Transcript Document 7846694

Physics 207, Lecture 18, Nov. 3
Goals:
• Chapter 14
 Interrelate the physics and mathematics of oscillations.
 Draw and interpret oscillatory graphs.
 Learn the concepts of phase and phase constant.
 Understand and use energy conservation in oscillatory systems.
 Understand the basic ideas of damping and resonance.
Phase Contrast Microscopy
Epithelial cell in brightfield (BF)
using a 40x lens (NA 0.75)
(left) and with phase contrast
using a DL Plan Achromat
40x (NA 0.65) (right).
A green interference filter is
used for both images.
Physics 207: Lecture 18, Pg 1
Physics 207, Lecture 18, Nov. 3
• Assignment
 HW8, Due Wednesday, Nov. 12th
 Wednesday: Read through Chapter 15.4
Physics 207: Lecture 18, Pg 2
Periodic Motion is everywhere
Examples of periodic motion
 Earth around the sun
 Elastic ball bouncing up an down
 Quartz crystal in your watch, computer
clock, iPod clock, etc.
Physics 207: Lecture 18, Pg 3
Periodic Motion is everywhere
Examples of periodic motion
 Heart beat
In taking your pulse, you count 70.0
heartbeats in 1 min.
What is the period, in seconds, of your
heart's oscillations?
Period is the time for one
oscillation
T= 60 sec/ 70.0 = 0.86 s
 What is the frequency?
f = 1 / T = 1.17 Hz
Physics 207: Lecture 18, Pg 4
A special kind of period oscillator: Harmonic oscillator
What do all “harmonic oscillators” have in common?
1. A position of equilibrium
2. A restoring force, which must be linear
[Hooke’s law spring F = -kx
(In a pendulum the behavior only linear for small
angles: sin θ where θ = s / L) ] In this limit we
have: F = -ks with k = mg/L)
3. Inertia
4. The drag forces are reasonably small
Physics 207: Lecture 18, Pg 5
Simple Harmonic Motion (SHM)
 In Simple Harmonic Motion the restoring force on the
mass is linear, that is, exactly proportional to the
displacement of the mass from rest position
 Hooke’s Law : F = -kx
If k >> m  rapid oscillations <=> large frequency
If k << m  slow oscillations <=> low frequency
Physics 207: Lecture 18, Pg 6
Simple Harmonic Motion (SHM)
 We know that if we stretch a spring with a mass on the end
and let it go the mass will, if there is no friction, ….do
something
1. Pull block to the right until x = A
2. After the block is released from x = A, it will
A: remain at rest
B: move to the left until it reaches
equilibrium and stop there
C: move to the left until it reaches
x = -A and stop there
D: move to the left until it reaches
x = -A and then begin to move to
the right
k
m
k
m
k
m
-A
0(≡Xeq) A
Physics 207: Lecture 18, Pg 7
Simple Harmonic Motion (SHM)
 We know that if we stretch a spring with a mass on the end
and let it go the mass will ….
1. Pull block to the right until x = A
2. After the block is released from x = A, it will
k
A: remain at rest
m
B: move to the left until it reaches
equilibrium and stop there
k
m
C: move to the left until it reaches
x = -A and stop there
k
D: move to the left until it reaches
m
x = -A and then begin to move to
-A
0(≡Xeq) A
the right
This oscillation is called Simple Harmonic Motion
Physics 207: Lecture 18, Pg 8
Simple Harmonic Motion (SHM)
 The time it takes the block to complete one cycle is called the
period.
Usually, the period is denoted T and is measured in seconds.
 The frequency, denoted f, is the number of cycles that are
completed per unit of time: f = 1 / T.
In SI units, f is measured in inverse seconds, or hertz (Hz).
 If the period is doubled, the frequency is
A. unchanged
B. doubled
C. halved
Physics 207: Lecture 18, Pg 9
Simple Harmonic Motion (SHM)
 An oscillating object takes 0.10 s to complete one
cycle; that is, its period is 0.10 s.
 What is its frequency f ?
Express your answer in hertz.
f = 1/ T = 10 Hz
Physics 207: Lecture 18, Pg 10
Simple Harmonic Motion
 Note in the (x,t) graph that the vertical axis represents the x
Position
coordinate of the oscillating object, and the horizontal axis
represents time.
Which points on the x axis are located a displacement A from the
equilibrium position ?
A. R only
B. Q only
C. both R and Q
time
Physics 207: Lecture 18, Pg 11
Simple Harmonic Motion
 Suppose that the period is T.
 Which of the following points on the t axis are separated by the
time interval T?
A. K and L
B. K and M
C. K and P
D. L and N
E. M and P
time
Physics 207: Lecture 18, Pg 12
Simple Harmonic Motion
 Now assume that the t coordinate of point K is 0.0050 s.
 What is the period T , in seconds?
T = 0.02 s
 How much time t does the block take to travel from the point of
maximum displacement to the opposite point of maximum
displacement
t = 0.01 s
time
Physics 207: Lecture 18, Pg 13
Simple Harmonic Motion
 Now assume that the x coordinate of point R is 0.12 m.
 What distance d does the object cover during one period of
oscillation?
d = 0.48 m
 What distance d does the object cover between the moments
labeled K and N on the graph?
d = 0.36 m
time
Physics 207: Lecture 18, Pg 14
SHM Dynamics
 At any given instant we know
that F = ma must be true.
 But in this case F = -k x
2
and ma = m
F = -k x
k
a
m
d x
dt 2
d 2x
 So: -k x = ma = m
dt 2
d 2x
k
 x
2
m
dt
x
a differential equation for x(t) !
“Simple approach”, guess a solution and see if it works!
Physics 207: Lecture 18, Pg 15
SHM Solution...
 Try either cos (  t ) or sin (  t )
 Below is a drawing of A cos (  t )
 where A = amplitude of oscillation
T = 2p/
A
p
A
p
p
 [with  = (k/m)½ and  = 2p f = 2p /T ]
 Both sin and cosine work so need to include both
Physics 207: Lecture 18, Pg 16
Combining sin and cosine solutions
B cos t + C sin t
= A cos ( t + )
= A (cos t cos  – sin t sin  )
=A cos  cos t – A sin  sin t)
Notice that B = A cos  C = -A sin   tan = -C/B
x(t) =
k
p
m
p

