Traveling Waves: Refraction
• Refraction: Zero Incident Angle
(speed of propagation is the frequency times the wavelength)
v1 = f1λ1
Crest B
v2 = f 2 λ2
∆t =
λ1
(time for crest B to reach medium 2)
v1
λ2 = v2 ∆t =
λ2
v2
Medium 1
Speed v1
=
v 2 λ1
v1
λ1
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
λ1
Crest A
λ2
v2
v2 < v 1
(Law of Refraction)
(frequency is the same)
PHY 2053
Page 1
Traveling Waves: Refraction
• Refraction: Incident Angle θ1
(speed of propagation is the frequency times the wavelength)
v1 = fλ1
f1 = f 2
v 2 = fλ 2
sin θ1 =
sin θ 2 =
sin θ 1
λ1
λ2
v2
=
=
λ1
L
λ2
λ1
(incident angle)
(refracted angle)
L
sin θ 2
λ2
λ1
v1
sin θ1 sin θ 2
=
v1
v2
R. Field 11/19/2013
University of Florida
v1
θ1
Medium 1
Speed v1
L
Medium 2
Speed v2
λ2
θ2
(Law of Refraction)
v2 < v1
v2
(Law of Refraction)
(Law of Refraction)
PHY 2053
Page 2
Standing Waves: Superposition
• Standing Waves:
A
y1 ( x, t ) = A sin(Φ1 ) = A sin( kx − ωt )
y2 ( x, t ) = A sin(Φ 2 ) = A sin( kx + ωt )
ω (Speed of the travelling waves)
v=
Φ 2= kx+ω
ωt
Φ1= kx-ω
ωt
k
(Right Moving)
(Left Moving)
Nodes fixed in position!
y
2AA
A
Standing Wave
(not rotating)
φ
X
∆Φ = Φ 2 − Φ1 = 2ωt
L
Add the two travelling waves together (right moving plus left moving) as follows:
y12 ( x, t ) = A sin(Φ1 ) + A sin(Φ 2 ) = 2 A sin( kx) cos(ωt )
L=n
λ
2
n = 1,2,3, L
ω vk v
f =
=
=
2π 2π λ
R. Field 11/19/2013
University of Florida
(Allowed wavelengths
of the Standing Wave)
(Allowed frequencies
of the Standing Wave)
λn =
nv
fn =
=
λn 2 L
PHY 2053
v
2L
n
(Standing Wave)
n = 1,2,3, L
n = 1,2,3, L
(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
µ
v
f1 =
2L
R. Field 11/19/2013
University of Florida
FT = µv 2
FT = 4 µL2 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
3λ
D=
2
f =
v
λ
=
2D
λ=
3
1
λ
R. Field 11/19/2013
University of Florida
FT
µ
=
3
2D
v=
FT
µ
=
FT
µ
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 µ = 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
FT
v=
fn =
v
λn
=
nv
2L
v
2v
f2 =
=
λ2 2 L
g
=
µL2 f 22
g
(0.0049kg / m)(2m) 2 (100 / s ) 2
=
9.8m / s 2
= 20kg
FT = mg = µv 2
µ
m=
µv 2
v 2 = (Lf 2 ) 2
(Second harmonic is n = 2)
R. Field 11/19/2013
University of Florida
PHY 2053
Page 6