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Two Masses Attached to a Spring

Date: 12/27/2000 at 11:14:20
From: Gary Cunningham
Subject: Behavior of springs and masses in a gravitational field

A friend proposed the following, but I can't seem to get the math to 
prove him wrong even though my intuition tells me he is wrong.

Two masses are attached to either end of an elastic spring and the 
whole system is in a gravitational field. One mass is held in a fixed 
position and the second mass is allowed to hang free below and stretch 
the spring. Now, let go of the upper mass.

He proposes that the spring force will pull the two masses toward each 
other in such a way that the lower mass will be pulled upward and thus 
remain stationary in space until all the potential elastic energy of 
the spring is exhausted (which will be when the masses are as close 
together as the spring allows), at which time the lower mass will 
begin to accelerate downward. (In contrast, the upper mass will 
accelerate downward at a rate dependent on the force of gravity plus 
the spring force.) Until the springs compress to a zero-force state, 
the lower mass will not move.

This makes no sense to me. The elastic force in the spring is EXACTLY 
EQUAL to the weight of the lower object. BUT, this spring force will 
act on BOTH masses to pull EACH TOWARD THE OTHER when the upper mass 
is released. This means the spring force (which was exactly equal only 
to the weight of the lower mass), will act on EACH mass attached at 
the ends. Therefore, since the spring force was exactly equal to the 
weight of the lower mass, and since that force is now divided between 
two masses, it cannot accelerate the lower mass upward to exactly 
counterbalance the gravitational force. Thus the lower mass must begin 
to accelerate downward the moment the upper mass is released (although 
more slowly than it would without the spring system).

In addition, the spring force is dependent on the degree of 
deformation of the spring, which declines as the spring returns to its 
neutral state. The spring will exert less force on the masses as it 
becomes less deformed. Since the force is lessened as the spring 
compresses, the force on the lower mass is progressively lessened. So 
even if the force were not divided between two masses to pull them 
together (if there were zero mass at the upper end of the stretched 
spring) the lower mass would still begin to fall instantly since the 
spring force is declining from its maximum.

That's it. Thanks for your help.

Date: 12/29/2000 at 12:59:35
From: Doctor Rick
Subject: Re: Behavior of springs and masses in a gravitational field

Hi Gary, thanks for writing to Ask Dr. Math.

The motion of a system of two masses attached by a spring can be 
broken into two parts. The system as a whole is acted on only by 
gravity, so the center of mass of the system accelerates downward with 
an acceleration of g. The motion of the masses relative to the center 
of mass is independent: gravity is not involved, only the spring 

Suppose you had this system on the Space Station. Gravity isn't 
operative, so the masses at rest are separated by the rest length of 
the spring. Stretch the spring until the separation is the same as you 
had in the earthbound scenario, and release them. Assuming that the 
spring is constrained to move along a straight line, the two masses 
will oscillate, first moving together, then apart, and so on.

If the spring is stretched a distance x beyond its rest length, and 
its spring constant is k, then the force that it exerts on each mass 
is kx, acting toward the center of mass. When you release the masses, 
they begin to accelerate toward each other with accelerations kx/m_1 
(for the first mass, m_1) and kx/m_2 (for the second mass, m_2).

What is that initial stretch length, x? Going back to earth, the 
hanging mass (m_2) experiences two forces: a gravitational force gm_2, 
and a spring force kx acting in the opposite direction. When the mass 
is at rest, these opposite forces must be equal:

     gm_2 = kx

        x = gm_2/k

So that's our starting length in the space-station experiment, and the 
acceleration on mass 2 is:

     kx/m_2 = k(gm_2/k)/m_2

            = g

That's interesting! When we drop the system in the earth-bound 
experiment, the center of mass accelerates with an acceleration of g, 
and the hanging mass accelerates upward (toward the center of mass) 
with an acceleration of g. The net acceleration of the mass is 0. In 
other words, at the moment you release the first mass, mass 2 is not 

Your friend was thus correct, if we consider only the moment at which 
the system is released. The source of your confusion is that the force 
is NOT divided between the masses: it acts solely on the hanging mass, 
while an equal and opposite force (Newton's third law) acts on the 
initially supported mass. Thus the initially supported mass is 
accelerated downward at twice the acceleration of gravity, while the 
hanging mass is pulled upward just as hard as gravity pulls it down, 
and it experiences no acceleration.

On the other hand, you are right in your second argument: the 
situation of zero acceleration will not persist. Your friend's 
statement that the mass will remain stationary until the spring 
reaches its rest state is incorrect. As the spring pulls the masses 
together, the force it exerts decreases, because it is proportional to 
the stretch length. Thus, a moment after the system is released, the 
upward force is no longer sufficient to suspend the mass motionless. 
By the time the rest-state of the spring is reached, both masses are 
already falling at a full 1 g acceleration.

What's interesting about the motion of the hanging mass is that, at 
the moment of release, its motion is described, not by a parabola 
(quadratic function of time), but by a cubic or third-power function 
of time. That's not as dramatic as being motionless for a period of 
time, but it is not the way we are accustomed to seeing things fall.

- Doctor Rick, The Math Forum   
Associated Topics:
College Physics
High School Physics/Chemistry

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