cos sin
0
x
Use “initial conditions” to determine phase  !
Physics 207: Lecture 18, Pg 17
Energy of the Spring-Mass System
We know enough to discuss the mechanical energy of the
oscillating mass on a spring.
x(t) = A cos ( t + )
If x(t) is displacement from equilibrium, then potential energy is
U(t) = ½ k x(t)2 = A2 cos2 ( t + )
v(t) = dx/dt  v(t) = A  (-sin ( t + ))
And so the kinetic energy is just ½ m v(t)2
K(t) = ½ m v(t)2 = (A)2 sin2 ( t + )
Finally,
a(t) = dv/dt = -2A cos(t + )
Physics 207: Lecture 18, Pg 18
Energy of the Spring-Mass System
x(t) = A cos( t +  )
v(t) = -A sin( t +  )
a(t) = -2A cos(t + )
Kinetic energy is always
K = ½ mv2 = ½ m(A)2 sin2(t+)
Potential energy of a spring is,
U = ½ k x2 = ½ k A2 cos2(t + )
And 2 = k / m or k = m 2
U = ½ m 2 A2 cos2(t + )
Physics 207: Lecture 18, Pg 19
Energy of the Spring-Mass System
x(t) = A cos( t +  )
v(t) = -A sin( t +  )
a(t) = -2A cos(t + )
And the mechanical energy is
K + U =½ m 2 A2 cos2(t + ) + ½ m 2 A2 sin2(t + )
K + U = ½ m 2 A2 [cos2(t + ) + sin2(t + )]
K + U = ½ m 2 A2 = ½ k A2
which is constant
Physics 207: Lecture 18, Pg 20
Energy of the Spring-Mass System
So E = K + U = constant =½ k A2
k
k
2

