Chapter 13
Vibrations and Waves
Hooke’s Law Reviewed
F  kx
 When x is positive
F is negative
;
,
 When at equilibrium (x=0),
F = 0 ;
 When x is negative
F is positive
;
,
Sinusoidal Oscillation
Pen traces a sine wave
Graphing x vs. t
A
T
A : amplitude (length, m)
T : period (time, s)
Some Vocabulary
x  A cos(t   )
 A cos( 2ft   )
2t


 A cos
 
 T

1
f 
T
2
  2f 
T
f = Frequency
 = Angular Frequency
T = Period
A = Amplitude
 = phase
Phases
Phase is related to starting point
2t


x  A cos
 
 T

2 (t  t0 ) 
2t0

 A cos
 if  
T
T


90-degrees changes cosine to sine


cos t    sin t 
2

Velocity and
Acceleration vs. time
x
 Velocity is 90
“out of phase” with x:
When x is at max,
v is at min ....
v
 Acceleration is 180°
“out of phase” with x
a = F/m = - (k/m) x
a
Find vmax with E conservation
v and a vs. t
x  A cos t
v  vmax sin t
a   amax cos t
1 2 1 2
kA  mvmax
2
2
k
vmax  A
m
Find amax using F=ma
 kx  ma
 kA cos t   mamax cos t
amax
k
A
m
What is ?
Requires calculus. Since
d
A cos t  A sin t
dt
k
vmax  A  A
m
k

m
Springs and masses
Formula Summary
1
f 
T
2
  2f 
T
x  A cos t
v  A sin t
a   A cos t
2
k

m
Example13.1
An block-spring system oscillates with an amplitude
of 3.5 cm. If the spring constant is 250 N/m and
the block has a mass of 0.50 kg, determine
(a) the mechanical energy of the system
(b) the maximum speed of the block
(c) the maximum acceleration.
a) 0.153 J
b) 0.783 m/s
c) 17.5 m/s2
Example 13.2
A 36-kg block is attached to a spring of constant
k=600 N/m. If the block is pulled 3.5 cm away from
its equilibrium positions and released from rest at
t=0, then at t=0.75 seconds,
a) what is the position of the block?
b) what is the velocity of the block?
a) -3.489 cm
b) -1.138 cm/s
Example 13.3
A 36-kg block is attached to a spring of constant
k=600 N/m. If the block is pulled 3.5 cm away from
its equilibrium position and is pushed so that is has an
initial velocity of 5.0 cm/s at t=0, then at t=0.75
seconds,
a) what is the position of the block?
a) 0.00179 cm
Pendulum Demo
Simple Pendulum
F   mg sin 
x
x
sin   2

2
x L L
mg
F 
x
L
Looks like Hooke’s law (k  mg/L)
Simple Pendulum
mg
F 
x
L
" k "  mg / L
" k"

m
g

L
   max cos(t   )
v   Lmax sin(t   )
Simple pendulum
g

L
Frequency is independent of mass and amplitude!
(for small amplitudes)
Example 13.4
A man enters a tall tower, needing to know its height h.
He notes that a long pendulum extends from the roof
almost to the ground and that its period is 15.5 s.
(a) How tall is the tower?
a) 59.7 m
(b) If this pendulum is taken to the Moon, where the
free-fall acceleration is 1.67 m/s2, what is the period
of the pendulum there?
b) 37.6 s
Damped Oscillations
In real systems,
friction slows motion
Longitudinal (Compression) Waves
Sound waves are longitudinal waves
Compression and Transverse Waves Demo
Transverse Waves
Elements move perpendicular to wave motion
Elements move parallel to wave motion
Snapshot of a Transverse Wave
x


y  A cos 2   
 

wavelength
x
Snapshot of Longitudinal Wave

x


y  A cos 2   
 

y could refer to pressure or density
Moving Wave
x  vt


y  A cos 2
 



moves to right with velocity v
Fixing x=0,
v


y  A cos  2 t   



v
f  , v  f

Moving Wave: Formula Summary
 x


y  A cos 2   ft    
 
 
v  f
Example 13.5
A wave traveling in the
positive x direction has a
frequency of f = 25.0 Hz as
shown in the figure. Find the
a) amplitude
b) wavelength
c) period
a) 9.0 cm
b) 20.0 cm
c) 0.04 s
d) speed of the wave.
d) 500 cm/s
Example 13.6
Consider the following expression for a pressure wave,
P  43.5  cos1.75 x  5.25t 
where it is assumed that x is in cm,t is in seconds and
P will be given in N/m2.
a) What is the amplitude?
b) What is the wavelength?
c) What is the period?
a) 43.5 N/m2
b) 3.59 cm
c) 1.197 s
d) What is the velocity of the wave?
d) -3.0 cm/s
Example 13.7
Which of these waves move in the positive x direction?
a)
b)
c)
d)
e)
y  21.3  cos( 4.2 x  3.4t )
y  21.3  sin( 4.2 x  3.4t )
y  21.3  cos( 4.2 x  3.4t )
y  21.3  sin( 4.2 x  3.4t )
y  21.3  cos( 4.2 x  3.4t )
Speed of a Wave in a Vibrating String
T
m
v
where  

L
For different kinds of waves: (e.g. sound)
• Always a square root
• Numerator related to restoring force
• Denominator is some sort of mass density
Example 13.9
A simple pendulum consists of a ball of mass
5.00 kg hanging from a uniform string of
mass 0.060 0 kg and length L. If the period
of oscillation for the pendulum is 2.00 s,
determine the speed of a transverse wave in
the string when the pendulum hangs
vertically.
28.5 m/s
Superposition Principle
Traveling waves can pass
through each other without
being altered.
y ( x, t )  y1 ( x, t )  y2 ( x, t )
Reflection – Fixed End
Reflected wave is inverted
Reflection – Free End
Reflected pulse not inverted
Standing Waves (Read 14.8)
Consider a wave and its reflection:
yright
yleft
yright  yleft
 x


 A sin 2   ft 

 
x
  x


 Asin  2  cos 2ft  cos 2  sin 2ft 
 
  

 x


 A sin 2   ft 

 
x
  x


 Asin  2  cos 2ft  cos 2  sin 2ft 
 
  

x

 2 A sin  2  cos 2ft
 
Standing Waves
yright  yleft
x

 2 A sin  2  cos 2ft
 
•Factorizes into x-piece and t-piece
•Always ZERO at x=0 or x=m/2
Resonances
Integral number of half
wavelengths in length L

n L
2
Fig 14.16, p. 442
Slide 18
Nodes and anti-nodes


A node is a minimum in the pattern
An antinode is a maximum
Fundamental, 2nd, 3rd... Harmonics
2nd harmonic

n L
2
3rd harmonic
Fundamental (n=1)
Fig 14.18, p. 443
Slide 25
Example 13.10
A cello string vibrates in its fundamental mode with a
frequency of 220 vibrations/s. The vibrating segment is
70.0 cm long and has a mass of 1.20 g.
a) Find the tension in the string
a) 163 N
b) Determine the frequency of the string when it
vibrates in three segments.
b) 660 Hz
Loose Ends
L
L  2n  1

4

4
L3
L5

4

(Organ pipes open at one end)
4
Example 13.11
An organ pipe of length 1.5 m is open at one end.
What are the lowest two harmonic frequencies?
DATA: Speed of sound = 343 m/s
57.2 Hz, 171.5 Hz