Q30E - University Physics with Modern Physics

Question

Singing in the Shower. A pipe closed at both ends can have standing waves inside of it, but you normally don’t hear them because little of the sound can get out. But you can hear them if you are inside the pipe, such as someone singing in the shower. (a) Show that the wavelengths of standing waves in a pipe of length L that is closed at both ends are ${\mathbf{\lambda }}_{\mathbf{0}}\mathbf{=}\mathbf{2}\mathbf{L}\mathbf{/}\mathbf{n}$ and the frequencies are given by ${\mathbf{f}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{nv}}{\mathbf{4}\mathbf{L}}{\mathbf{nf}}_{\mathbf{1}}$, where n = 1, 2, 3, c.(b) Modelling it as a pipe, find the frequency of fundamental and the first two overtones for a shower 2.50 m tall. Are these frequencies audible?

Step-by-Step Solution

Answer

The three fundamental frequencies are 68.8 Hz, 137.6Hz, 206.4Hz

STEP 1 Concept of the frequency of wave in an open pipe

The frequency of wave in an open pipe is given as $\mathbf{f}\mathbf{=}\mathbf{n}\frac{\mathbf{v}}{\mathbf{2}\mathbf{L}}$ were, were, f is the frequency of nth harmonic, v is the velocity of the wave, $\mathbf{n}\mathbf{\to }\mathbf{nth}$ harmonic (n — 1, 3, 5, ...), L is the length of the pipe.

STEP 2 To prove L = nλ n 2 and f = nv 2 L

The distance between two nodes equals $\frac{\lambda }{2}$ at any standing wave both ends are closed, so the molecules at this end can't move Therefore, each end is considered a node. 

When the length of the pipe = $\frac{{\lambda }_1}{2}$ we have two nodes at the ends of the pipe 

and to add additional node, we add $\frac{{\lambda }_2}{2}$ 

$\mathrm{L}=\frac{{\mathrm{\lambda }}_1}{2}=\frac{{\mathrm{\lambda }}_2}{2}=\frac{{\mathrm{\lambda }}_2}{2}=\frac{2{\mathrm{\lambda }}_2}{2}={\mathrm{\lambda }}_2$ and so on…

 

$\mathrm{L}=\frac{{\mathrm{n\lambda }}_{\mathrm{n}}}{2}$ hence, proved

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{\lambda }}{\mathrm{T}}={\mathrm{\lambda f}}_1 \\ \mathrm{f}=\frac{\mathrm{v}}{\mathrm{\lambda }} \\ {\mathrm{f}}_{\mathrm{n}}=\frac{\mathrm{v}}{{\mathrm{\lambda }}_{\mathrm{n}}}=\frac{\mathrm{v}}{2\mathrm{L}/\mathrm{n}}=\frac{\mathrm{n\lambda }}{2\mathrm{L}} \end{gathered}$$ 

 

Hence proved.

Step 3 Calculate the fundamental frequency

The fundamental frequency IS n=1, n=2, n=3

Use the formula, ${\mathrm{f}}_{\mathrm{n}}=\frac{\mathrm{n\lambda }}{2\mathrm{L}}$

$$\begin{gathered} {\mathrm{f}}_1=\frac{\mathrm{n\lambda }}{2\mathrm{L}}=\frac{1\times 344\mathrm{m}/\mathrm{s}}{2\left(2.5 \mathrm{m}\right)}=68.8\mathrm{Hz} \\ {\mathrm{f}}_2=\frac{\mathrm{n\lambda }}{2\mathrm{L}}=\frac{2\times 344\mathrm{m}/\mathrm{s}}{2\left(2.5\mathrm{m}\right)}=137.6\mathrm{Hz} \\ {\mathrm{f}}_3=\frac{\mathrm{n\lambda }}{2\mathrm{L}}=\frac{3\times 344\mathrm{m}/\mathrm{s}}{2\left(2.5\mathrm{m}\right)}=206.4\mathrm{Hz} \end{gathered}$$

Therefore, the three fundamental frequencies are 68.8 Hz, 137.6Hz, 206.4Hz