 
m
m
At maximum displacement K = 0 and U = ½ k A2
and acceleration has it maximum (or minimum)
At the equilibrium position K = ½ k A2 = ½ m v2 and U = 0
E = ½ kA2
U~cos2
K~sin2
p
p 
Physics 207: Lecture 18, Pg 21
SHM So Far
 The most general solution is x = A cos(t + )
where A = amplitude
 = (angular) frequency
 = phase constant
k

 For SHM without friction,
m
 The frequency does not depend on the amplitude !
 This is true of all simple harmonic motion!
 The oscillation occurs around the equilibrium point where the
force is zero!
 Energy is a constant, it transfers between potential and kinetic
Physics 207: Lecture 18, Pg 22
Physics 207, Lecture 18, Nov. 3
• Assignment
 HW8, Due Wednesday, Nov. 12th
 Wednesday: Read through Chapter 15.4
The rest are for Wednesday, plus damping, resonance and
part of Chapter 15.
Physics 207: Lecture 18, Pg 26
The shaker cart
 You stand inside a small cart attached to a heavy-duty spring, the
spring is compressed and released, and you shake back and forth,
attempting to maintain your balance. Note that there is also a
sandbag in the cart with you.
 At the instant you pass through the equilibrium position of the spring,
you drop the sandbag out of the cart onto the ground.
 What effect does jettisoning the sandbag at the equilibrium position
have on the amplitude of your oscillation?
It increases the amplitude.
It decreases the amplitude.
It has no effect on the amplitude.
Hint: At equilibrium, both the cart and the bag
are moving at their maximum speed. By
dropping the bag at this point, energy
(specifically the kinetic energy of the bag) is
lost from the spring-cart system. Thus, both the
elastic potential energy at maximum displacement
and the kinetic energy at equilibrium must decrease
Physics 207: Lecture 18, Pg 27
The shaker cart
 Instead of dropping the sandbag as you pass through equilibrium, you
decide to drop the sandbag when the cart is at its maximum distance
from equilibrium.
 What effect does jettisoning the sandbag at the cart’s maximum
distance from equilibrium have on the amplitude of your oscillation?
It increases the amplitude.
It decreases the amplitude.
It has no effect on the amplitude.
 Hint: Dropping the bag at maximum
distance from equilibrium, both the cart
and the bag are at rest. By dropping the
bag at this point, no energy is lost from
the spring-cart system. Therefore, both the
elastic potential energy at maximum displacement
and the kinetic energy at equilibrium must remain constant.
Physics 207: Lecture 18, Pg 28
The shaker cart
 What effect does jettisoning the sandbag at the cart’s maximum
distance from equilibrium have on the maximum speed of the cart?
It increases the maximum speed.
It decreases the maximum speed.
It has no effect on the maximum speed.
Hint: Dropping the bag at maximum distance
from equilibrium, both the cart and the bag
are at rest. By dropping the bag at this
point, no energy is lost from the spring-cart
system. Therefore, both the elastic
potential energy at maximum displacement
and the kinetic energy at equilibrium must
remain constant.
Physics 207: Lecture 18, Pg 29
Exercise Simple Harmonic Motion
A mass oscillates up & down on a spring. It’s position as a
function of time is shown below. At which of the points
shown does the mass have positive velocity and negative
acceleration ?
Remember: velocity is slope and acceleration is the curvature

y(t)
(a)
(c)
t
(b)
Physics 207: Lecture 18, Pg 30
Example
 A mass m = 2 kg on a spring oscillates with amplitude
A = 10 cm. At t = 0 its speed is at a maximum, and is v=+2 m/s
 What is the angular frequency of oscillation  ?
 What is the spring constant k ?
General relationships E = K + U = constant,  = (k/m)½
So at maximum speed U=0 and ½ mv2 = E = ½ kA2
thus k = mv2/A2 = 2 x (2) 2/(0.1)2 = 800 N/m,  = 20 rad/sec
k
m
x
Physics 207: Lecture 18, Pg 31
Home Exercise
Simple Harmonic Motion


