A massless spring has unstretched length /0 and force constant k. One end is now attached to the ceiling and a mass m is hung from the other. The equilibrium length of the spring is now /1. (a) Write down the condition that determines 11. Suppose now the spring is stretched a further distance x beyond its new equilibrium length. Show that the net force (spring plus gravity) on the mass is F = kx. That is, the net force obeys Hooke's law, when x is the distance from the equilibrium position a very useful result, which lets us treat a mass on a vertical spring just as if it were horizontal. (b) Prove the same result by showing that the net potential energy (spring plus gravity) has the form U (x) = const ikx2.
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Textbook Solutions for Classical Mechanics
Question
The mass shown from above in Figure 5.27 is resting on a frictionless horizontal table. Each of the two identical springs has force constant k and unstretched length \(l_{\mathrm{o}}\). At equilibrium the mass rests at the origin, and the distances a are not necessarily equal to \(l_{\mathrm{o}}\). (That is, the springs may already be stretched or compressed.) Show that when the mass moves to a position (x, y), with x and y small, the potential energy has the form (5.104) (Problem 5.14) for an anisotropic oscillator. Show that if \(a<l_{0}\) the equilibrium at the origin is unstable and explain why.
Solution
Step 1 of 4
When the mass is at (x, y) let the length of the springs are \(I_{1}\) and \(I_{2}\).
These lengths are shown in the figure below:
From the figure the length,
\(\begin{aligned}I_{1} & =\sqrt{(a+x)^{2}+y^{2}} \\ & =a\left(1+\frac{2 x}{a}+\frac{x^{2}+y^{2}}{a^{2}}\right)^{\frac{1}{2}} \\ & \approx a\left[1+\frac{1}{2}\left(\frac{2 x}{a}+\frac{x^{2}+y^{2}}{a^{2}}\right)-\frac{1}{8}\left(\frac{2 x}{a}\right)^{2}\right] \\ & =a+x+\frac{y^{2}}{2 a}\end{aligned}\)
full solution