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Mass Suspended by a Vertical Spring |
Another case of Simple Harmonic Motion, this time with gravity (near-earth) 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.
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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
[ ka - mg = 0 ]\end
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
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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 Simple Harmonic Motion
\begin
[ x = A sin(\omega t) + B cos(\omega t) ]\end
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
[A = {\frac{v_{i}}{\omega}} ]\end
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
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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_
= \frac
ka^
-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
+ x_
\right)^
- m g \left(\frac
+ x_
\right) ]\end
Expanding this, then consolidating terms yields
\begin
[ \Delta E_
= \frac{(mg)^{2}}
+ m g x_
+ \frac
k {x_{i}}^
- \frac{(mg)^
)}
- m g x_
]\end
\begin
[ \Delta E_
= \frac
k {x_{i}}^
- \frac{(mg)^{2}}
]\end
Now consider the energy when the mass has rebounded and is at the original location a. It is now moving at velocity v and the total energy, potential and kinetic, is
\begin
[ \Delta E_
= \frac
k \left(\frac
\right)^
+ \frac
mv^
-mg\left(\frac
\right) ]\end
or
\begin
[ \Delta E_
= \frac
mv^
- \frac{(mg)^{2}}
]\end
Comparing these two equations, we have that, aside from a constant term of * - (mg)2/2k* , the kinetic energy at the "equilibrium" position is simply * (1/2) m v2 * and the potential energy at full extension (aside from that same constant term) is (1/2) k {xi}2 , so the emergies are, except for this constant offset term, the same as those for a horizontal spring and mass. Choosing the position at which we calculate zero energy properly would completely eliminate this superfluous term. The vertical spring and mass behaves exactly like a horizontal one in both force and energy relationships, despite the presence of gravity (interaction).