Download Physics 231 Topic 7: Oscillations Wade Fisher October 5-10 2012

Physics 231
Topic 7: Oscillations
Wade Fisher
October 5-10 2012
MSU Physics 231 Fall 2012
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Key Concepts: Oscillations
Periodic Motion
Frequency & Period
Simple Harmonic Motion
Amplitude & Period
Angular Frequency
SHM Concepts
Energy & SHM
Uniform Circular Motion
Simple Pendulum
Damped & Driven Oscillations
Covers chapter 7 in Rex & Wolfson
MSU Physics 231 Fall 2012
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Hooke’s Law Redux
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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The spring constant k
When the object hanging from
the spring is not moving:
∑F = ma = 0
Fspring + Fgravity = 0
-k(-d) - mg = 0
k = mg/d
x=0
x=-d
x=-d
Fspring = -kx
Fg = -mg
k is a constant, so if we
hang twice the amount
of mass from the spring,
d becomes twice larger:
k=(2m)g/(2d)=mg/d
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Concepts of SHM
x=+A
x=-A
x=0
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Concepts of SHM
Frequency vs Period
Time to go from +A to –A and back is the period. IE, the period is
the time it takes to complete one full cycle. Period = T
Frequency: f = 1/T. Eg, if T = 2 sec  f = ½ Hz
Hz = 1/s
If f is large, period is small: f = 1kHz = 1000 cycles per second
T = 1/(1000 Hz) = 0.001 sec
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Angular Frequency
Earth orbits the sun with a period of 365 days.
R

T=365 days(24 h/day)(3600 s/h) = 3.15107 sec
Frequency: f = 1/T = 1/(3.15107 sec)
= 3.1710-8 Hz
Earth travels at constant speed throughout its orbit: x/t = S.
It must traverse the circumference of the orbit: D = 2 π R
Thus, the speed S = D/T = 2 π R / T
We can also express this in terms of an angular frequency:
The angular frequency  =  /  t = 2 π / T
 = the speed at which the angle is changing
Units = 1/s or Radians/s
MSU Physics 231 Fall 2012
f = 1/T
=2πf
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Clicker Question
A hummingbird with mass 4g flies northwest at a speed of
10 m/s and at an angle of 30o relative to the horizontal. It
beats its wings once every 0.014 sec. Assuming it flies at
constant velocity, what is the beat frequency of its wings?
a)
b)
c)
d)
e)
0.14 Hz
7.1 Hz
14 Hz
71 Hz
None of the above
f = 1/T
T = 0.014 sec
f = 1/0.014 sec
Frequency: f = 71 Hz
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Simple harmonic motion
displacement x
A
a)
b)
time (s)
c)
(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
5cm
2
4
6
-5cm
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: 5cm (0.05 m)
b) period: time to complete one full oscillation: 4s
c) frequency: number of oscillations per second
f = 1/T = ¼ sec = 0.25 Hz
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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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Concepts of SHM
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
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Energy and Velocity
Ekin(½mv2)
x=+A
x=0
x=-A
Epot,spring(½kx2)
0
½mv2
0
Sum
½kA2
½kA2
0
½mv2
½k(-A)2
½kA2
conservation of ME: ½m[v(x=0)]2 = ½kA2 so v(x=0) = ±A(k/m)
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An Example
The maximum velocity and acceleration of mass connected to a
spring are v=0.95 m/s and a=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!
Using forces:
F = ma = -kx
F(-A) = kA = ma  m = kA/a or k = ma/A
Using energy conservation:
(½mv2 + ½ kx2)initial = (½mv2 + ½ kx2)final
½kA2 = ½ mv2
 A2 = mv2/k
Putting them together:
A2 = (m)(v2/k) = (kA/a)(v2/k)
A2 = Av2/a
A = v2/a = (0.95)2/(1.56) = 0.58 m
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Velocity 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]
position X velocity V acceleration
+A
0
0
±A(k/m) 0
-A
0
MSU Physics 231 Fall 2012
-kA/m
kA/m
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If oscillation is in vertical direction:
also add gravitational PE
ME = KE + PEspring + PEgravity
= ½mv2 + ½kx2 + mgh
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Period of a Spring
Not really trivial to derive, so we
won’t do so here. But we can
use concepts to get to the right
answer!
Period has units of [s]
Spring constant has units:
[N/m] = [kg/s2]
Mass has units of [kg]
m/k = [kg] / [kg/s2] = [s2]
√(m/k) = [s]
𝐓𝐬𝐩𝐫𝐢𝐧𝐠
𝐦
= 𝟐𝛑
𝐤
Increase m:
increase T
MSU Physics 231 Fall 2012
Increase k:
decrease T
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Another Example
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 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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Another Example
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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Harmonic oscillations vs circular motion
v0=r=A
=t
v0
vx 

