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
Another interpretation of the \(Q\) of a resonance comes from the following: Consider the motion of a driven damped oscillator after any transients have died out, and suppose that it is being driven close to resonance, so you can set \(\omega=\omega_{\mathrm{o}}\). (a) Show that the oscillator's total energy (kinetic plus potential) is \(E=\frac{1}{2} m \omega^{2} A^{2}\). (b) Show that the energy \(\Delta E_{\text {dis }}\) dissipated during one cycle by the damping force \(F_{\mathrm{dmp}}\) is \(2 \pi m \beta \omega A^{2}\). (Remember that the rate at which a force does work is \(Fv\).) (c) Hence show that \(Q\) is \(2 \pi\) times the ratio \(E / \Delta E_{\text {dis }}\).
Solution
Step 1 of 7
(a)
The displacement of a driven damped oscillator is given as,
\(x=A \cos (\omega t-\delta)\)
Here, A is the amplitude, \(\omega\) is the angular frequency, t is the time and \(\delta\) is the phase constant.
The amplitude of an oscillator driven by a sinusoidal force with variable frequency is given as,
\(A=\frac{f_{\mathrm{o}}}{\sqrt{\left(\omega_{\mathrm{o}}^{2}-\omega^{2}\right)+4 \beta^{2} \omega^{2}}}\)
Here, \(\omega_{\mathrm{o}}\) is the natural angular frequency, \(\beta\) is the decay parameter and \(f_{\mathrm{o}}\) is the Fourier coefficient.
The velocity of the particle is given as,
\(v=\frac{d x}{d t}\)
Substitute \(A \cos (\omega t-\delta)\) for x,
\(\begin{aligned} v & =\frac{d}{d t}[A \cos (\omega t-\delta)] \\ & =-A \omega \sin (\omega t-\delta) \end{aligned}\)
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