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|!Spring-pendulum.jpg!|
|Mass Suspended by a Vertical Spring
Photo courtesy of Wikimedia Commons|



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{excerpt}Another case of [Simple Harmonic Motion], this time with [gravity 

...

(near-earth)] thrown in.

...

{excerpt}

 

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h2.Part A

Consider first the _static_ case with the mass hanging from the spring and not moving.
h4. Solution

{toggle-cloak:id=sysA} *System:*  {cloak:id=sysA}   [Simple Harmonic Motion].{cloak}

{toggle-cloak:id=intA} *Interactions:*  {cloak:id=intA} The forces due to the compression or extension of the two springs acting as the [restoring force] and the force of [gravity (near-earth)].{cloak}

{toggle-cloak:id=modA} *Model:* {cloak:id=modA} [Simple Harmonic Motion].{cloak}

{toggle-cloak:id=appA} *Approach:*  

{cloak:id=appA}

{toggle-cloak:id=diagA} {color:red} *Diagrammatic Representation* {color}

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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.

|!Vertical Mass on Spring Static Extension.PNG!|

|!Vertical Mass on Spring Static Extension Forces.PNG!|
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{toggle-cloak:id=mathA} {color:red} *Mathematical Representation* {color}

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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 *F{~}g{~} = mg* and the spring force pulling upwards with force *F{~}s{~} = ka* . Since the mass is stationary we must have

{latex}\begin{large}\[ ka - mg = 0 \]\end{large}{latex}

we can solve for the displsacement at equilibrium:

{latex}\begin{large}\[ a = \frac{mg}{k}  \]\end{large}{latex}


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=Part

 B}

h2. Part B

Analyze the forces on the mass as it oscillates up and down and give its equation of motion.

h4. Solution

{toggle-cloak:id=sysB} *System:*  {cloak:id=sysB}   [Simple Harmonic Motion].{cloak}

{toggle-cloak:id=intB} *Interactions:*  {cloak:id=intB} The forces due to the compression or extension of the spring acting as the [restoring force] and the force of [gravity (near-earth)].{cloak}

{toggle-cloak:id=modB} *Model:* {cloak:id=modB} [Simple Harmonic Motion].{cloak}

{toggle-cloak:id=appB} *Approach:*  

{cloak:id=appB}

{toggle-cloak:id=diagB} {color:red} *Diagrammatic Representation* {color}

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|!Vertical Mass on Spring Dynamic Extension Forces.PNG!|

The mass *m* is now displaced from its equilibrium position (extended by a distance *a*) by an additional distance *x* . 


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{toggle-cloak:id=mathB} {color:red} *Mathematical Representation* {color}

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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

{latex}\begin{large}\[ F_{\rm total} = k(a + x) - mg \]\end{large}{latex}

But we know from the previous section that *ka = mg*, so the first and last terms will cancel, leaving

{latex}\begin{large}\[ F_{\rm toal} = kx  \]\end{large}{latex}

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]

{latex}\begin{large}\[ x = A sin(\omega t) + B cos(\omega t) \]\end{large}{latex}

where

{latex}\begin{large}\[ \omega = \sqrt{\frac{k}{m}} \]\end{large}{latex}

and *A* and *B* are determined by the initial conditions, the initial position *x{~}i{~}* and the initial velocity *v{~}i{~}*

{latex}\begin{large} \[ B = x_{i} \]\end{large}{latex}

\\

{latex}\begin{large} \[A = {\frac{v_{i}}{\omega}} \]\end{large}{latex}

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).

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{card:label=Part C}

h2. 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?

h4. Solution

{toggle-cloak:id=sysC} *System:*  {cloak:id=sysC}   [Simple Harmonic Motion].{cloak}

{toggle-cloak:id=intC} *Interactions:*  {cloak:id=intC} The forces due to the compression or extension of the two springs acting as the [restoring force] and the force of [gravity (near-earth)].{cloak}

{toggle-cloak:id=modC} *Model:* {cloak:id=modC} [Simple Harmonic Motion].{cloak}

{toggle-cloak:id=appC} *Approach:*  

{cloak:id=appC}

{toggle-cloak:id=diagC} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diagC}

|!Vertical Mass on Spring Dynamic Energy.PNG!|


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{toggle-cloak:id=mathC} {color:red} *Mathematical Representation* {color}

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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 {*}ΔE{~}1{~}{*}:


{latex}\begin{large}\[ \Delta E_{1} = \frac{1}{2}ka^{2} -m g a \]\end{large}{latex}

If you now displace the mass downward by a distance *x{~}i{~}* 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 *x{~}i{~}*, and you've also lost gravitational potential energy in moving downwards). The new energy relative to the neutral position is {*}ΔE{~}2{~}{*}


{latex}\begin{large}\[ \Delta E_{2} = \frac{1}{2}k(a + x_{i})^{2} - m g (a + x_{i})  \]\end{large}{latex}

From *Part A* we know that *a = mg/k* , so, substituting:

{latex}\begin{large}\[ \Delta E_{2} = \frac{1}{2}k\left(\frac{mg}{k} + x_{i} \right)^{2} - m g \left(\frac{mg}{k} + x_{i}\right) \]\end{large}{latex}

\\
Expanding this, then consolidating terms yields

{latex}\begin{large}\[ \Delta E_{2} = \frac{(mg)^{2}}{2k}+ m g x_{i} + \frac{1}{2} k {x_{i}}^{2} - \frac{(mg)^{2})}{k} - m g x_{i} \]\end{large}{latex}

\\

{latex}\begin{large} \[ \Delta E_{2} = \frac{1}{2}k {x_{i}}^{2} - \frac{(mg)^{2}}{2k} \]\end{large}{latex}

\\
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

{latex}\begin{large} \[ \Delta E_{2} = \frac{1}{2}k \left(\frac{mg}{k}\right)^{2} + \frac{1}{2}mv^{2} -mg\left(\frac{mg}{k}\right) \]\end{large}{latex}

or

{latex}\begin{large} \[ \Delta E_{2} = \frac{1}{2}mv^{2} - \frac{(mg)^{2}}{2k} \]\end{large}{latex}

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 v{^}2{^} * and the [Potential Energy] at full extension (aside from that same constant term) is *(1/2) k {x{~}i{~}}{^}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)].




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