Question

A PVC pipe has a length of 53.594 centimeters.a. What are the frequencies of the first three harmonics when the pipe is open at both ends? b. What are the frequencies of the first three harmonics when the pipe is closed at one end and open at the other?

Answer 1

Given:

Length of pipe, L = 53.594 cm

Let's solve for the following:

• (a). What are the frequencies of the first three harmonics when the pipe is open at both ends?

To find the frequency of the first harmonics, apply the formula:

$f_1=\frac{v}{2l}$

Where:

v is the speed of sound = 343 m/s

l is the length of the pipe in meters.

Where:

100 cm = 1 m

53.594 cm = 0.53594 m

Hence, for the frequency, we have:

$\begin{gathered} f_1=\frac{343}{2*0.53594} \\ \\ f_1=319.99\approx320\text{ Hz} \end{gathered}$

For the frequency of the second harmonics, we have:

$\begin{gathered} f_2=2f_1 \\ \\ f_2=2*320 \\ \\ f_2=640\text{ Hz} \end{gathered}$

For the frequency of the third harmonics:

$\begin{gathered} f_3=3f_1 \\ \\ f_3=3*320 \\ \\ f_3=960\text{ Hz} \end{gathered}$

• Part B.

,

• What are the frequencies of the first three harmonics when the pipe is closed at one end and open at the other?

For the first harmonics when the pipe is closed at one end, apply the one end open harmonics equation.

$\begin{gathered} f_1=\frac{v}{4l} \\ \\ f_1=\frac{343}{4*0.53594} \\ \\ f_1=160\text{ Hz.} \end{gathered}$

The next 2 harmonics will be the odd multiples of the first.

Hence, we have:

$\begin{gathered} f_3=3f_1 \\ f_3=3*160 \\ f_3=480\text{ Hz.} \\ \\ \\ f_5=5f1 \\ f_5=5*160 \\ f_5=800\text{ Hz.} \end{gathered}$

Therefore, the frequencies of of the first three harmonics when the pipe is closed at one end and open at the other are:

160 Hz, 480 Hz, 800 Hz.

ANSWER:

(A). 320Hz, 640 Hz, 960 Hz.

(B). 160 Hz, 480 Hz, 800 Hz.