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Spring SHM System
Name: Brian S.
Status: educator
Age: 30s
Location: N/A
Country: N/A
Date: 2/26/2003
Question:
I need some help with the energy equation for the SHM
f a vertical spring.
I would like to know what the equation would look like for the max
ocity, velocity at equilibrium, of a mass hanging from a vertical spring.
The COE equation gives us this
DKE + DPEgravity + DPEspring = 0
We cannot cancel PE gravity because if you pull down on a mass and let
it go, as the mass reaches equilibrium, the velocity should be a maximum.
Is this correct?
Vmax = SQRT(k/m*X^2 - 2gh)
where h = X the displacement of the spring....
I am also wondering if I need to take in account the energy the spring
has when the mass reaches equilibrium?
Anyway, I just need to know is this equation for V at equilibrium for
a mass hanging from a vertical spring is correct?
Replies:
Yes, if h=X is the displacement of the mass downward from its equilibrium
position. In principle, you should take the energy of the spring into
account, but it's better to start students off with an idealized, massless
spring, and anyway, it is usually a pretty good approximation.
Tim Mooney
This is a rather confusing problem partly because gravity can be completely
ignored if we measure displacements from the equilibrium position of the
mass hanging on the spring.
Notice that the force acting on the mass is F = -ky - mg if y is measured
upwards from the position of the mass when the spring is not extended and so
the spring is exerting no force on the mass. k is the spring constant and
mg is the weight of the mass. If you now hang the mass on the spring, it
will come to an equilibrium when F = 0 which happens when y = -mg/k. Right?
Now do a change of variable to z = y + d, where d = mg/k, so z is measured
from the mass position when it is hanging on the spring at rest. Then
F = -ky - mg becomes F = -k(z-d) - mg = -kz (use d=mg/k) and we can ignore
gravity.
In the energy description, which you used, we can show the same thing,
though it needs a little more algebra. If the mass is pulled down to
y = -d -h and released from rest, its speed at y = -d, which is the
equilibrium position is v = sqrt((k/m)*h^2). Using z the mass is pulled
down to z = -h. Then its speed at z = 0 is the same.
E = mv^2/2 + mgy + ky^2/2 = mv^2/2 + kz^2/2 - kd^2/2 (using d=mg/k)
Since kd^2/2 is a constant, it has no effect. Gravity has disappeared!
I recommend you draw a careful diagram and go through the algebra carefully.
If it does not work out and you want to see it, send me an email directly
via NEWTON (SYSOP@newton.dep.anl.gov).
Best, Dick Plano, Professor of Physics emeritus, Rutgers University
So if you measure the displacement of the mass from its equilibrium
position, the
effect of gravity vanishes! What really happens, of course, is that the
extra spring tension just exactly balances the weight of the mass.
Brian,
The first factor you seem to have missed is the initial stretch factor as
it relates to the equilibrium position. Initial stretch=X, where zero
stretch is zero spring force and zero spring potential energy. Let initial
position be zero height, zero gravitational potential energy Initial
speed=0, initial KE=0. Final position,x,is at equilibrium: kx=mg,
x=mg/k. Note that x < X. v=final velocity
final height=X-x, initial stretch-final stretch
{ (1/2)mv^2-0 } + { mg(X-x)-0 } + { (1/2)kx^2-(1/2)kX^2 } = 0
(1/2)mv^2 = -mg(X-x) - (1/2)kx^2 + (1/2)kX^2
v^2 = -2g(X-x) - (k/m)x^2 + (k/m)X^2
v^2 = 2gx - 2gX - (k/m)x^2 + (k/m)X^2, substitute x=mg/k
v^2 = 2mg^2/k - 2gX - mg^2/k + (k/m)X^2
v^2 = mg^2/k + (k/m)X^2 - 2gX
Dr. Ken Mellendorf
Physics Instructor
Illinois Central College
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