Question
What is the length of the pipe?
Answer:
(a) 0.52m
(b) f2=640 Hz and f3=960 Hz
(c) 352.9 Hz
Explanation:
For an open pipe, the velocity is given by
[tex]v=\frac {2Lf}{n}[/tex]
Making L the subject then
[tex]L=\frac {nV}{2f}[/tex]
Where f is the frequency, L is the length, n is harmonic number, v is velocity
Substituting 1 for n, 320 Hz for f and 331 m/s for v then
[tex]L=\frac {1*331}{2*320}=0.5171875\approx 0.52m[/tex]
(b)
The next two harmonics is given by
f2=2fi
f3=3fi
f2=3*320=640 Hz
f3=3*320=960 Hz
Alternatively, [tex]f2=2\times \frac {v}{2L}[/tex] and [tex]f3=3\times \frac {v}{2L}[/tex]
[tex]f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz[/tex]
(c)
When v=367 m/s then
[tex]f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz[/tex]
