Oscillations
Monday, November 19
Lecture 30
Workbook problems due Wednesday
• WB 10.5, problems 14-25
Workbook Problems due Friday
• Problems 14-1 through 8, pages 14-1 -- 5
Power
• Power is the rate of transformation of energy
E
P
t
• Unit is 1 Watt=1W = 1 J/s
• If energy being transformed is work, W then
W F x
x
P

F
 Fv
t
t
t
Is the work done by F + or - ?
1. Positive
2. negative
50%
50%
e
at
iv
ne
g
Po
sit
i
d
ve
F
Is the work done by F + or - ?
1. Positive
2. negative
50%
50%
F
e
at
iv
ne
g
Po
sit
i
ve
d
Is the work done by F + or - ?
1. Positive
2. negative
50%
50%
e
at
iv
ne
g
Po
sit
i
d
ve
F
Problem 10:20
• A pendulum is made by tying a 500g ball to a
75-cm-long string. The pendulum is pulled 300
to one side and then released.
• A) What is the ball’s speed at the lowest point
in its trajectory?
• B) To what angle does the pendulum swing on
the other side.
Problem 10:20
• Use energy conservation
300
L=0.75 m
Δy=L-Lcos 300
Problem 10:20 cont
• Set y=0 at lowest point of swing
U gi  Ki  U gf  K f
1 2
mg y  0  0  mv
2
1 2
mgL(1  cos  )  mv
2
v  2 gL(1  cos  )  1.4 m/s
Problem 10:24
• A student places her 500g physics textbook on
a frictionless table. She pushes the book
against a spring 4.00cm and then releases the
book. What is the book’s speed as it slides
away? The spring constant is k = 1250 N/m.
Problem 10:24
• A student places her 500g physics textbook on
a frictionless table. She pushes the book
against a spring 4.00cm and then releases the
book. What is the book’s speed as it slides
away? The spring constant is k = 1250 N/m.
Problem 10:24
• Using the initial position as the compressed
spring, final after book leaves spring:
U Si  Ki  U sf  K f
1 2 1
2
U Si  kxi  1250 N / m  (.04m)
2
2
Ki  0
U Sf  0
1 2
K f  mv f
2
Problem 10:24
• Finally
k 2
1250 N/m
2
vf 
xi 
(.0400cm) 
m
0.500 kg
 2.00 m/s
Equilibrium and Oscillation
• Frequency and Period
1
f 
T
  2 f
Simple Harmonic Motion
• Linear restoring force—
– Example, mass on a spring
FNET , y  ky
k

m
– Set y=0 at equilibrium point:
–
 2 t 
y(t )  A cos(t )  A cos(2 f )  A cos 

 T 
Simple Harmonic Motion
• If restoring force is linearly proportional to
displacement (e.g. F=-kx) then we will have
simple harmonic motion.
• In lab last week you experimented with a
simple pendulum. Was its motion simple
harmonic?
Fnet ,tangential  mg sin   mg
θ
T
w cos θ
w sin θ
w
Simple Pendulum
Find the angular frequency is

g
L
1
1
T

2 2
L
g
Description of motion
x(t )  A cos(t )
dx(t )
v(t ) 
  A sin(t )  vMAX sin(t )
dt
dv(t )
a(t ) 
  2 A cos(t )   2 x(t )
dt
x(t) vs. t
5
4
3
x(t) meters
2
1
x(t)
0
0
2
4
6
8
-1
v(t)
-2
a(t)
-3
-4
-5
t sec
x(t) vs. t
4
3
2
x(t) meters
1
0
0
2
4
6
8
-1
-2
-3
-4
t sec
a) At what time(s) is particle moving right at
maximum speed?
b) At what time(s) is particle moving right at
maximum speed?
c) At what time(s) is the speed zero?
Problem 14:7
An air-track glider is attached to a spring. The
glider is pulled to the right and released from rest
at t=0s. It then oscillates with T=2.0s and vmax =
40cm/s
a) A=?
b) x(t=0.25s) = ?
Problem 14.7
2
T  2s  

T
vMAX   A   A  .4m/s
A=
.4

 0.127m
x(t )  A cos(t )  0.127 cos( t )
x(.25)  .127 cos( (.25))  .0898
Wednesday
Oscillations continued
Problems CQ3,CQ9,MC18,MC19, 1, 4, 6, 7, 10