SME
University Physics III: Waves, Optics & Modern · Problem type
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.”
Diagram
λ = 2L/n = 2 × 0.8/3 = 0.533 m
f = v/λ = 100/0.533 = 188 Hz
4 nodes (N, no motion) and 3 antinodes (A, the most motion).
The fixed ends are nodes.
This is an example. Type your own values to start a new problem.
fₙ (
)
=
×
v (m/s)
2
×
L (
)
FT
Tension
example
μ
Linear density
example · kg/m
λₙ
Wavelength
calculated · m
f₁
Fundamental
calculated
Formulas
v
=
√(
FT
μ
)
100
=
√(
100
0.01
)
λₙ
=
2
×
L
n
0.5333
=
2
×
0.8
3
v
=
fₙ
×
λₙ
100 = 187.5 × 0.5333
fₙ
=
n
×
f₁
187.5 = 3 × 62.5
Assumptions
•
Both ends are fixed, so both are nodes: harmonic n fits n half wavelengths in the length, λₙ = 2L ÷ n.
•
The tension and the mass per meter set the speed, v = √(FT ÷ μ), the same for every harmonic.
•
Each harmonic rings at fₙ = v ÷ λₙ = nv ÷ (2L), a whole multiple of the fundamental f₁.
Step-by-step
WE KNOW
FT = 100 N, μ = 0.01 kg/m, L = 0.8 m, n = 3
FIND
wave speed (v), wavelength (λₙ), frequency (fₙ), fundamental (f₁)
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 (λₙ)
Use λₙ = 2 × L ÷ n
Harmonic
n
fits
n
half
wavelengths
between
the
fixed
ends.
λₙ
=
2
×
0.8
3
λₙ
=
1.6
3
λₙ = 0.5333 m
Step 3 · Find frequency (fₙ)
Use v = fₙ × λₙ
Waves
of
length
λₙ
pass
each
point
at
v
:
v
λₙ
of
them
each
second.
fₙ
=
v
λₙ
fₙ
=
100
0.5333
fₙ = 187.5 Hz
Step 4 · Find fundamental (f₁)
Use fₙ = n × f₁
The
fundamental
has
one
half
wavelength:
fₙ
is
n
times
it.
f₁
=
fₙ
n
f₁
=
187.5
3
f₁
=
62.5
Hz
Check
Put the numbers back in. Both sides should match.
100
=
√(
100
0.01
)
✓
0.5333
=
2
×
0.8
3
✓
100 = 187.5 × 0.5333 ✓
187.5 = 3 × 62.5 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.