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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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We ignore the distribution of tensions in the upper cables, and simply view the pendulum as a simple pendulum along either the plane of the drawing or perpendicular to it. In the plane perpendicular to the drawing (where the mass oscillates toward and away from the reader) the pendulum length is *L{~}2{~}* and the angular frequency of oscillation is given by the formula for the Simple Pendulum (see [Simple Harmonic Motion].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}\[ \omega_m \frac{d^{2} = \sqrt{\frac{gx}{L_dt^{\rm 2}}} + 2kx = 0 \]\end{large}{latex}

Along
This is the planefamiliar lyingequation inof the page, where[simple harmonic oscillator] (although the massusual movesmultiplier leftfor and right, the pendulum length is*x* is now *2k* instead of the shorterusual *L{~}1{~}* and the angular frequency isk*, because there are two springs). This has the solution:

{latex}\begin{large}\[ \omega_{1} = \sqrt{\frac{g}{L_{\rm 1}}} \]\end{large}{latex}

the ratio of frequencies is thus:

{latex}\begin{large}\[ \frac{\omega_{2}}{\omega_{1}}= \frac{\sqrt{\frac{g}{L_{2}}}}{\sqrt{\frac{g}{L_{1}}}} = \sqrt{\frac{L_{1}}{L_{2}}} \]\end{large}{latex}

in order to have a ratio of 1:2, one thus needs pendulum lengths of ratio 1:4. In order to get a ratio of 3:4 (as in the figure at the top of the page), the lengths must be in the ration 9:16.

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