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

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


{cloak:mathA}
{cloak:appA}

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

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

]

{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}

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 {

Solution

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:

*  {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}\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 ΔE₂

 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 = 

From Part A we know that a = mg/k

,

so,

substituting:

}\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

}\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}

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{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}

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).