2. (35 pts) Consider an unstretched spring hanging from a fixed support. When a mass mis attached
to the end of the spring and gently lowered, it reaches equilibrium when the spring is stretched by an
amount ℓ0.
a) When the mass is pulled an additional distance xbeyond equilibrium and released, show that the
equation of motion only depends on the amount of stretch beyond equilibrium xand not ℓ0.
At equilibrium, the net force is gravity plus the spring force (taking down to be positive)
F=mg −kℓ0= 0
The equation of motion is then
F=mg −k(ℓ0+x) = ma
Given the equilibrium condition above, this reduces to
mg −kℓ0−kx =mg −mg −kx =ma
¨x+k
mx= 0
The equation of motion thus does not depend on the equilibrium stretch ℓ0.
b) Now consider the same spring and mass but in a system that has a velocity dependent friction,
~
f=−b~v. Modify your equation of motion from (a) to include this frictional force and show that the
solution of this equation gives rise to three distinct cases. You only need to show how to mathematically
distinguish the three cases in the solution to the equation of motion, you do not need to name nor
discuss the physical differences in the resulting motion.
Now we add the friction force to the equation of motion above, giving
¨x+b
m˙x+k
mx= 0
If we make the substitutions 2β=b/m and ω2=k/m we have
¨x+ 2β˙x+ω2x= 0
This has exponential solutions of the form
x(t) = XAierit
where riare the roots of the characteristic equation associated with the differential equation of motion,
r2+ 2βr +ω2= 0
The roots of this equation are
r=−2β±p4β2−4ω2
2
r=−β±pβ2−ω2
