SME
University Physics III: Waves, Optics & Modern
Waves
Use this for “A rope under 180 N tension has a linear density of 0.2 kg/m. A 3 Hz shake with an amplitude of 4 cm sends waves along it. Find v, λ, k, ω and the power carried.”
Diagram
Crest to crest is one wavelength, 2 m.
The crest rises 0.005 m above the middle.
50 waves pass each second.
This is an example. Type your own values to start a new problem.
v (m/s)
=
FT (
)
μ (kg/m)
f
Frequency
example
λ
Wavelength
calculated · m
k
Wave number
calculated · rad/m
ω
Angular frequency
calculated · rad/s
A
Amplitude
example · m
P
Power carried
calculated
Formulas
v
=
√(
FT
μ
)
100
=
√(
100
0.01
)
v
=
f
×
λ
100 = 50 × 2
k
=
2π
λ
3.1416
=
2π
2
ω
=
2π
×
f
314.1593 = 2π × 50
P
=
1
2
×
μ
×
v
×
ω
2
×
A
2
1.2337
=
1
2
×
0.01
×
100
×
314.1593
2
×
0.005
2
Assumptions
•
The wave y(x, t) = A sin(kx − ωt) moves toward +x: k = 2π ÷ λ in radians per meter, ω = 2πf in radians per second.
•
The string sets the speed, v = √(FT ÷ μ); the source sets the frequency, so the wavelength λ = v ÷ f follows.
•
Small amplitudes: the tension stays FT as the string moves, and the wave carries the power P = ½μvω²A².
Step-by-step
WE KNOW
FT = 100 N, μ = 0.01 kg/m, f = 50 Hz, A = 0.005 m
FIND
wave speed (v), wavelength (λ), wave number (k), angular frequency (ω), power carried (P)
Step 1 · Find wave speed (v)
Use v = √(FT ÷ μ)
Tension
pulls
each
piece
back;
its
mass
per
meter
resists.
The
speed
is
the
square
root
of
FT
over
μ
.
v
=
√(
100
0.01
)
v
=
√
10
,000
v
=
100
m/s
Step 2 · Find wavelength (λ)
Use v = f × λ
Undo
v
=
fλ:
the
distance
one
wave
covers
in
one
period,
v
f
.
λ
=
v
f
λ
=
100
50
λ
=
2
m
Step 3 · Find wave number (k)
Use k = 2π ÷ λ
The phase kx grows by 2π radians over one wavelength.
k
=
2π
2
k
=
3.1416
rad/m
Step 4 · Find angular frequency (ω)
Use ω = 2π × f
The
phase
ωt
grows
by
2π
radians
each
period,
f
times
a
second.
ω
=
2π
×
50
ω
=
314.1593
rad/s
Step 5 · Find power carried (P)
Use P = ½ × μ × v × ω² × A²
Each
meter
of
string
carries
energy
1
2
μω²
A
2
;
the
wave
moves
it
along
at
v
meters
each
second.
P
=
1
2
×
0.01
×
100
×
314.1593
2
×
0.005
2
P
=
1
2
×
0.01
×
100
×
98,696
×
(2.5000
×
10
-5
)
P
=
0.005
×
100
×
98,696
×
(2.5000
×
10
-5
)
P
=
0.5
×
98,696
×
(2.5000
×
10
-5
)
P
=
49,348
×
(2.5000
×
10
-5
)
P
=
1.2337
W
Check
Put the numbers back in. Both sides should match.
100
=
√(
100
0.01
)
✓
100 = 50 × 2 ✓
3.1416
=
2π
2
✓
314.1593 = 2π × 50 ✓
1.2337
=
1
2
×
0.01
×
100
×
314.1593
2
×
0.005
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Standing waves on a string
Use this for “A 0.65 m guitar string under 72 N tension has 0.4 g of mass per meter. Find its fundamental and the wavelength of its third harmonic.”
Sound intensity and decibels at a distance
Use this for “A loudspeaker sends out 2.5 W of sound evenly in all directions. What are the intensity and the sound level 15 m away?”