Question

What are the three longest wavelengths for standing sound waves in a 121-cm-long tube that is (a) open at both ends and (b) open at one end, closed at the other?

Short answer

Part a

The three longest wavelengths for standing sound waves in a tube open at both ends are $2.42m,1.21m,$and $0.81m$respectively.

Part b

The three longest wavelengths for standing sound waves in a tube are open at one end, closed at the other are $4.84m,1.61m,$and $0.97m$ respectively,

Step-by-step solution

Given information

The length of the tube is$L=121cm=1.21m$.

Part a

For the open-open tube, the two open ends exhibit antinodes of a standing wave. The possible wavelengths for this case are ${\lambda }_m=\frac{2L}{m}$where $m=1,2,3,\cdots$

The three longest wavelengths are as follows

(1)

$$\begin{gathered} {\lambda }_1=\frac{2\times 1.21}{1} \\ =2.42m \end{gathered}$$

(2)

$$\begin{gathered} {\lambda }_2=\frac{2\times 1.21}{2} \\ =1.21m \end{gathered}$$

(3)

$$\begin{gathered} {\lambda }_3=\frac{2\times 1.21}{3} \\ =0.81m \end{gathered}$$

Therefore, the three longest wavelengths for standing sound waves in a tube open at both ends are $2.42m,1.21m,$and $0.81m$respectively.

Part b

In the case of an open-closed tube, wavelengths are

${\lambda }_m=\frac{4L}{m}$where $m=1,2,3,\dots$

The three longest wavelengths are as follows

(1)

$$\begin{gathered} {\lambda }_1=\frac{4\times 1.21}{1} \\ =4.84m \end{gathered}$$

(2)

$$\begin{gathered} {\lambda }_2=\frac{4\times 1.21}{2} \\ =1.61m \end{gathered}$$

(3)

$$\begin{gathered} {\lambda }_3=\frac{4\times 1.21}{3} \\ =0.97m \end{gathered}$$

Therefore, the three longest wavelengths for standing sound waves in a tube open at one end, closed at the other are$4.84m,1.61m,$and$0.97m$respectively.$0.97m$