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.
The
projection
of the linear velocity of the
PHY
231
rotating object on the x-axis is the velocity of
the mass on the spring.
MSU Physics 231 Fall 2012
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Circular Motion & SHM
x=+a
A particle moves in a circular orbit
with angular velocity , corresponding
to a linear velocity v0 = r = A
x=0
The horizontal position as a function
of time: x(t) = Acos = Acos(t) (=t)
x=-a
vx
A

The horizontal velocity as a function
of time: sin = -vx/v0
vx(t) = -v0sin = -Asin(t)
v0

t=0
x=0
Time to complete one circle
(I.e. one period T):
T = 2A/v0 = 2A/A = 2/
 = 2/T = 2f (f: frequency)
: angular frequency
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Circular Motion & SHM
The simple harmonic motion can be described by the
projection of circular motion on the horizontal axis.
xharmonic(t) = Acos(t)
vharmonic(t) = -Asin(t)
•
•
•
•
A is the amplitude of the oscillation
 = 2/T = 2f
T is the period of the harmonic motion
f = 1/T the frequency.
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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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For the case of a spring
position
velocity
acceleration
+A
0
-kA/m
0
±A(k/m)
0
-A
0
kA/m
1) velocity is maximum if v=±A(k/m)
2) circular motion: vspring(t)=-Asint maximal if vspring=±A
 combine 1) & 2) =(k/m)
Acceleration: a(t) = -(kA/m)cos(t) = -2Acos(t)
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A
x
time (s)
-A
velocity v
A(k/m)
time (s)
-A(k/m)
kA/m
time (s)
a
-kA/m
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A
x
xharmonic(t)=Acos(t)
-A
velocity v
time (s)
=2f=2/T=(k/m)
A(k/m)
vharmonic(t)=-Asin(t)
time (s)
-A(k/m)
kA/m
a
-kA/m
time (s)
aharmonic(t)=-2Acos(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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Another simple harmonic
oscillation: the pendulum
Restoring force: F=-mgsin
The force pushes the mass m
back to the central position.
L

sin if  is small (<150) radians!!!
T
F=-mg also =s/L
so: F=-(mg/L)s
s
mgsin

mgcos
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Another
simple
harmonic
pendulum vs spring
oscillation: the pendulum
parameter
spring
pendulum
restoring
force F
F=-kx
F=-(mg/L)s
period T
T=2(m/k)
frequency f
f=(k/m)/(2) f=(g/L)/(2)
angular
frequency
=(k/m)
*
T=2(L/g)
=(g/L)
*
m
m
L
T  2
 2
 2
k
(mg / L)
g
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Example
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
circular 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) xharmonic(t)=Acos(t)=0.1cos(22.4t)
vharmonic(t)=-Asin(t)=-2.24sin(22.4t)
aharmonic(t)=-2Acos(t)=-50.2cos(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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example: a pendulum clock
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
m
moon
elevator
faster
L
T  2
g
same
slower
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pendulum
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) 0.5T
L
T  2
B) 1T
g
C) 2T
4L
L
T   2
 4
 2T
D) 4T
g
g
E) 8T
Mass does not matter
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Damped & Driven Oscillations
With driven oscillation, the
amplitude grows!
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.
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 in sync with the motion,
this is known as resonance.
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For Next Week
Chapter 8:
Rotational Motion
Homework Set 5
Due Wednesday Oct 12!
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An Example
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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example
h
d
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)
ME=½mv2+½kx2+mgh must be conserved.
Initial: v=0, x=-0.05m h=0
0.5x0.2x02 + 0.5x250x0.052 + 0.2*9.8*0 = 0.3125
Final: v=0 (at highest point mass does not move), x=0 (spring no longer
compressed), h=???
0.5x0.2x02 + 0.5x250x0.02 + 0.2*9.8*h = 1.96 x h
Conservation of ME: 0.3125 = 1.96 x h h=0.16 m
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