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
A massless spring is hanging vertically and unloaded, from the ceiling. A mass is attached to the bottom end and released. How close to its final resting position is the mass after 1 second, given that it finally comes to rest 0.5 meters below the point of release and that the motion is critically damped?
Solution
Step 1 of 5
The following are given by the question:
The distance where mass comes to rest potion, \(h=0.5 \mathrm{~m}\) from the releasing point.
Time for mass to come in rest, \(t=1 \ \mathrm{sec}\)
At the equilibrium, the gravitational force on the mass and upward restraining force would be equal to each other.
The gravitational force, \(F_{g}=m g\)
Here, m is mass and g is the acceleration due to gravity.
The restraining force, \(F_{r}=k h\)
Here, k is the spring constant and h is the stretched length.
Equate both the forces
\(\begin{aligned}
m g & =k h \\
\frac{k}{m} & =\frac{g}{h}
\end{aligned}\)
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