Periodic oscillatory motion Assignment No HW this week Wednesday Read through Chapter 154 Periodic Motion is everywhere Examples of periodic motion Earth around the sun Elastic ball bouncing up and down ID: 783188
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Slide1
Lecture 19
Goals:
Chapter 14 Periodic (oscillatory) motion.
Assignment
No HW this week.
Wednesday: Read through Chapter 15.4
Slide2Periodic Motion is everywhere
Examples of periodic motionEarth around the sun
Elastic ball bouncing up and downQuartz crystal in your watch, computer clock, iPod clock, etc.
Slide3Periodic 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
Slide4Simple 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 = A2. After the block is released from x = A, it willA: remain at restB: 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
-A
A
0
(
≡X
eq
)
Slide5Simple 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
Slide6Simple 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
Slide7Simple 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
Slide8Simple Harmonic Motion
Now assume that the t coordinate of point K is 0.25 s.What is the period T , in seconds?
T = 1 sHow much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement?
time
Slide9Simple Harmonic Motion
Now assume that the t coordinate of point K is 0.25 s.What is the period T , in seconds?
T = 1 sHow 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
Slide10k
m
-A
A
0
(
≡X
eq
)
T = 1 s
k
2m
-A
A
0
(
≡X
eq
)
T
is
:
T > 1 s
T < 1 s
T=1 s
Slide11k
m
-A
A
0
(
≡X
eq
)
T = 1 s
2k
m
-A
A
0
(
≡X
eq
)
T
is
:
T > 1 s
T < 1 s
T=1 s
Slide12SHM Dynamics: Newton
’
s Laws still applyAt any given instant we know
that
F
=
m
a
must be true.
But in this case
F
=
-k x
and
ma
=
F
So:
-k x = ma =m d2
x/dt
2
k
x
m
F
= -k x
a
a differential equation for
x(t)
!
“
Simple approach
”
, guess a solution and see if it works!
d
2
x/dt
2
=-(k/m)x
Slide13SHM Solution...
Try
cos ( t )
Below is a drawing of
A
cos (
t
)
where
A
=
amplitude of oscillation
[with
w
= (k/m)
½
and w
= 2p
f = 2
p /T ]
T=2
(m/k)
½
T = 2
/
A
A
Slide14Use
“initial conditions”
to determine phase !The general solution is: x(t) = A cos ( wt + f)
SHM Solution...
k
m
-A
A
0
(
≡X
eq
)
x(t) = A cos (
w
t +
0
)
k
m
-A
A
0
(
≡X
eq
)
x(t) = A cos (
w
t +
π
)
at
t
=0
at
t
=0
Slide15Energy of the Spring-Mass System
We know enough to discuss the mechanical energy of the oscillating mass on a spring
.
x(t) =
A cos (
w
t +
f
)
If x(t) is displacement from equilibrium, then potential energy is
U(t) = ½ k
x(t)
2
=
½ k
A
2 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 = ½ m (
Aw)2 sin
2 (
wt +
f)
Finally,a(t) = dv/dt = -
2
A cos(t +
)
Slide16Energy of the Spring-Mass System
Potential energy of the spring is
U = ½ k x
2
= ½ k A
2
cos
2
(
t +
)
The Kinetic energy is
K = ½ mv2
= ½ m(
A)2 sin
2(t+
f)And
w2
= k / m or k = m w
2 K = ½ k A
2 sin2
(t+
f)
x(t) =
A cos(
t + )
v(t) = -
A sin(
t + )
a(t) = -2
A cos(
t + )
Slide17Energy of the Spring-Mass System
So E = K + U = constant =
½ k A2
U~cos
2
K~sin
2
E = ½ kA
2
At maximum displacement
K = 0
and
U = ½ k A
2
and acceleration has it maximum
At the equilibrium position
K = ½ k A
2
= ½ m v
2
and
U = 0
Slide18SHM 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
Slide19The
“
Simple” PendulumA 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
F
y
= ma
y
= T – mg cos(
q
)
≈
m v
2/L
S
Fx = ma
x = -mg sin(
q)
If
q small then x
L q
and sin(
q
)
q dx/dt = L d
q
/dtax = d
2x/dt2 = L d
2q
/dt2so a
x = -g
q = L d
2q
/ dt2
and
q
= q
0 cos(
wt +
f)
with
w
= (g/L)½
L
m
mg
z
y
x
T
Slide20The 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
Slide21The 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
.
Slide22The 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.