You are sitting on a swing. A friend gives you a small push
and you start swinging back & forth with period T1.
Suppose you were standing on the swing rather than sitting.
When given a small push you start swinging back & forth with
period T2.
Which of the following is true recalling that  = (g/L)½
(A) T1 = T2
(B) T1 > T2
(C) T1 < T2
Physics 207: Lecture 18, Pg 32
Home Exercise
Simple Harmonic Motion


You are sitting on a swing. A friend gives you a small
push and you start swinging back & forth with period T1.
Suppose you were standing on the swing rather than
sitting. When given a small push you start swinging back
& forth with period T2.
If you are standing, the center of mass moves towards the
pivot point and so L is less,  is bigger, T2 is smaller
(A) T1 = T2
(B) T1 > T2
(C) T1 < T2
Physics 207: Lecture 18, Pg 33
BTW: The Rod Pendulum
(not tested)
 A pendulum is made by suspending a thin rod of length L
and mass M at one end. Find the frequency of oscillation
for small displacements.
S tz = I a = -| r x F | = (L/2) mg sin()
z
(no torque from T)
-[ mL2/12 + m (L/2)2 ] a  L/2 mg 
T
-1/3 L d2/dt2 = ½ g 

xCM
L
mg
Physics 207: Lecture 18, Pg 34
BTW: General Physical Pendulum
(not tested)
 Suppose we have some arbitrarily shaped
solid of mass M hung on a fixed axis, that we
know where the CM is located and what the
moment of inertia I about the axis is.
 The torque about the rotation (z) axis for small
 is (sin    )
t=
-MgR sin  -MgR 
d 2
 MgR  I
dt 2
d 
dt 2
  2 
where

R

xCM
a
t
2
z-axis
MgR
I
Mg
 = 0 cos(t + )
Physics 207: Lecture 18, Pg 35
Torsion Pendulum
 Consider an object suspended by a wire
attached at its CM. The wire defines the
rotation axis, and the moment of inertia I
about this axis is known.
 The wire acts like a “rotational spring”.
 When the object is rotated, the wire
is twisted. This produces a torque
that opposes the rotation.
 In analogy with a spring, the torque
produced is proportional to the
displacement: t = - k 
where k is the torsional spring
constant
wire

t
I
  = (k/I)½
Physics 207: Lecture 18, Pg 36
Home Exercise
Period
 All of the following torsional pendulum bobs have the same
mass and  = (k/I)½
 Which pendulum rotates the fastest, i.e. has the longest
period? (The wires are identical)
R
R
(A)
(B)
R
R
(C)
(D)
Physics 207: Lecture 18, Pg 37
Reviewing Simple Harmonic Oscillators
 Spring-mass system
d2x
2



x
dt 2
k
where  
F = -kx
m
a
k
m
x
z-axis
x(t) = A cos( t + )
 Pendula
d 2
dt
2

  2 
 General physical pendulum

xCM
Mg
 = 0 cos( t + )
 Torsion pendulum
R

k
I
MgR
I
wire

t
I
Physics 207: Lecture 18, Pg 38
Energy in SHM
 For both the spring and the pendulum, we can derive
the SHM solution using energy conservation.
 The total energy (K + U) of a
U
system undergoing SMH will
always be constant!
E
 This is not surprising since
K
U
there are only conservative
-A
forces present, hence energy is conserved.
0
A
x
Physics 207: Lecture 18, Pg 39
SHM and quadratic potentials
 SHM will occur whenever the potential is quadratic.
 For small oscillations this will be true:
 For example, the potential between
H atoms in an H2 molecule looks
something like this:
U
E
U
x
K
U
-A
0
A
x
Physics 207: Lecture 18, Pg 40