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,
[tex]l=\dfrac{\lambda}{4}[/tex]
[tex]\lambda=4l[/tex]
Also, [tex]v = f\lambda[/tex]
where f is the frequency and v is the velocity.
[tex]f = \dfrac{v}{\lambda}[/tex]
Substitute the value of [tex]l[/tex].
[tex]f = \dfrac{343}{4\times1.5}=57.2 \text{ Hz}[/tex]
For the second harmonic,
[tex]l=\dfrac{3\lambda}{4}[/tex]
[tex]\lambda=4l/3[/tex]
[tex]f = \dfrac{v}{\lambda}[/tex]
[tex]f = \dfrac{3v}{4l}[/tex]
[tex]f = \dfrac{3\times343}{4\times1.5}=171.5\text{ Hz}[/tex]
For the third harmonic,
[tex]l=\dfrac{5\lambda}{4}[/tex]
[tex]\lambda=4l/5[/tex]
[tex]f = \dfrac{v}{\lambda}[/tex]
[tex]f = \dfrac{5v}{4l}[/tex]
[tex]f = \dfrac{5\times343}{4\times1.5}=285.8\text{ Hz}[/tex]
