mass kept over a spring

chandran

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

Hootenanny

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.

HallsofIvy

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.

Doc Al

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.


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.)