A pipe has a length of 1.5m. determine the frequencies of the first three harmonics if the pipe is open
Answers
Answer:
57.2 Hz, 171.5 Hz and 285.8 Hz
Explanation:
In a pipe, an open end has antinodes while closed ends have nodes.
The distance is between a node and an antinode is an odd multiple of one-quarter of a wavelength.
For the first harmonic,
l=\dfrac{\lambda}{4}l=
4
λ
\lambda=4lλ=4l
Also, v = f\lambdav=fλ
where f is the frequency and v is the velocity.
f = \dfrac{v}{\lambda}f=
λ
v
Substitute the value of ll .
f = \dfrac{343}{4\times1.5}=57.2 \text{ Hz}f=
4×1.5
343
=57.2 Hz
For the second harmonic,
l=\dfrac{3\lambda}{4}l=
4
3λ
\lambda=4l/3λ=4l/3
f = \dfrac{v}{\lambda}f=
λ
v
f = \dfrac{3v}{4l}f=
4l
3v
f = \dfrac{3\times343}{4\times1.5}=171.5\text{ Hz}f=
4×1.5
3×343
=171.5 Hz
For the third harmonic,
l=\dfrac{5\lambda}{4}l=
4
5λ
\lambda=4l/5λ=4l/5
f = \dfrac{v}{\lambda}f=
λ
v
f = \dfrac{5v}{4l}f=
4l
5v
f = \dfrac{5\times343}{4\times1.5}=285.8\text{ Hz}f=
4×1.5
5×343
=285.8 Hz
Explanation:
hope it helps u