Transcript ch7

Physics 231
Topic 7: Oscillations
Alex Brown
October 14-19 2015
MSU Physics 231 Fall 2015
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Key Concepts: Springs and Oscillations
Springs
Periodic Motion
Frequency & Period
Simple Harmonic Motion (SHM)
Amplitude & Period
Angular Frequency
SHM Concepts
Energy & SHM
Uniform Circular Motion
Simple Pendulum
Damped & Driven Oscillations
Covers chapter 7 in Rex & Wolfson
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Springs
xo=0
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Hooke’s Law
xo=0
Fs = -kx
Hooke’s law
k = spring constant in units
of N/m.
x = displacement from
the equilibrium point xo=0
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How to find the spring constant k
When the object hanging from
the spring is not moving:
∑F = ma = 0
x
Fspring + Fgravity = 0
x=0
-k(d) + mg = 0
Fspring = -kx
x=d
k = mg/d
Fgravity = mg
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Springs
Hooke’s Law:
Fs = -kx
k: spring constant (N/m)
x: distance the spring is stretched from
equilibrium
W = area under F vs x plot = ½kx2
The energy stored in a spring depends
on how far the spring has been
stretched: elastic potential energy
PEspring = ½ k x2
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PINBALL!
The ball-launcher spring has a
constant k = 120 N/m.
A player pulls the handle 0.05 m.
The mass of the ball is 0.1 kg.
What is the launching speed?
PEspring = ½ k x2 PEgravity = mgh KE = ½ mv2 ME = PE + KE
= Constant
(PEgravity+PEspring+KEball)pull = (PEgravity+PEspring+KEball)launch
mghpull + ½kxpull2 + ½mvpull2 = mghlaunch + ½kxlaunch2 + ½mvlaunch2
½120(0.05)2 = 0.1 (9.81) ( 0.05*sin(10o) ) + ½(0.1)vlaunch2
0.15 = 8.5x10-3 + 0.05v2
v =1.7 m/s
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Momentum p = mv
Impulse
p = pf - pi = Ft
Conservation of momentum (closed system) p1i + p2i = p1f + p2f
Collisions in one dimension given
given m1 and m2 or c = m2 /m1
v1i and v2i
Inelastic (stick together, KE lost)
vf = [v1i + c v2i] / (1+c)
Elastic (KE conserved)
v1f = [(1-c) v1i + 2c v2i] / (1+c)
v2f = [(c-1) v2i + 2v1i] / (1+c)
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Collisions in one dimension given
given m1 = m2 or c = 1
v1i and v2i
Inelastic (stick together, KE lost)
vf = [v1i + v2i] / 2
Elastic (KE conserved)
v1f = v2i
v2f = v1i
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Carts on a spring track
m1
v
spring
m2
|||||||||||
k = 50 N/m
v = 5.0 m/s
m1 = m2 = 0.25 kg
What is the maximum compression of the spring
if the carts collide a) elastically and b) perfectly inelastic?
A) Conservation of momentum and KE with c = 1
v1f = v v2f = 0
:: v1f = 0
v2f = v = 5 m/s
Conservation of energy: ½mv2 = ½kx2
(0.5) (0.25) (5)2 = (0.5)(50)x2
x = 0.35 m
B) Conservation of momentum only with c = 1
vf = (v1f + v2f)/2 = (v + 0)/2= 2.5 m/s
Conservation of energy: ½mvf2 = ½kx2
(0.5)(0.5)(2.5)2 = (0.5)(50)x2
x=0.25 m
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Hooke’s Law
Fs = -kx
Hooke’s law
If there is no friction, the
mass continues to oscillate
back and forth.
If a force is proportional to the displacement x, but
opposite in direction, the resulting motion of the object is
called: simple harmonic oscillation
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Simple harmonic motion
displacement x
A
c)
b)
time (s)
a)
(b)
Amplitude (A): maximum distance from
equilibrium (unit: m)
Period (T): Time to complete one full
oscillation (unit: s)
Frequency (f): Number of completed
oscillations per second. f=1/T
(unit: 1/s = 1 Herz [Hz])
(a)
Angular frequency  = 2πf = 2π/T
(c)
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Concept Quiz
displacement x
5m
2
4
6
-5 m
8
10 time (s)
a) what is the amplitude of the harmonic oscillation?
b) what is the period of the harmonic oscillation?
c) what is the frequency of the harmonic oscillation?
a) Amplitude: 5 m
b) period: time to complete one full oscillation: T = 4s
c) frequency: number of oscillations per second
f = 1/T = 0.25/s = 0.25 Hz
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What the equation?
displacement x
5m
2
4
6
8
-5 m
10 time (s)
x = A cos(t)
A=5
T = 2π
 = 2π/T = 2π/4 = 1.57 rad/s
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Clicker Quiz!
A mass on a spring in SHM has amplitude A and period T. What is
the total distance traveled after a time interval of 2T?
A)
B)
C)
D)
E)
0
A/2
A
4A
8A
In one cycle covering period T, the mass
moves from +A to -A and back: 2A+2A = 4A
In two periods, the mass moves twice this
distance: 8A
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Energy and Velocity
Ekin(½mv2)
x=+A
x=0
x=-A
conservation of ME:
Epot,spring(½kx2)
0
½kA2
½kA2
0
½mv2max
½mv2max
0
½mv2max = ½kA2
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Sum
½k(-A)2
so
½kA2
vmax = ±A(k/m)
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Velocity and acceleration in general
Total ME at any displacement x:
½mv2 + ½kx2
Total ME at max. displacement A: ½kA2
Conservation of ME: ½kA2 = ½mv2 + ½kx2
So: v = ±[(A2-x2)k/m]
also F = ma = -kx so a = -kx/m
position x
velocity v
Acceleration a
+A
0
-kA/m
0
±A(k/m)
0
-A
0
kA/m
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An Example
The maximum velocity and acceleration of mass connected to a
spring are vmax=0.95 m/s and amax=1.56 m/s2, respectively.
Find the oscillation amplitude.
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An Example
The maximum velocity and acceleration of mass connected to a
spring are vmax=0.95 m/s and amax=1.56 m/s2, respectively.
Find the oscillation amplitude.
We do not know the mass of the object or the spring constant. So we
must find a way to eliminate these variables!
so
vmax = ±A(k/m)
so (vmax )2 = A2 (k/m)
so amax = A (k/m)
A = vmax2/amax = (0.95)2/(1.56) = 0.58 m
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Harmonic oscillations vs circular motion
x(t) = A cos
The projection of the position of the circulating
object on the x-axis as a function of time is the
same as the position of the oscillating spring.
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r=A

