Q29E - University Physics with Modern Physics

Question

The longest pipe found in most medium-size pipe organs is 4.88 m (16 ft) long. What is the frequency of the note corresponding to the fundamental mode if the pipe is (a) open at both ends, (b) open at one end and closed at the other?

Step-by-Step Solution

Answer

a) The frequency of the standing wave in an open pipe is  $35.25\text{\hspace{0.17em}Hz}$ 

b) The frequency of the standing wave in a pipe closed at one end is   $17.623\text{\hspace{0.17em}Hz}$

Step 1 Concept of the frequency of standing wave in an open pipe and closed pipe

The frequency of the standing wave in an open pipe is given as $\mathbf{f}\mathbf{=}\mathbf{n}\frac{\mathbf{v}}{\mathbf{2}\mathbf{L}}$  and the frequency of the standing wave in a closed pipe is $\mathbf{f}\mathbf{=}\mathbf{n}\frac{\mathbf{v}}{\mathbf{4}\mathbf{L}}$  where, f is the frequency of nth harmonic, v is the velocity of the wave, $\mathbf{n}$ represents  $\mathbf{nth}$ harmonic (n — 1, 3, 5, ...), L is the length of the pipe.

Step 2 Calculation of the frequency in an pipe

The frequency of standing wave in an open pipe is given as 

 ${\mathrm{f}}_1=\mathrm{n}\frac{\mathrm{v}}{2\mathrm{L}}=\frac{1\times 344\text{\hspace{0.17em}m}/\text{s}}{2\times 4.88\text{\hspace{0.17em}m}}=35.25\text{\hspace{0.17em}Hz}$

The frequency of standing wave in a pipe closed at one end is given as 

 ${\mathrm{f}}_1=\mathrm{n}\frac{\mathrm{v}}{4\mathrm{L}}=\frac{1\times 344\text{\hspace{0.17em}m}/\text{s}}{4\times 4.88\text{\hspace{0.17em}m}}=17.623\text{\hspace{0.17em}Hz}$

The frequencies are $35.25\text{\hspace{0.17em}Hz}$  and $17.623\text{\hspace{0.17em}Hz}$  respectively.