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Physics 207, Lecture 18, Nov' 3

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Title: Physics 207, Lecture 18, Nov' 3


1
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.
2
Physics 207, Lecture 18, Nov. 3
  • Assignment
  • HW8, Due Wednesday, Nov. 12th
  • Wednesday Read through Chapter 15.4

3
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.

4
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

5
A special kind of period oscillator Harmonic
oscillatorWhat do all harmonic oscillators
have in common?
  • 1. A position of equilibrium
  • 2. A restoring force, which must be linear
  • Hookes law spring F -kx
  • (In a pendulum the behavior only linear for
    small angles sin ? where ? s / L) In
    this limit we have F -ks with k mg/L)
  • 3. Inertia
  • 4. The drag forces are reasonably small

6
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
  • Hookes Law F -kx
  • If k gtgt m ? rapid oscillations ltgt large
    frequency
  • If k ltlt m ? slow oscillations ltgt low frequency

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, 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

8
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
  • 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
  • This oscillation is called Simple Harmonic Motion

9
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

10
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

11
Simple Harmonic Motion
  • Note in the (x,t) graph that the vertical axis
    represents the x 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

Position
time
12
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
13
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
14
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
15
SHM Dynamics
  • At any given instant we know that F ma must be
    true.
  • But in this case F -k x and
    ma
  • So -k x ma

a differential equation for x(t) !
Simple approach, guess a solution and see if it
works!
16
SHM Solution...
  • Try either cos ( ? t ) or sin ( ? t )
  • Below is a drawing of A cos ( ? t )
  • where A amplitude of oscillation
  • with w (k/m)½ and w 2p f 2p /T
  • Both sin and cosine work so need to include both

17
Combining sin and cosine solutions
x(t) B cos wt C sin wt A cos
( wt f) A (cos wt cos f sin wt sin
f ) A cos f cos wt A sin f sin
wt) Notice that B A cos f C -A sin f ?
tan f -C/B
?
?
??
sin
cos
Use initial conditions to determine phase ? !
18
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 ( wt f) If x(t) is
displacement from equilibrium, then potential
energy is U(t) ½ k x(t)2 A2 cos2 ( wt
f) v(t) dx/dt ? v(t) A w (-sin ( wt
f)) And so the kinetic energy is just ½ m
v(t)2 K(t) ½ m v(t)2 (Aw)2 sin2 ( wt
f) Finally, a(t) dv/dt -?2A cos(?t ?)
19
Energy of the Spring-Mass System
Kinetic energy is always K ½ mv2 ½
m(?A)2 sin2(?tf) Potential energy of a spring
is, U ½ k x2 ½ k A2 cos2(?t ?) And w2
k / m or k m w2 U ½ m w2 A2 cos2(?t
?)
20
Energy of the Spring-Mass System
And the mechanical energy is K U ½ m w2 A2
cos2(?t ?) ½ m w2 A2 sin2(?t ?) K U ½
m w2 A2 cos2(?t ?) sin2(?t ?) K U ½
m w2 A2 ½ k A2 which is constant
21
Energy of the Spring-Mass System
So E K U constant ½ k A2
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
22
SHM So Far
  • The most general solution is x A cos(?t ?)
  • where A amplitude
  • ? (angular) frequency
  • ? phase constant
  • For SHM without friction,
  • 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

23
What about Vertical Springs?
  • For a vertical spring, if y is measured from the
    equilibrium position
  • Recall force of the spring is the negative
    derivative of this function
  • This will be just like the horizontal case-ky
    ma

j
k
y 0
F -ky
m
Which has solution y(t) A cos( ?t ?)
where
24
The Simple Pendulum (In class torques, t Ia,
were used instead but the results are the same)
  • A pendulum is made by suspending a mass m at the
    end of a string of length L. Find the frequency
    of oscillation for small displacements.
  • S Fy mac T mg cos(q) m v2/L
  • S Fx max -mg sin(q)
  • If q small then x ? L q and sin(q) ? q
  • dx/dt L dq/dt
  • ax d2x/dt2 L d2q/dt2
  • so ax -g q L d2q / dt2 ? L d2q / dt2 - g q
    0
  • and q q0 cos(wt f) or q q0 sin(wt
    f)
  • with w (g/L)½

z
y
?
L
x
T
m
mg
25
Velocity and Acceleration
Position x(t) A cos(?t ?) Velocity v(t)
-?A sin(?t ?) Acceleration a(t) -?2A
cos(?t ?)
26
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.

27
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

28
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
    carts 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.

29
The shaker cart
  • What effect does jettisoning the sandbag at the
    carts 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.

30
Exercise Simple Harmonic Motion
  • A mass oscillates up down on a spring. Its
    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)
31
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 v2 m/s
  • What is the angular frequency of oscillation ? ?
  • What is the spring constant k ?
  • General relationships E K U constant, w
    (k/m)½
  • So at maximum speed U0 and ½ mv2 E ½ kA2
  • thus k mv2/A2 2 x (2) 2/(0.1)2 800 N/m, w
    20 rad/sec

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.
  • Which of the following is true recalling that w
    (g/L)½

(A) T1 T2 (B) T1 gt T2 (C) T1 lt T2
33
Home ExerciseSimple 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, w is
    bigger, T2 is smaller

(A) T1 T2 (B) T1 gt T2 (C) T1 lt T2
34
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(q)
  • (no torque from T)
  • - mL2/12 m (L/2)2 a ? L/2 mg q
  • -1/3 L d2q/dt2 ½ g q

z
T
?
x
CM
L
mg
35
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 ? ? ? )


    ? -MgR sinq ? -MgR???

z-axis
R
?
x
CM
Mg
36
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 ? - k ?
    where k is the torsional spring constant
  • w (k/I)½

37
Home Exercise Period
  • All of the following torsional pendulum bobs have
    the same mass and w (k/I)½
  • Which pendulum rotates the fastest, i.e. has the
    longest period? (The wires are identical)

(A)
(B)
(C)
(D)
38
Reviewing Simple Harmonic Oscillators
  • Spring-mass system
  • Pendula
  • General physical pendulum
  • Torsion pendulum

where
z-axis
x(t) A cos( ?t ?)
R
?
x
CM
Mg
39
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 system undergoing
    SMH will always be constant!
  • This is not surprising since there are only
    conservative forces present, hence energy is
    conserved.

40
SHM and quadratic potentials
  • SHM will occur whenever the potential is
    quadratic.
  • For small oscillations this will be true
  • For example, the potential betweenH atoms in an
    H2 molecule lookssomething like this

U
x
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