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Harmonic oscillations vs circular motion
vx(t) = -v sin
v
vx


The vx projection of the linear velocity of the
rotating object is the velocity of the mass on the
spring.
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r=A
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x=+A
But remember for circular motion
with constant speed around the
circle  = t and v = r = A
where  is the “angular frequency”
x=0
x=-A
So x(t) = A cos = A cos(t)
vx(t) = -v sin = - A sin(t)
r=A

time to complete one circle
= T = one period
so T = 2
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or  = 2/T
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Harmonic oscillations vs circular motion
The simple harmonic motion can be described by the
projection of circular motion on the horizontal axis.
x(t) = A cos(t)
v(t) = -A sin(t)
•
•
•
•
A
 = 2/T = 2f
T
f = 1/T
amplitude of the oscillation
angular frequency
period
frequency.
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x = A cos(t)
A
x
time (s)
-A
A
velocity v = -A  sin(t)
v
time (s)
-A 
Remember
vmax
k
A
m
thus
k

m
The angular frequency does not depend on A
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Displacement vs Acceleration
displacement x
A
time (s)
-A
Newton’s second law: F=ma  -kx=ma  a=-kx/m
acceleration(a)
kA/m
-kA/m
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x = A cos(t)
A
x
time (s)
-A
A
v = -A  sin(t)
v
time (s)
-A 
A 2
acceleration a = -A 2 cos(t)
time (s)
a
-A 2
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Period of a Spring
2
m
T
 2

k
Increase m:
increase T
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Increase k:
decrease T
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Clicker
Quiz!
C
A mass is oscillating horizontally while attached to a spring
with spring constant k. Which of the following is true?
a) When the magnitude of the displacement is largest,
the magnitude of the acceleration is also largest.
b) When the displacement is positive, the acceleration is
also positive
c) When the displacement is zero, the acceleration is
non-zero
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The pendulum
x=+A
x=-A
x=0
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The pendulum
 = max = A
y=h
Conservation of Energy!
At x=+A: PE = mgh
KE = ½mv2 = 0
At x=0: PE = mg(0) = 0 KE = ½mv2
½mv2 = mgh
v2 = 2gh velocity at the bottom
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The pendulum


Tension at max angle  = max
T = mg cosmax
L
Tension at bottom  = 0
T = mg + mv2/L
T
s
mg sin

mg cos
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The pendulum
Restoring force: F=-mg sin
The force pushes the mass m
back to the central position.
L

T
s
mg sin

mg cos
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The pendulum
Restoring force: F=-mg sin
The force pushes the mass m
back to the central position.
L
sin   (radians) if  is small

