Quiz #4

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This quiz is due at the beginning of class on the 5th of December (Friday). You are on your honor not to consult others or reference materials in the completion of this quiz. This includes textbooks and notes. Please review chapter 12 material before beginning this quiz. If you can't figure out an answer to one part, make a guess, and use that answer in the following parts.

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1.) (25 Points) A 1.5 kg object that stretches a particular spring 2.8 cm from its natural length when hanging at rest oscillate with an amplitude of 2.2 cm.

General Comments:
I accepted a lot of answers on this quiz becuase the wording of the question aparently confused the majority of the class. Below is the intended solution. Please come see me if you are still unsure of why this is the correct solution.
 

a) What is the total energy of the system?

Solution:
The answer to this depends on the fact that gravitational potential energy of the system is zero at the equillibrium of the oscillations. In other words, when the spring is hanging motionless at the end of the spring (at equillibrium), the gravitational potential energy of the system is zero. At that point, there is potential energy in the spring because it is stretched a distance of .028 m.
First, we need to find the spring constant, k. When hanging at equillibrium, the sum of the forces is equal, so
mg = kx
$k =\frac{mg}{.028 m} ~=~ 525 N/m$
Now, consider the object while its oscillating. At the lowest point of its oscillation, the total energy of the system is a sum of the potential energy stored in the spring and the gravitational potential energy. At this point, the kinetic energy is zero.
E_tot = U_spring + U_gravity
$E_{tot} = \frac{1}{2}kx^{2} + mg\Delta x$
$E_{tot} = \frac{1}{2}(525 N/m)(.028 m + .022 m)^{2} + (1.5 kg)(g)(-.022 m)$
Note that the x used in the spring force is the TOTAL stretch of the spring and is measured from the point when the spring was unstretched, or x = 2.8 cm + 2.2 cm.
E_tot = .656 J - .324 J
E_tot = .333 J
 

b) If the gravitational potential energy of the system is defined to be zero when the object is in equillibrium, then at maximum downward displacement find the gravitational potential energy.

Solution:
If the zero of the potential energy is zero at equillibrium, then if the object is oscillating with an amplitude of 2.2 cm about equillibrium, then
$U_{gravity} = mg\Delta x$
U_gravity = mg(-.022 m)
U_gravity = -.324 J
 

c) Also at maximum downward displacement, find the potential energy stored in the spring.

Solution:
This one fooled almost everyone. The total amount that the spring is stretched is 2.8 cm + 2.2 cm = 5.0 cm, so the potential energy in the spring is
$U_{spring} = \frac{1}{2} k (.05 m)^{2}$
U_spring = .656 J
 

d) Find the kinetic energy of the object when displaced halfway from equillibrium, or 1.1 cm.

Solution:
I didn't state whether this was above or below the equillibrium position, so either answer was acceptable. Below is the case for the object 1.1 cm BELOW equillibrium.
The easiest way to do this is with conservation of energy.
E_tot = KE + U_spring + U_gravity
so the kinetic energy at x = 1.1 cm is
KE = E_tot - U_spring - U_gravity
$KE = .333 J - \frac{1}{2}k(.028m + .011m)^{2} - mg(-.011m)$
KE = .333 J - .399 J - (-.162 J)
KE = .096 J
 

e) Write an equation for the position and for the velocity of the object as a function of time t, and the values given in the problem. Choose t=0 when the object is at the equillibrium position. Recall that

$\omega^{2} = \frac{k}{m}$
$v = \frac{dx}{dt}$
$U_{spring} = \frac{1}{2} k x^{2}$

Solution:
The general form of harmonic motion as a function of time is
$x(t) = A \cos (\omega t + \phi)$
Where we know that $\omega$ is given by
$\omega = \sqrt{\frac{k}{m}}$
$\omega = \sqrt{\frac{525}{1.5}}$
$\omega = 18.7 rad/sec$
We also realize that the constant A is the amplitude of oscillation, here
A = .022 m
The problem left is finding the constant $\phi$. Plugging in when t=0, we find,
$x(0) = .022 \cos (0 + \phi)$
and at t=0, we were told that the object was at the equillibrium position, or
x(0) = 0
So, $\cos \phi = 0$
which implies that
$\phi = \frac{\pi}{2}$
So, the solution for the position of our of object as a function of time is
$x(t) = .022 m \cos (18.7 t + \frac{\pi}{2})$
 
Now, to find the velocity as a function of time, we take the derivative of the position.
$v(t) = -(.022)(18.7) \sin (18.7 t + \frac{\pi}{2})$
$v(t) = -.412 m/s ~\sin (18.7 t + \frac{\pi}{2})$