Question

The fundamental frequency of a pipe that is open at both ends is 524 Hz. (a) How long is this pipe? If one end is now closed, find (b) the wavelength and (c) the frequency of the new fundamental.

Short answer

a)the length of the pipe is$0.328\text{\hspace{0.17em}m}$

b) The wavelength is $1.312\text{\hspace{0.17em}m}$

c) The frequency of new fundamental is $262.2\text{\hspace{0.17em}Hz}$.

Step-by-step solution

Concept of the frequency of the open and closed pipe.

The frequency of standing wave in an open pipe is $\mathbf{f}\mathbf{=}\mathbf{n}\frac{\mathbf{v}}{\mathbf{2}\mathbf{L}}$ and the frequency of standing wave in a closed pipe is$\mathbf{f}\mathbf{=}\frac{\mathbf{v}}{\mathbf{\lambda }}\mathbf{=}\mathbf{n}\frac{\mathbf{v}}{\mathbf{4}\mathbf{L}}$ where,f is The frequency of nth harmonic, v is the velocity of the wave,$\mathbf{n}\mathbf{,}\mathbf{}\mathbf{nth}$ harmonic (n — 1, 2, 3, ... ) , L is the length of the pipe.

Calculate the length of the pipe Fundamental frequency is the first harmonic n=1

$\mathrm{L}=\frac{\mathrm{nv}}{2{\mathrm{f}}_1}=\frac{1\times 344\text{\hspace{0.17em}m}/\text{s}}{2\times 524\text{\hspace{0.17em}Hz}}=0.328\text{\hspace{0.17em}m}$

Therefore, the length of the pipe is $0.328\text{\hspace{0.17em}m}$.

Calculate the wavelength

The wavelength is given as

$\mathrm{l}=\frac{4\mathrm{L}}{\mathrm{n}}=\frac{4\times 0.328\text{\hspace{0.17em}m}}{1}=1.312\text{\hspace{0.17em}m}$

Calculate the frequency of new fundamental

The new fundamental frequency is given as

$\mathrm{f}=\mathrm{n}\frac{\mathrm{v}}{4\mathrm{L}}=1\times \frac{344\text{\hspace{0.17em}m}/\text{s}}{4\times 0.328\text{\hspace{0.17em}m}}=262.2\text{\hspace{0.17em}Hz}$

The new fundamental frequency and wavelength are$262.2\text{\hspace{0.17em}Hz}$ and $1.312\text{\hspace{0.17em}m}$respectively.