T
F = - mg
s
mg sin

so:
also s = L 
F=-(mg/L)s
mg cos
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pendulum
vs spring
The pendulum
k
mg / L



m
m
spring
g
L
Does not depend on mass
pendulum
F  kx
F  (mg / L) s
k

m
g

L
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Damped Oscillations
Newton’s 1st Law: An object will stay in
motion unless acted upon by a force.
SHM is the same: oscillations will last
forever in the absence of additional
forces.
Common force of friction often
“damps” oscillations and brings them
to a stop.
In homework A is reduced by some
factor f after one period
After one oscillation A1 = f A
So for n oscillations An = fn A
mechanical energy ME = (1/2) k A2
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Driven Oscillations
With driven oscillation, the
amplitude grows!
The same can be applied in reverse:
An object NOT in motion can be put
into oscillation by applying a force
This is known as driven oscillations.
If the force acts with the same
frequency as the as that of the object
this is known as resonance.
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Spring problem
A bungee jumper with height h=2 m and mass m = 80 kg leaps from a
bridge with height H = 30 m. A bungee cord with spring constant k =
100 N/m attached to his legs. What is the maximum length the cord
needs to be if he is to avoid hitting the water below?
Define A = extended “amplitude” of the bungee cord
Total extension = jumper’s height + nominal length of the cord
+ extended length of the cord = h+L+A
Must stop before h+L+A = H = 30m, or A = H-L-h
Use conservation of ME: ½mv2 + ½kx2 + mgH
On top of the bridge: ME = mgH
Maximum extension of the cord:
ME = ½ kA2 = ½ k(H-L-h)2
ME(bridge) = ME(ground)
mgH = ½k(H-L-h) 2
(80 kg)(9.81 m/s2)(30m) = ½(100 N/m)(30m-L-2m) 2
23544 = 50 (28m-L) 2
(28m-L) = sqrt(23544/50) = 21.7 m
L = 6.3 m
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Spring problem
A h=2m tall, m=80 kg bungee jumper leaps from a
H=30m bridge with a bungee cord with spring
constant k = 100 N/m attached to his legs.
What is the frequency of his subsequent SHM?
𝐓𝐬𝐩𝐫𝐢𝐧𝐠
= 𝟐𝛑
Frequency
𝐦
= 𝟐𝛑
𝐤
𝟖𝟎
𝟏𝟎𝟎
= 5.62 sec
f = 1/T
= 1/5.62s = 0.18 Hz
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Spring problem
A mass of 0.2 kg is attached to a spring with k=100 N/m.
The spring is stretched over 0.1 m and released.
a) What is the angular frequency () of the corresponding
motion?
b) What is the period (T) of the harmonic motion?
c) What is the frequency (f)?
d) What are the functions for x,v and t of the mass
as a function of time? Make a sketch of these.
a)  =(k/m) = (100/0.2) = 22.4 rad/s
b)  = 2/T
T= 2/ = 0.28 s
c)  = 2f
f=/2 = 3.55 Hz (=1/T)
d) x(t) = Acos(t)
= 0.1 cos(22.4t)
v(t) = -Asin(t) = -2.24 sin(22.4t)
a(t) = -2Acos(t) = -50.2 cos(22.4t)
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0.1
x
-0.1
0.28
0.56
0.28
0.56
0.28
0.56
time (s)
velocity v
2.24
-2.24
50.2
a
-50.2
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pendulum problem
The machinery in a pendulum clock is kept
in motion by the swinging pendulum.
Does the clock run faster, at the same speed,
or slower if:
a) The mass is hung higher
b) The mass is replaced by a heavier mass
c) The clock is brought to the moon
d) The clock is put in an upward accelerating
elevator?
L

g
L
m
Moon
Elevator
Faster(A)
Same(B)
Slower(C)
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Pendulum Quiz
A pendulum with length L of 1 meter and a mass
of 1 kg, is oscillating harmonically. The period of
oscillation is T.
If L is increased to 4 m, and the mass replaced
by one of 0.5 kg, the period becomes:
A) 0.5 T
L
T  2
B) 1 T
g
C) 2 T
4L
L
T   2
 4
 2T
D) 4 T
g
g
E) 8 T
Mass does not matter
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Spring problem
A mass of 1 kg is hung from a spring. The spring stretches
by 0.5 m. Next, the spring is placed horizontally and fixed
on one side to the wall. The same mass is attached and the
spring stretched by 0.2 m and then released. What is
the acceleration upon release?
1st step: find the spring constant k
Fspring =-Fgravity or -kd =-mg
k = mg/d = (1)(9.8)/0.5 = 19.6 N/m
2nd step: find the acceleration upon release
Newton’s second law: F=ma  -kx = ma 
a = -kx/m
a = -(19.6)(0.2)/1 = -3.92 m/s2
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Sprng problem
A block with mass of 200 g is placed
over an opening. A spring is placed under the
opening and compressed by a distance d=5
cm. Its spring constant is 250 N/m. The
spring is released and launches the block.
How high will it go relative to its rest
position (h)? (the spring can be assumed
massless)
x
d
ME =
½mv2
Initial: v = 0
+ ½kd2
+ mgx
d = -0.05 m
must be conserved.
x=0
m = 0.2 kg
0 + (0.5)(250)(0.05)2 + 0
Final: v = 0
0 +
d=0
0
Conservation of ME:
= 0.3125
x=h
+ (0.2)(9.8)(h) = 1.96 h
0.3125 = 1.96 h
so h = 0.16 m
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