Question 14.9: HARMONICS OF A PIPE GOAL Find frequencies of open and closed...

(a) Find the frequencies if the pipe is open at both ends.

Substitute into Equation 14.18, with n=1 :

f_{1}=\frac{v}{2 L}=\frac{343 \mathrm{~m} / \mathrm{s}}{2(2.46 \mathrm{~m})}=69.7 \mathrm{~Hz}

Multiply to find the second and third harmonics:

f_{2}=2 f_{1}=139 \mathrm{~Hz} \qquad f_{3}=3 f_{1}=209 \mathrm{~Hz}

(b) How many harmonics lie between 20 \mathrm{~Hz} and 20  000 \mathrm{~Hz} for this pipe?

Set the frequency in Equation 14.18 equal to 2.00 \times 10^{4} \mathrm{~Hz} and solve for n :

\begin{aligned}&f_{n}=n \frac{v}{2 L}=n \frac{343 \mathrm{~m} / \mathrm{s}}{2(2.46 \mathrm{~m})}=2.00 \times 10^{4} \mathrm{~Hz} \end{aligned}

This works out to n=286.88, which must be truncated down \left(n=287\right. gives a frequency over \left.2.00 \times 10^{4} \mathrm{~Hz}\right) :

n=286

(c) Find the frequencies for the pipe closed at one end.

Apply Equation 14.19 with n=1

\begin{aligned}&f_{1}=\frac{v}{4 L}=\frac{343 \mathrm{~m} / \mathrm{s}}{4(2.46 \mathrm{~m})}=34.9 \mathrm{~Hz} \\&f_{3}=3 f_{1}=105 \mathrm{~Hz} \quad f_{5}=5 f_{1}=175 \mathrm{~Hz}\end{aligned}

REMARKS For a pipe closed at one end, referring to the “second harmonic” can be confusing, because that corresponds to f_{3}. Calling it the first overtone, however, is unambiguous.