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
Consider the mass attached to four identical springs, as shown in Figure 5.7(b). Each spring has force constant k and unstretched length \(l_{\mathrm{o}}\), and the length of each spring when the mass is at its equilibrium at the origin is a (not necessarily the same as \(l_{\mathrm{o}}\)). When the mass is displaced a small distance to the point (x, y), show that its potential energy has the form \(\frac{1}{2} k^{\prime} r^{2}\) appropriate to an isotropic harmonic oscillator. What is the constant \(k^{\prime}\) in terms of k? Give an expression for the corresponding force.
Solution
Step 1 of 3
The two springs are attached along the x-axis and two along the y-axis.
The potential energy of the two springs along x-axis is,
\(U_{x}=k\left[x^{2}+\left(1-\frac{I_{0}}{a}\right) y^{2}\right]\)
The potential energy of the two springs along y-axis is,
\(U_{y}=k\left[y^{2}+\left(1-\frac{I_{0}}{a}\right) x^{2}\right]\)
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