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Image Added

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.

Deck of Cards

id	bigdeck

{excerpt} {deck:id=bigdeck} {card:label=Part A} 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} {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. |!Vertical Mass on Spring Static Extension.PNG!| |!Vertical Mass on Spring Static Extension Forces.PNG!| {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}

Card

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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id	sysA

System:

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id	sysA

.

Toggle Cloak

id	intA

Interactions:

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

Toggle Cloak

id	modA

Model:

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id	modA

.

Toggle Cloak

id	appA

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.

Image Added

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	diagA
	diagA

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id	mathA

Mathematical Representation

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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

\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 (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} {cloak:id=diagB} |!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* . {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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	mathA

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	appA

Card

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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id	sysB

System:

Cloak

id	sysB

.

Toggle Cloak

id	intB

Interactions:

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id	intB

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

Toggle Cloak

id	modB

Model:

Cloak

id	modB

.

Toggle Cloak

id	appB

Approach:

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Diagrammatic Representation

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Image Added

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

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	diagB
	diagB

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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

[

Simple

Harmonic

Motion

] {latex}

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}\


\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 (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!| {cloak:diagC} {toggle-cloak:id=mathC} {color:red} *Mathematical Representation* {color} {cloak:id=mathC} 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}

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Card

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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id	sysC

System:

Cloak

id	sysC

.

Toggle Cloak

id	intC

Interactions:

Cloak

id	intC

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

Toggle Cloak

id	modC

Model:

Cloak

id	modC

.

Toggle Cloak

id	appC

Approach:

Cloak

id	appC

Toggle Cloak

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Diagrammatic Representation

Cloak

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Image Added

Cloak

	diagC
	diagC

Toggle Cloak

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

is ΔE₂

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 =

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

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}

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}

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}

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)

]. {cloak:mathC} {cloak:appC} {card} {deck}

.

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Key

Part A

Solution

Part B

Solution

Part C

Solution