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Mass on a frictionless surface between two springs |
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
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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.
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Mathematical Representation
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 L2 and the angular frequency of oscillation is given by the formula for the Simple Pendulum (see Simple Harmonic Motion.
\begin
[ \omega_
= \sqrt{\frac
{L_
}} ]\end
Along the plane lying in the page, where the mass moves left and right, the pendulum length is the shorter L1 and the angular frequency is
\begin
[ \omega_
= \sqrt{\frac
{L_
}} ]\end
the ratio of frequencies is thus:
\begin
[ \frac{\omega_{2}}{\omega_{1}}= \frac{\sqrt{\frac
{L_{2}}}}{\sqrt{\frac
{L_{1}}}} = \sqrt{\frac{L_{1}}{L_
}} ]\end
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.