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Mass Suspended by a Vertical Spring |
Another case of Simple Harmonic Motion, this time with [gravity] thrown in.
Part A
Consider first the static case with the mass hanging from the spring and not moving.
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
The mass m is suspended from a perfect spring with force constant k . Attaching the mass stretches the spring a distance a from its equilibrium length. Draw the force diagram and determine what a must be.
Force Diagram of Mass on Vertical Spring |
Mathematical Representation
We first consider the case of a stationary mass on a vertical spring. Adding the mass causes the spring to extend a distance a beyond its "neutral", unstretched length. After the mass has come to rest, what are the forces?
From the above diagram, we have the force of gravity pulling downwards with Fg = mg and the spring force pulling upwards with force Fs = ka . Since the mass is stationary we must have
\begin
we can solve for the displsacement at equilibrium:
\begin
[ a = \frac
]\end
Part B
Analyze the forces on the mass as it oscillates up and down and give its equation of motion.
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
Force Diagram of Displaced Vertical Spring |
The mass m is now displaced from its equilibrium position (extended by a distance a) by an additional distance x .
Mathematical Representation
Displacing the mass a distance x downwards results in restoring force from the spring. We calculated the total force on the mass, due to the original extension and this additional extension, as well as by gravity. we obtain the upward force
\begin
[ F_
= k(a + x) - mg ]\end
But we know from the previous section that ka = mg, so the first and last terms will cancel, leaving
\begin
[ F_
= kx ]\end
just as in the case of a horizontal spring and mass on a frictionless surface (see, for example, the worked example Mass Between Two Springs)The solution for the equation of motion is, as in the case of any [Simple Harmonic Oscillator]
\begin
where
\begin
[ \omega = \sqrt{\frac
{m}} ]\end
and A and B are determined by the initial conditions, the initial position xi and the initial velocity vi
\begin
[ B = x_
]\end
\begin
These are exactly the same angular velocity and coefficients A and B as in the case of the horizontal mass on a spring (in the absence of gravity).
Part C
What if we look at the problem from the point of view of energy. How can the energy be conserved in the same way as a horizontal mass on a spring if we have to consider Gravitational Potential Energy as well?
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
Mass on Horizontal Spring with Energy Indicated |
Mathematical Representation
Consider the total energy relative to the spring's neutral position, where it is neither extended nor comprseed. Even for the static situation of Part A, then, the system has stored energy, since you have extended the spring. On the other hand, the mass has fallen, and so you have lost some potential energy due to gravity. The total energy change is ΔE1:
\begin
[ \Delta E_
-m g a ]\end
If you now displace the mass downward by a distance xi and hold onto it you change the overall energy because you have done work on the system (you've exerted force against the spring over a distance xi, and you've also lost gravitational potential energy in moving downwards). The new energy relative to the neutral position is ΔE2
\begin
[ \Delta E_
= \frac
k(a + x_
)^
- m g (a + x_
) ]\end
From Part A we know that a = mg/k , so, substituting:
\begin
[ \Delta E_
= \frac
k\left(\frac
\right)^
- m g \left(\frac
+ x_
\right) ]\end
\begin
[ A = x_
]\end
\begin
We have assumed above for simplicity that the springs are in their relaxed state when the mass is at x = 0, but this is not necessary. The springs can both be under tension or under compression. as long as they have not gone past their ranges of motion, the above expressions will still hold. Assume that each spring is compressed by the same amount Δ at the start. (Since the spring constants are equal, if the springs are not compressed by the same amount when first attached to the mass, they will automatically do so after you release them. The mass will settle into an equilibrium position that we will define as x = 0 )Now both springs are compressed by an amount Δ , and each therefore pushes with a force F = kΔ against both the wall and against the mass m. But since the Forces are of equal magnitude and opposite direction, the sum of the forces is zero, and the mass remains in its equilibrium position.
Unable to render embedded object: File (Mass With Two Springs Displaced counter forces.PNG) not found. |
If we now displace mass m to the right by a distance x, we compress the spring more on that side, so the total force pushing to the left becomes Fleft = - k ( x + Δ ) , while we compress the spring on the left by a smaller amount (and possibly even extend it), so that the total force to the right becomes Fright = - k( - x - Δ ) . The total force is then
\begin
[ F_
= F_
- F_
]\end
or
\begin
[ F_
= - k (x + \Delta ) - (- k ( x - \Delta) = -2 k x ]\end
This is the same as before. The equation of motion is the same, and so the results will be the same. Note that all of this still holds if both springs are under tension instead of compression at the start.
Question
What would happen if we were to make the springs have different spring constants? Call them k1 and k2.
Answer
The mass will re-adjust to a new equilibrium position, where the compressions of the two springs are such that the forces from each spring are of equal magnitude and opposite direction. The restoring force of the system will be
\begin
[ F_
= - ( k_
+ k_
)x ]\end
So you should replace 2k in the above expressions with ( k1 + k2 ) . This makes the frequency of oscillation
\begin
[ \omega_
= \sqrt{\frac{k_
+ k_{2}}{m}} ]\end