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Mass Suspended by a Vertical Spring
Photo courtesy of Wikimedia Commons

Composition Setup

Excerpt
Another case of Simple Harmonic Motion, this time with gravity (near-earth) thrown in.

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{table:cellspacing=0|cellpadding=8|border=1|frame=void|rules=cols} {tr:valign=top} {td:width=310|bgcolor=#F2F2F2} {live-template:Left Column} {td} {td} |!Spring-pendulum.jpg!| |Mass Suspended by a Vertical Spring Photo courtesy of Wikimedia Commons| {composition-setup}{composition-setup} {excerpt}Another case of [Simple Harmonic Motion], this time with [gravity] thrown in.{excerpt} {deck:id=bigdeck} {card:label=Part A} h2.Part A Consider first the _statis_ 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].{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} {cloak:id=diagA} 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| {cloak:diagA} {toggle-cloak:id=mathA} {color:red} *Mathematical Representation* {color} {cloak:id=mathA} 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}

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label	Part A

Part A

Consider first the static case with the mass hanging from the spring and not moving.

Solution

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System:

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Interactions:

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The forces due to the compression or extension of the two springs acting as the and the force of .

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Model:

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Approach:

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Diagrammatic Representation

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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.

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Mathematical Representation

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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} {cloak:mathA} {cloak:appA} {card} {card:label=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].{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} {cloak:id=diagB} |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* . {cloak:diagB} {toggle-cloak:id=mathB} {color:red} *Mathematical Representation* {color} {cloak:id=mathB} 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}

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label	Part B

Part B

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

Solution

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System:

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Interactions:

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The forces due to the compression or extension of the spring acting as the and the force of .

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Model:

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Approach:

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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 .

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Mathematical Representation

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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

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

[

]

)The

solution

for

the

equation

of

motion

is,

as

in

the

case

of

any [

Simple

Harmonic Oscillator] {latex}

Harmonic Motion

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

{latex} where {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}

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}

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).

{cloak:mathB} {cloak:appB} {card} {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].{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} |Mass on Horizontal Spring with Energy Indicated| {cloak:diagC} {toggle-cloak:id=mathC} {color:red} *Mathematical Representation* {color} {cloak:id=mathC} {latex}

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label	Part C

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

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The forces due to the compression or extension of the two springs acting as the and the force of .

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Model:

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Approach:

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Diagrammatic Representation

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Mathematical Representation

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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 ΔE₁:

Latex
\begin{large}\[

m

\Delta E_{1} = \frac{

d^

1}{2}

x}{dt^

ka^{2

}

+

-m

2kx

=

0

\]\end{large}

{latex} This is the familiar equation of [simple harmonic motion] (although the multiplier for *x* is now *2k* instead of the usual *k*, because there are two springs). This has the solution: {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 ΔE₂

Latex
\begin{large}\[

x

\Delta E_{2} =

A sin(\omega t) + B cos(\omega t

\frac{1}{2}k(a + x_{i})^{2} - m g (a + x_{i})  \]\end{large}

{latex} where: {latex}

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

Latex
\begin{large}\[ \

omega

Delta E_{2} = \

sqrt{\

frac{

2k

1}{

m

2}

}

k\

]

left(\

end

frac{

large

mg}{

latex

k}

and *A* and *B* are determined by the initial conditions, the initial position *x{~}i{~}* and the initial velocity *v{~}i{~}* {latex}\begin{large} \[ A =

 + x_{i} \right)^{2} - m g \left(\frac{mg}{k} + x_{i}\right) \]\end{large}

{latex} \\ {latex}\

Expanding this, then consolidating terms yields

Latex
\begin{large}\[ \

[B

Delta E_{2} =

-{

\frac{

v_{i

(mg)^{2}}{

\omega}} \]\end{large}{latex} 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. |!Mass With Two Springs Displaced counter forces.PNG!| 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 *F{~}left{~} = - 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 *F{~}right{~} = - k( - x - Δ )* . The total force is then {latex}\begin{large} \[ F_{\rm total} = F_{\rm right}- F_{\rm left

2k}+ m g x_{i} + \frac{1}{2} k {x_{i}}^{2} - \frac{(mg)^{2})}{k} - m g x_{i} \]\end{large}

{latex} or {latex}

Latex
\begin{large} \[

F

\Delta E_{

\rm total

2} =

-

k (x + \Delta ) - (- k ( x - \Delta) = -2 k x \]\end{large}{latex} 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. {info:title=Question} What would happen if we were to make the springs have _different_ spring constants? Call them *k{~}1{~}* and *k{~}2{~}*.{info} {toggle-cloak:id=answerC} *Answer* {cloak:id=answerC} 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 {latex}

\frac{1}{2}k {x_{i}}^{2} - \frac{(mg)^{2}}{2k} \]\end{large}

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

F

E_{

\rm total

2} =

- ( k_

\frac{1}

+ k_

{2}

)x

k \

]

left(\

end

frac{

large

mg}{

latex} So you should replace *2k* in the above expressions with *(

{~}1{~

}\right)^{2} +

k

\frac{

~

1}{2}mv^{

~

2}

)* . This makes the frequency of oscillation {latex}

-mg\left(\frac{mg}{k}\right) \]\end{large}

or

Latex
\begin{large} \[ \

omega

Delta E_{

\rm new

2} = \

sqrt

frac{

\frac{k_{1} + k_

1}{2}mv^{2} - \frac{(mg)^{2}}{

m

2k}

}

 \]\end{large}

{latex} {cloak:answerC} {cloak:mathC} {cloak:appC} {card} {deck} {td} {tr} {table} {live-template:RELATE license}

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 {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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Key

Part A

Solution

Part B

Solution

Part C

Solution