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
An undamped oscillator has period \(\tau_{\mathrm{o}}=1.000 \mathrm{~s}\), but I now add a little damping so that its period changes to \(\tau_{1}=1.001 \mathrm{~s}\). What is the damping factor \(\beta\)? By what factor will the amplitude of oscillation decrease after 10 cycles? Which effect of damping would be more noticeable, the change of period or the decrease of the amplitude?
Solution
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The angular frequency change due to damping can be expressed as,
\(\begin{aligned} \omega_{1} & =\sqrt{\omega_{0}^{2}-\beta^{2}} \\ \omega_{1}^{2} & =\omega_{0}^{2}-\beta^{2} \\ \beta^{2} & =\omega_{0}^{2}-\omega_{1}^{2} \\ \beta & =\omega_{0} \sqrt{1-\frac{\omega_{1}^{2}}{\omega_{0}^{2}}} \end{aligned}\) ….. (1)
The ratio of periods of undamped oscillator to the little damping (damped oscillator) can be expressed as,
\(\frac{\omega_{1}^{2}}{\omega_{0}^{2}}=\frac{\tau_{0}^{2}}{\tau_{1}^{2}}\)
For \(\tau_{0}=1.000 \mathrm{~s}\), \(\tau_{1}=1.001 \mathrm{~s}\),
\(\begin{array}{l} \frac{\omega_{1}^{2}}{\omega_{0}^{2}}=\frac{(1.000)^{2}}{(1.001)^{2}} \\ \frac{\omega_{1}^{2}}{\omega_{0}^{2}}=0.998 \\ \omega_{1}^{2}=0.998 \omega_{0}^{2} \end{array}\)
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