SME
Grade 11 Science · Problem type
A pipe closed at one end
Use this for “A 0.25 m tube is closed at the bottom. What is its lowest note in 343 m/s air, and the next one it can play?”
Diagram
λ = 4L/n = 4 × 0.25/1 = 1 m
f = v/λ = 343/1 = 343 Hz
1 node (N, no motion) and 1 antinode (A, the most motion).
The closed end is a node, the open end an antinode.
This is an example. Type your own values to start a new problem.
λ (m)
=
4
×
L (m)
n
v (m/s)
=
f (Hz)
×
λ (m)
Formulas
λ
=
4
×
L
n
1 = 4 × 0.25/1
v
=
f
×
λ
343 = 343 × 1
Assumptions
•
The closed end is a node and the open end an antinode: odd numbers of quarter wavelengths fit, λ = 4L/n.
•
Only odd harmonics sound: n = 1, 3, 5, …
•
The next note after n = 1 is n = 3: three times the lowest frequency.
Step-by-step
WE KNOW
n = 1, L = 0.25 m, v = 343 m/s
FIND
wavelength (λ), frequency (f)
Step 1 · Find wavelength (λ)
Use λ = 4 × L/n
Harmonic
n
fits
n
quarter
wavelengths
in
the
length.
λ
=
4
×
L
λ
=
4
×
0.25
λ
=
1
m
Step 2 · Find frequency (f)
Use v = f × λ
Waves
pass
at
v
;
each
is
λ
long:
v
λ
of
them
each
second.
f
=
v
λ
f
=
343
1
f
=
343
Hz
Check
Put the numbers back in. Both sides should match.
1 = 4 × 0.25/1 ✓
343 = 343 × 1 ✓
Topic check
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