Transcript Physics 207: Lecture 2 Notes

Lecture 19
Goals:
• Chapter 14
 Periodic motion.
• Assignment
 No HW this week.
 Wednesday: Read through Chapter 15.4
Physics 207: Lecture 19, Pg 1
Periodic Motion is everywhere
Examples of periodic motion
 Earth around the sun
 Elastic ball bouncing up and down
 Quartz crystal in your watch, computer clock, iPod clock, etc.
Physics 207: Lecture 19, Pg 2
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
Physics 207: Lecture 19, Pg 3
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
-A
0(≡Xeq)
m
A
Physics 207: Lecture 19, Pg 4
Simple Harmonic Motion (SHM)
 The time it takes the block to complete one cycle is the period 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 19, Pg 5
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 19, Pg 6
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 19, Pg 7
Simple Harmonic Motion
 Now assume that the t coordinate of point K is 0.25 s.
 What is the period T , in seconds?
 How much time t does the block take to travel from the point of
maximum displacement to the opposite point of maximum
displacement?
time
Physics 207: Lecture 19, Pg 8
Simple Harmonic Motion
 Now assume that the t coordinate of point K is 0.25 s.
 What is the period T , in seconds?
T=1s
 How much time t does the block take to travel from the point of
maximum displacement to the opposite point of maximum
displacement?
time
Physics 207: Lecture 19, Pg 9
Simple Harmonic Motion
 Now assume that the t coordinate of point K is 0.25 s.
 What is the period T , in seconds?
T=1s
 How much time t does the block take to travel from the point of
maximum displacement to the opposite point of maximum
displacement?
t = 0.5 s
time
Physics 207: Lecture 19, Pg 10
Simple Harmonic Motion
 Now assume that the x coordinate of point R is 1 m.
 What total distance d does the object cover during one period of
oscillation?
 What distance d does the object cover between the moments
labeled K and N on the graph?
time
Physics 207: Lecture 19, Pg 11
Simple Harmonic Motion
 Now assume that the x coordinate of point R is 1 m.
 What total distance d does the object cover during one period of
oscillation?
d=4m
 What distance d does the object cover between the moments
labeled K and N on the graph?
time
Physics 207: Lecture 19, Pg 12
Simple Harmonic Motion
 Now assume that the x coordinate of point R is 1 m.
 What total distance d does the object cover during one period of
oscillation?
d=4m
 What distance d does the object cover between the moments
labeled K and N on the graph?
d=3m
time
Physics 207: Lecture 19, Pg 13
k
-A
0(≡Xeq)
k
-A
0(≡Xeq)
m
T=1s
A
2m
A
T is:
A)T > 1 s
B)T < 1 s
C) T=1 s
Physics 207: Lecture 19, Pg 14
k
-A
0(≡Xeq)
2k
-A
0(≡Xeq)
m
T=1s
A
m
A
T is:
A)T > 1 s
B)T < 1 s
C)T=1 s
Physics 207: Lecture 19, Pg 15
SHM Dynamics: Newton’s Laws still apply
 At any given instant we know
that F = ma must be true.
k
 But in this case F = -k x
2
and ma = Fm
F = -k x
a
m
d x
dt
2
 So: -k x = ma =m d2x/dt2
d2x/dt2=-(k/m)x
x
a differential equation for x(t) !
“Simple approach”, guess a solution and see if it works!
Physics 207: Lecture 19, Pg 16
SHM Solution...
 Try cos (  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 ]
 T=2p (m/k)½
Physics 207: Lecture 19, Pg 17
SHM Solution...
The general solution is: x(t) = A cos ( t + )
Use “initial conditions” to determine phase  !
k
at t=0
-A
at t=0
m
0(≡Xeq)
x(t) = A cos ( t + 0)
A
k
m
-A
x(t) = A cos ( t + )
0(≡Xeq)
A
Physics 207: Lecture 19, Pg 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 ( t + )
If x(t) is displacement from equilibrium, then potential energy is
U(t) = ½ k x(t)2 = ½ k 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 = ½ m (A)2 sin2 ( t + )
Finally,
a(t) = dv/dt = -2A cos(t + )
Physics 207: Lecture 19, Pg 19
Energy of the Spring-Mass System
x(t) =
A cos( t +  )
v(t) = -A sin( t +  )
a(t) = -2A cos( t +  )
Potential energy of the spring is
U = ½ k x2 = ½ k A2 cos2(t + )
The Kinetic energy is
K = ½ mv2 = ½ m(A)2 sin2(t+)
And 2 = k / m or k = m 2
K = ½ k A2 sin2(t+)
Physics 207: Lecture 19, Pg 20
Energy of the Spring-Mass System
So E = K + U = constant =½ k A2
k
k

 
m K = 0 and
m U = ½ k A2
At maximum displacement
and acceleration has it maximum
2
At the equilibrium position K = ½ k A2 = ½ m v2 and U = 0
E = ½ kA2
U~cos2
K~sin2
Physics 207: Lecture 19, 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 19, Pg 22
The “Simple” Pendulum
 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 = may = T – mg cos(q) ≈ m v2/L
S Fx = max = -mg sin(q)
z
y
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
and q = q0 cos(t + )
with  = (g/L)½
q L
x
T
m
mg
Physics 207: Lecture 19, Pg 23
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 19, Pg 24
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 19, Pg 25
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 19, Pg 26