If one end of the cord, say at
x= ,
is a loose end,
then at that position one must have an anti-node.
Hence, 2A sin(k) must be maximum
2A sin(k) = ±2A
That can be achieved for
k = π/2, 3π/2, 5π/2, 7π/2, ...
With k=2π/λ
and v=fλ,
one arrives at
f = v/4,
3v/4,
5v/4,
7v/4, ...
The above formula describes the relation
for the lengths of tubes and the tones produced,
for wind instruments like
a pipe
organ,
a trumpet
or a saxophone.
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