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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
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:
\begin
[ m \frac{d^
x}{dt^{2}} + 2kx = 0 ]\end
This is the familiar equation of the [simple harmonic oscillator] (although the usual multiplier for x is now 2k instead of the usual k, because there are two springs). This has the solution:
\begin
[ x = A sin(\omega t) + B cos(\omega t) ]\end
where:
\begin
[ \omega = \sqrt{\frac
{m}} ]\end
and A and B are determined by the initial conditions, the initial position xi and the initial velocity vi
\begin
[ A = x_
]\end
\begin
[B = -{\frac{v_{i}}{\omega}} ]\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 figu
re at the top of the page), the lengths must be in the ration 9:16.