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Clicker Question
Room Frequency BA
A mass is oscillating back and forth on a spring as shown. Position 0
is the equilibrium position and position E is the extreme position (its
amplitude).
x
At which position is the magnitude of the acceleration of the mass
maximum?
A) 0
B) M
C) E
D) a is constant everywhere
Acceleration = (Net Force/mass) = -(k/m)x
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Announcements
• CAPA assignment #13 is due on Friday at 10 pm.
• CAPA assignment #14 is in the distribution boxes and is
due Friday December 2nd at 10pm.
• Next week (11/28-12/2) in Section: Lab #6 (with
prelab)
• Finish reading Chapter 11 on Vibrations and Waves
• 10 am Lecture was videotaped by mistake!!! Video files
were deleted.
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Clicker Question
Room Frequency BA
A mass is oscillating back and forth on a spring as shown. Position 0
is the equilibrium position and position E is the extreme position (its
amplitude).
x
At what position is the magnitude of the net force on the mass a
maximum?
A) 0
B) M
C) E
D) force is constant everywhere.
Net force = -kx
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Clicker Question
Room Frequency BA
A mass is oscillating back and forth on a spring as shown. Position 0
is the equilibrium position and position E is the extreme position (its
amplitude).
x
At what position is the total mechanical energy (PE + KE) a maximum?
A) 0
B) M
C) E
D) energy is constant everywhere.
No friction or dissipative forces! Energy is conserved.
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Energy Analysis of Horizontal Spring and Mass
ET  PE  KE
Total Mechanical Energy
PE for Spring with equilibrium position at x = 0 is
As usual, KE =
1
2
1
2
kx 2
mv 2 , so
ET  12 kx 2  12 mv2
Once oscillator is set in motion, total energy is a constant,
hence we can always determine x from v or v from x if we
know ET
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Special Points of
Horizontal Spring and Mass Oscillation
x = -A, v = 0
x = 0, v = ±vmax
x = A, v = 0
Amplitude A = E
ET  kx  mv
1
2
2
1
2
2
At turning points x = ±A, v = 0, ET = only PE= 12 kA2 .
2
At equilibrium point x = 0, v =±vmax, ET = only KE= 12 mvmax
.
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Amplitude - Max speed Relation
x = -A, v = 0
x = 0, v = ±vmax
x = A, v = 0
Amplitude A = E
Total energy is conserved!!! So we can conclude
Related to
acceleration kx/m
2
2
1
1
ET  2 kA  2 mvmax
Solving for vmax in terms of A gives the result
vmax 
k
A.
m
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Room Frequency BA
Clicker Question
A stiff spring and a floppy spring have potential energy diagrams
shown below. Which is the stiff spring? A) PINK B) YELLOW
PE
PE
PE = ½kx2
x
x
Pink
Yellow
A
m
vmax
k
I take two identical masses; one is attached to the stiff spring; the
other to the floppy spring. Both are positioned at x = 0 and given the
same initial speeds. Which spring produces the largest amplitude
motion?
A) The stiff spring
B) The floppy spring
C) Same!
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Room Frequency BA
Clicker Question
A stiff spring and a floppy spring have potential energy diagrams
shown below.
PE
PE
x
x
Pink
PE = ½kx2
A
m
vmax
k
Yellow
Now the identical masses on the two different springs are pulled to
the side and released from rest with the same initial amplitude.
Which spring produces the largest maximum speed of its mass?
A) The stiff spring
vmax 
k
A
m
B) The floppy spring
C) Same!
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Example of Energy Analysis I
You are given that a mass m attached to a spring with spring
constant k was released from rest from position x0 (the spring’s
equilibrium position is x = 0). What is the speed at position x1 in
between 0 and x0?
Yes! A = x0 because it was
Are we given the amplitude?
released from rest.
Total energy = ½kA2 and it is a constant!
ET  12 kx 2  12 mv2  12 kA2  12 kx02
Solve for v and find
k 2
v
(A  x 2 )
m
For final result, plug in x1
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Example of Energy Analysis II
Note what we can do with a little algebraic manipulation…
2
2
k 2
k
x
x
2
v
(A  x )  A
1 2  vmax 1 2 .
m
m
A
A
Units are correct and at x = 0, x = A, we get the correct
results!
If you are given v at x = 0, you know vmax, and you can
find a formula for x given some v with magnitude less
than vmax…
If given a v at a particular x you can always find A and vmax
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Room Frequency BA
Clicker Question
A mass is oscillating back and forth on a spring as shown. Position 0
is the equilibrium position and position E is the extreme position (its
amplitude).
x
What is the speed at the position x = E/2 in terms of vmax?
A) vmax
x2
B) vmax /2
v  vmax 1 2
C) vmax /Sqrt(2)
A
D) vmax Sqrt(3/2)
E) vmax Sqrt(3)/2 v = v sqrt[1-(E2/4E2)] = v sqrt[1-(1/4)]
max
max
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Harmonic Time Dependence of SHM I
To find the time dependence of SHM we use an amazing fact:
The time dependence of one space component of circular
motion is exactly the same as the time dependence of SHM!
Consider a particle moving around a circle of radius A in
the x-y plane, with constant speed vm
vm
The angular velocity ω =Δθ/Δt
A
For constant speed, ω is constant
and so θ = ωt
As usual x = A cos θ so
x = A cos ωt

at t = 0
x
–A
+A
0
at t = 0
x = A sin ωt
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Harmonic Time Dependence of SHM II
How do we relate ω to vm?
One cycle takes time T, called the period.
Time to go around circle is 2πA/vm = T = 2π rad/ω
Solving yields ω = vm/A
vm
SHM has same time dependence,
so x = A cos [(vm/A)t]
A
But for SHM, we derived
vm

A
k
m
k
So we have x  A cos( t)
m

x
–A
+A
0
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SHM Period and Frequencies
We just found that T = 2π rad/ω = 2πA/vm . For SHM then
T  2
Frequency f = 1/T so we find
m
k
1
f 
2
k
m
Independent of amplitude!
Note even for back and forth motion we have an angular frequency ω!

k
in radians/sec for k and m in SI units
m
Cosines and sines must be functions of angles; here we will
always use radians
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Simple Harmonic Oscillator with x = 0 at t = 0
x  Asin  t
= A sin(2 ft)
= A sin(2 t / T )
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Room Frequency BA
Clicker Question
A mass on a spring oscillates with a certain amplitude and a certain
period T. If the mass is doubled, the spring constant of the spring is
doubled, and the amplitude of motion is doubled, THEN the period ...
A) increases
B) decreases
T  2
C) stays the same.
m
k
Ratio unchanged if m and k both doubled and
the period is independent of amplitude
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