you write v times v-hat, the v does not have a sign, it is always just a magnitude. That difference means that in the one-dimensional form, both x and v have a sign, whereas only x did in the vector notation.

ETA: I see Orodruin has answered similarly, and note that in his interpretation, your x-hat always points in the direction of the displacement, it is not an x coordinate vector. That is indeed more consistent with what you wrote, though the notation can be ambiguous so it may be better to use r-hat than x-hat for that purpose. Anyway, if x-hat means the unit vector in the direction of the displacement, and not the x-direction unit vector, then both x and v are magnitudes, and neither has a sign in the vector notation, but both acquire signs in the one-dimensional version.
 
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  • #3
Put in another way, when you write
Auror said:
restoring force,##\vec F_r=-kx\hat{x}##
and damping force,##\vec F_d=-bv\hat{v}##
your ##x## and ##v## are the magnitudes of the displacement and velocity, respectively. You can write it like this and yes, if you do you need to account for the sign change manually. However, doing so is just introducing complication as you can just use the velocity and actual displacement instead which will lead you to an equation where you (a) do not need to worry about the sign change and (b) get more information out of.
 

1. What is an ideal mass-spring system?

An ideal mass-spring system is a simplified model used in physics to study the behavior of a mass attached to a spring. It assumes that there is no external force acting on the mass and that the spring is massless and obeys Hooke's law.

2. What is damping in an ideal mass-spring system?

Damping in an ideal mass-spring system refers to the resistance or friction that opposes the motion of the mass. It is caused by various factors such as air resistance, internal friction in the spring, and external forces.

3. What is the equation for damping in an ideal mass-spring system?

The equation for damping in an ideal mass-spring system is given by: F = -bv, where F is the damping force, b is the damping coefficient, and v is the velocity of the mass. This equation is also known as the damping equation.

4. How does damping affect the motion of a mass-spring system?

Damping affects the motion of a mass-spring system by reducing the amplitude of the oscillations and causing the system to reach equilibrium faster. It also changes the frequency of the oscillations, which becomes lower as the damping increases.

5. How can damping be controlled in a mass-spring system?

Damping can be controlled in a mass-spring system by changing the properties of the system, such as the mass or the spring constant. Other methods include adding a damper or shock absorber to the system or adjusting the external forces acting on the mass.

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