Download Chapter 8 File

CHAPTER 8
VIBRATIONS AND WAVES
8.1
HOOKE’S LAW
-The motion that repeat itself over and over is referred as periodic motion.
-The motion of an oscillating object depends on the restoring forces that make it
go back and forth.
Eg: the beating of your heart, the ticking of a clock, the movement of a child on a
swing.
-If the spring is stretched or compressed a small distance, x, from its unstretched
(equilibrium) position, and then released, it exerts a force on the mass.
The spring force ( or restoring force),
FS  kx
Hooke's law (spring-mass system) …(8.1)
where k = spring constant,
1
x = displacement of the mass from its unstretched (not too large)
(-) sign: the force exerted by the spring is always directed opposite the
displacement of the mass.
-The value of k is a measure of the stiffness of the spring.
-Stiff spring large k
-Soft spring  small k
-A particular type of periodic motion is Simple Harmonic Motion (SHM).
-Simple harmonic motion occurs when the net force along the direction of motion is a
Hooke’s law type of force-that is, when the net force is linearly proportional to the
magnitude of the system's displacement from equilibrium and in the opposite direction.
Eg: A clock pendulum moving back & forth
SHM is described by the following parameters:
 Amplitude (A)
: the magnitude of the maximum displacement (+A or-A) of a mass
from its equilibrium position,
 Period (T) : the time needed to complete one cycle of oscillation.
SI unit: second (s)
 Frequency (f)
: the number of cycles (oscillations) per second
SI unit
: Hertz (Hz) or 1/s or cycle per second (cycle/s)
-How to find acceleration???
-By combining Newton's 2nd law with Hooke's law:
2
-kx = ma
a
8.2
F
k
 x
m
m
…(8.2)
ELASTIC POTENTIAL ENERGY
-Mechanical energy – 4 types:
1.
2.
3.
4.
Gravitational Potential Energy
Translational Kinetic Energy
Rotational Kinetic Energy
Elastic Potential Energy
-An object has potential energy by virtue of its shape or position.
-Energy is stored in a spring only when it is either stretched or compressed.
-The energy stored in a stretched or compressed spring or other elastic material is called
elastic potential energy (PE).
PE of a spring of the system:
KE of a spring:
PEs 
1 2
kx
2
KE 
1
m 2
2
…(8.3)
…(8.4)
-The elastic potential energy is stored in a spring is zero when the spring is unstretched or
uncompressed (x=0)
-The elastic potential energy is a maximum when a spring has reached its maximum
compression or extension.
-PE is always positive because it proportional to x2.
3
Energy
E=1/2 kA2
PE=1/2kx2
KE
KEmax=E
KE=1/2 mv2
-A
x=0
x1
+A
Figure shows the energy transfers as the spring-mass system oscillates
-In SHM, the total energy is proportional to the square of the amplitude of the motion.
-From conservation of mechanical energy
Ei  E f
…(8.5)
KE  PE
g
 PEs i  KE  PEg  PEs f
…(8.6)
-If nonconservative forces (eg friction) are present, the final mechanical energy dos not
equal the initial mechanical energy.
From work-energy theorem:
Wnc  KE  PEg  PEs  f  KE  PEg  PEs i
…(8.7)
Because the spring force is conservative and since there are no external forces that can
do work on the system, the total mechanical energy of the system consisting the block and
spring remains constant.
4
Example:
A 0.50-kg object is attached to a spring of spring constant 20N/m along a horizontal,
frictionless surface. The object oscillates in SHM and has a speed of 1.5m/s at the
equilibrium position.
a.
What is the total energy of the system?
b.
What is the amplitude
c.
At what locations are the valued for the PE and KE the same?
Solution:
a.
At x=0, PE=0, So total E=KE
EK
b.
1
1
2
m max
 (0.50kg)(1.5m / s) 2  0.56 J
2
2
At x=  A, v=0, so total E =PE
E  PE 
1 2 1 2
kx  kA
2
2
5
A
c.
2E

k
2(0.56 J )
 0.24m
20 N / m
1
2
When KE = PE, E = KE+PE = PE + PE = 2 kx2 = kx2
So
x
E

k
0.56 J
 0.17m
20 N / m
8.3
VELOCITY AS A FUNCTION OF POSITION
-Refer to Fig, as the mass moves toward the origin to some new position, x, part of this
energy is transformed into KE, and the PE stored in the spring is reduced to 1/2kx2.
Figure (a): A mass attached to a spring on a frictionless surface is released from rest with
the spring extended a distance of A. Just before released, E=1/2kx2.
Figure (b) : when the mass reaches position x, it has KE and PE (decreased)
Total energy, E=KE + PE
6
1 2 1
1
kA  m 2  kx2
2
2
2
Solving for v,   

k 2
A  x2
m

At x=0, v=vmax
At v=  A , v=0
8.4
COMPARING SIMPLE HARMONIC MOTION WITH UNIFORM CIRCULAR
MOTION
-As the turntable rotates with constant angular speed, the shadow of the ball moves back
and forth with SHM.
7
-Figure shows the ball moving with constant speed, v0, in a direction tangent to circular
path.
  0 sin 

sin  

0
…(8.8)
A2  x 2
A
-Equating the right-hand sides of the last two expressions.

0
A
A2  x 2  C A 2  x 2
…(8.9)
Period and Frequency
-Period, T=time required for one complete trip back and forth and T=the time it takes the
ball to make one complete circular trip on the turntable.
-For the ball moves through a distance 2  A, the speed of the ball around the circular
path is
0 
Period
T 
2A
T
2A
0
…(8.10)
For a mass-spring system moving with SHM
8
Period
:
T  2
m
k
…(8.11)
Unit : second (s)
Frequency
f 
:
1
1

T 2
k
m
…(8.12)
Unit : s-1 or hertz (Hz)
Angular frequency
8.5
:
  2f 
k
m
…(8.13)
POSITION, VELOCITY AND ACCELERATION AS A FUNCTION OF TIME
-See back the relationship between SHM and uniform circular motion.
x  A cos  A cos(t )
…(8.14)
9
x  Acost
Displacement:
Velocity
:
Acceleration :
…(8.15)
  Asin t
..(8.16)
a   2 A cos t   2 x
…(8.17)
Where A = amplitude

= angular frequency of the motion
f = frequency of revolution/rotation
The maximum speed
:
The maximum acceleration:
max  A
amax  A 2
10
8.6
MOTION OF A PENDULUM
Restoring force
Ft  mg sin 
…(8.18)
For motion with a small angles
Ft  mg
So
where
s  L
 mg 
Ft  
s
L


T  2
L
g
..(8.19)
Eg : Pendulum clocks
Comparison :
11
 For a simple pendulum
T  2
L
g
1
f 
2
g
L

g
L
where L = length of the pendulum, g = acc due to gravity.
-The period is independent of the suspended mass.
 For a mass-spring system:
m
T  2
k
f 
1
2
k
m

k
m
m = mass of the object, k = spring constant
-The period is dependent of the suspended mass
Example:
An object with a mass of 1.0kg is attached to a spring with a spring constant of 10N/m.
The object is displaced by 3.0cm from the equilibrium position and let go.
a.
What is the amplitude A?
b.
What is the period T?
c.
What is the frequency f?
Solution:
a.
b.
c.
The initial stretch is the max displacement
So A=3.0cm
T  2
f 
m
1.0kg
 2
 2.0s
k
10 N / m
1
1

 0.50 Hz
T 2.0s
12