Worksheet 13 Key: Simple Harmonic Motion
A crate of cabbages with a mass of 750kg slides along a level, frictionless track with a speed
of 5.0m/s until it meets a spring with spring constant 12000N/m. As the spring compresses,
it slows the crate to a momentary stop.
+x
1. Think the process starting when the crate encounters the spring to when it stops as part of
a cycle of a simple harmonic oscillation x = Acos(t+).
a. What is the angular frequency of the oscillator?
2 = k/m = (12000/750)/s2 = 16/s2, so = 4 rad/s.
b. What is the amplitude A of the oscillation?
A = x02 + v02/2 = 0 + v02/2, so A = v0/ = (5.0 m/s)/(4/s) = 1.25 m.
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c. What fraction of a full cycle does this stopping process occupy?
It runs from x = 0 to x = maximum, so that is 1/4 of a cycle.
d. At what phase angle does this stopping process begin?
Cos() =0 at = /2 and 3/2. At /2 it is moving in the negative direction, and at
3/2 it is moving in the positive direction. So the initial phase must be = 3/2.
v. What is the maximum velocity A in the oscillation?
A = (1.25 m)(4 rad/s) = 5 m/s.
e. What is the maximum acceleration A2 in the oscillation?
A2 = (1.25 m)(16 rad2/s2) = 20 m/s2.
f. What is the period T of one oscillation?
Since = 2f and T = 1/f, T = 2/ = 2/(4/s) = 1.571 s.
g. Once the crate encounters the spring, how much time does it take to stop?
We have 1/4 of a cycle, so the time is T/4 = 0.393 s.
2. Let’s analyze this using the tools we knew before studying oscillations.
a. What is the initial kinetic energy of the crate?
Ki = 1/2 mv02 = (1/2)(750 kg)(5.0 m/s)2 = 9375 J.
b. What is the final potential energy of the spring?
By conservation of energy, Ki = Uf, so Uf = 9375 J.
c. How far does the crate compress the spring in stopping?
Ki = Uf
1/2 mv02 = 1/2 kxf2
mv02/k = xf2
xf = v0 m k = (5 m/s)
(750 kg) (12000 N/m) =(5 m/s)(0.25 s) = 1.25 m.
Hey! That’s what we found was the amplitude of an oscillation!
d. What is the magnitude of the maximum force the spring exerts on the crate?
Force F = –kx, maximum force will be at the maximum x of 1.25 m. Thus F =
(12000 N/m)(1.25 m) = 15000 N.
e. What is the crate’s maximum acceleration?
By Newton’s second law, a = F/m, so a = (15000 N)/(750 kg) = 20 m/s.
f. What is the total energy of the oscillator?
In this instance, E = Ki = Uf = 9375 J.
3. Now let’s make the spring a damped oscillator, which applies a force F = – bv – kx to the
crate.
a. What must the damping coefficient b be to provide critical damping?
Critical damping occurs when k/m – b2/4m2 = 0, so b = 2 km =
2 (12000 N/m)( 750 kg) = 6000 kg/s.
b. If b were half the critical value, what would be the oscillation angular frequency '?
2
Here b = 300 kg/s, so '=
k
b2
12000 N/m ( 3000 kg/s)
/s =
=
2 = 16 4
2
m 4m
750 kg
4 ( 750 kg)
3.464 rad/s.
PHYS 1210 Worksheet 13 Key
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