Understanding the Relationship Between Mass, Spring Deformation and Velocity

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Discussion Overview

The discussion revolves around the dynamics of a mass placed on a spring, specifically exploring the relationship between mass, spring deformation, and velocity. Participants examine whether the problem can be treated as static or if it involves dynamic motion, particularly when the mass is dropped onto the spring versus being lowered gently.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants propose that when a mass is placed on a spring, the spring will deform until the force exerted by the spring equals the weight of the mass (mg).
  • Others argue that once the mass descends, it acquires velocity and continues to move downward until it comes to rest, suggesting that this is not merely a static problem.
  • A participant mentions that the system will oscillate around the equilibrium point and will eventually come to rest when the forces balance.
  • Another participant introduces the concept of friction, noting that it will eventually stop the motion of the mass.
  • One participant provides a mathematical formulation of the motion of the spring, indicating that the equation of motion can be simplified by considering the rest position of the spring.
  • Some participants discuss the implications of lowering the mass slowly versus dropping it, highlighting the differences in energy dynamics and resulting motion.

Areas of Agreement / Disagreement

Participants express differing views on whether the problem can be treated as static or if it inherently involves dynamics. There is no consensus on the treatment of the system, as some emphasize the oscillatory nature of the motion while others focus on the static equilibrium condition.

Contextual Notes

Limitations include assumptions about friction and the nature of the initial conditions (e.g., whether the mass is dropped or lowered). The discussion also reflects varying interpretations of equilibrium and motion in the context of spring dynamics.

chandran
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when a mass(mg) is kept over a spring according to statics the spring will deform till the force supplied by the spring(kx) is equal to mg.


But when thought carefully even when the mass descends by some distance(x) when the equilibrium is reached,the mass must have acquired some velocity and it will continue to move downwards till the velocity is zero.

is my thinking correct . So this is not a static problem
 
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chandran said:
when a mass(mg) is kept over a spring according to statics the spring will deform till the force supplied by the spring(kx) is equal to mg.But when thought carefully even when the mass descends by some distance(x) when the equilibrium is reached,the mass must have acquired some velocity and it will continue to move downwards till the velocity is zero.

is my thinking correct . So this is not a static problem
You are correct, it will oscillate about this stationary point. However, it will constantly be accelerating towards this stationary point and will eventually come to rest at the point when the force supplied by the spring is equal to the weight of the mass. Therefore, to calculate the compression of the spring the problem can be treated as a statics problem, as eventually the system will come to rest.
 
Last edited:
Of course, in a real spring, there will be friction which will fairly quickly bring the motion to a stop.

By the way, if the mass is m, the "weight" or force downward on the spring will be -mg and the spring will be compressed until -kx= -mg (k is the spring constant) or x= mg/k. The equation of motion of the spring is
[tex]m\frac{d^2 x}{dt^2}= -kx- mg[/tex]
where x is measured from the "rest position" of the spring without force on it.
If we let y= x- m/k, then the equation becomes
[tex]m\frac{d^2 y}{dt^2}= -k(y+ mg/k)- mg= -ky[/tex]
In other words, we can always ignore gravity if we use the "rest position" of the spring with gravity as our 0 position.
 
chandran said:
when a mass(mg) is kept over a spring according to statics the spring will deform till the force supplied by the spring(kx) is equal to mg.
If you slowly lower the mass onto the spring (using an external system, such as your hand, to take up any extra energy) then the spring will compress to the equilibrium point as you state.


But when thought carefully even when the mass descends by some distance(x) when the equilibrium is reached,the mass must have acquired some velocity and it will continue to move downwards till the velocity is zero.
Of course, if you drop the mass onto the spring it will oscillate about the equilibrium point, since there is "extra" energy that remains in the system. (Until friction brings it to rest as Halls explained.)
 

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