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Solution

The mass m is subjected to force from the springs on each side. If you assume that the springs are in their relaxed state when the mass is at rest between them, then displacement of the mass to the right (as shown) compresses the spring on the right and extends the spring on the left. this results in restoring force to the left from both springs.

Displacing the mass a distance x to the right results in restoring force from both springs. The spring to the right is compressed while the spring on the left is extended. The force from the spring on the right will be -kx (assuming, as we have, that the spring is at its neutral position, neither compressed nor extended, at the start). The force from the spring on the left will also be -kx, and will be in the same direction as the force from the first spring. The sum of these two forces is -2kx. Equating this to the force on the mass, which is equal to the mass times the acceleration, gives us the equation:

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|!Mass Between Two Springs.PNG!|
|Mass on a frictionless surface between two springs|



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{excerpt:hidden=true}A case of [Simple Harmonic Motion].{excerpt}

 



h4. Solution

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

{toggle-cloak:id=int} *Interactions:*  {cloak:id=int} The forces due to the compression or extension of the two springs acting as the [restoring force].{cloak}

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

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

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

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|!Mass With Two Springs Displaced.PNG!|

The mass *m* is subjected to force from the springs on each side. If you assume that the springs are in their relaxed state when the mass is at rest between them, then displacement of the mass to the right (as shown) compresses the spring on the right and extends the spring on the left. this results in [restoring force] to the left from _both_ springs. 

|!Mass With Two Springs Displaced with forces.PNG!|
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{toggle-cloak:id=math} {color:red} *Mathematical Representation* {color}

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Displacing the mass a distance *x* to the right results in restoring force from _both_ springs. The spring to the right is compressed while the spring on the left is extended. The force from the spring on the right will be {*}-kx* (assuming, as we have, that the spring is at its neutral position, neither compressed nor extended, at the start). The force from the spring on the left will also be {*}-kx*, and will be in the same direction as the force from the first spring. The sum of these two forces is {*}-2kx*.  Equating this to the force on the mass, which is equal to the mass times the acceleration, gives us the equation:


{latex}\begin{large}\[ m \frac{d^{2}x}{dt^{2}} + 2kx = 0 \]\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}

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:

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

where:

}\begin{large}\[ \omega = \sqrt{\frac{2k}{m}} \]\end{large}{latex}

and *A* and *B

and A and B are determined by the initial conditions, the initial position x_i and the initial velocity v_i

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

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

\}

\\

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

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.

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} \[ F_{\rm total} = F_{\rm right}- F_{\rm left} \]\end{large}

or

{latex}
or
{latex}\begin{large} \[ F_{\rm total} = - k (x + \Delta ) - (- k ( x - \Delta) = -2 k x \]\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.

...

...

...

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}\begin{large} \[ F_{\rm total} = - ( k_{1} + k_{2} )x \]\end{large}{latex}

So

...

you

...

should

...

replace

...

2k

...

in

...

the

...

above

...

expressions

...

with

...

(

...

k

...

₁

...

+

...

k

...

₂

...

)

...

.

...

This

...

makes

...

the

...

frequency

...

of

...

oscillation

}\begin{large}\[ \omega_{\rm new} = \sqrt{\frac{k_{1} + k_{2}}{m}} \]\end{large}{latex}

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