Traveling Waves: Refraction
• Refraction: Zero Incident Angle
(speed of propagation is the frequency times the wavelength)
v1  f11
v 2  f 2 2
t 
1
2  v2 t 
v2
Medium 1
Speed v1
(time for crest B to reach medium 2)
v1
2
Crest B

v2 1
v1
1
v1
f1  f 2
R. Field 11/19/2013
University of Florida
(distance traveled by crest A in time t)
Medium 2
Speed v2
v1
1
Crest A
2
v2
v2 < v 1
(Law of Refraction)
(frequency is the same)
PHY 2053
Page 1
Traveling Waves: Refraction
• Refraction: Incident Angle q1
(speed of propagation is the frequency times the wavelength)
v1  f1
f1  f 2
v2  f2
sin q1 
sin q 2 
sin q1
1
2
v2


1
L
2
1
(incident angle)
(refracted angle)
L
sin q 2
2
1
v1
sin q1 sin q 2

v1
v2
R. Field 11/19/2013
University of Florida
v1
q1
Medium 1
Speed v1
L
Medium 2
Speed v2
2
q2
(Law of Refraction)
v2 < v1
v2
(Law of Refraction)
(Law of Refraction)
PHY 2053
Page 2
Standing Waves: Superposition
• Standing Waves:
y1 ( x, t )  A sin( 1 )  A sin( kx  t )
y2 ( x, t )  A sin( 2 )  A sin( kx  t )
A
v
2= kx+t
1= kx-t

k
(Right Moving)
(Left Moving)
(Speed of the travelling waves)
Nodes fixed in position!
y
2AA
A
Standing Wave
(not rotating)
f
X
  2  1  2t
L
Add the two travelling waves together (right moving plus left moving) as follows:
y12 ( x, t )  A sin( 1 )  A sin( 2 )  2 A sin( kx) cos(t )
Ln

2
n  1,2,3, 
 vk v
f 


2 2 
R. Field 11/19/2013
University of Florida
(Allowed wavelengths
of the Standing Wave)
(Allowed frequencies
of the Standing Wave)
n 
2L
n
nv
fn 

n 2 L
PHY 2053
v
(Standing Wave)
n  1,2,3, 
n  1,2,3, 
(Fundamental Frequency is n = 1 )
Page 3
Example Problem: Standing Waves
• A string in a grand piano is 2 m long and has a mass density of 1 g/m. If the
fundamental frequency of oscillations of the string is 440 Hz, what is the tension in the
string (in N)?
Answer: 3097.6
v
FT

v
f1 
2L
R. Field 11/19/2013
University of Florida
FT  v 2
FT  4L2 f12  4(0.001kg / m)( 2m) 2 (440 / s) 2
 3097.6 N
v  2Lf1
PHY 2053
Page 4
Example Problem: Standing Waves
• A nylon guitar string has a linear density of 5 g/m and is under a tension of 200 N. The
fixed supports are D = 60 cm apart. The string is oscillating in the standing wave pattern
shown in the figure. What is the frequency of the traveling waves whose superposition
gives this standing wave?
Answer: 500 Hz
3
D
2
f 
v


2D

3
1

R. Field 11/19/2013
University of Florida
FT
3

 2D
v
FT

FT
3
200 N

 500 Hz
 2(0.6m) (0.005kg / m)
PHY 2053
Page 5
Example Problem: Standing Waves
• A string, which is tied to a sinusoidal oscillator at P
and which runs over a support Q, is stretched by a
block of mass m. The distance L = 2.0 m, the linear
mass density of the string  = 4.9 g/m, and the
oscillator frequency f = 100 Hz. The motion at P is in
the vertical direction, and its amplitude is small
enough for that point to be considered a node. A
node also exists at Q. What mass allows the
oscillator to set up the second harmonic on the
string?
Answer: 20 kg
v
f2 
FT

v
2
m
FT  mg  v 2

2v
2L
v 2
g

L2 f 22
g
(0.0049kg / m)( 2m) 2 (100 / s ) 2

9.8m / s 2
 20kg
v 2  (Lf 2 ) 2
(Second harmonic is n = 2)
R. Field 11/19/2013
University of Florida
PHY 2053
Page 6