Question

Organ pipe $\mathbf{A}$, with both ends open, has a fundamental frequency of$\mathbf{\text{300Hz}}$. The third harmonic of organ piperole="math" localid="1661418890848" $\mathbf{\text{B}}$, with one end open, has the same frequency as the second harmonic of pipe$\mathbf{\text{A}}$. How long are (a) pipe$\mathbf{\text{A}}$and (b) pipe$\mathbf{\text{B}}$?

Short answer

1. The length of pipe A is$0.572\mathrm{m}$ .
2. The length of pipe B is$0.429\mathrm{m}$ .

Step-by-step solution

Given data

• Fundamental frequency ,${\mathrm{f}}_0=300\text{Hz}$ ( for pipe A)
• Second harmonics of A = 3^rd harmonics of B

Determining the concept

Apply the formula for harmonics for a closed and open organ pipe.

The expression for the frequency is given by,

${\mathrm{f}}_0=\frac{\mathrm{n}}{2{\mathrm{L}}_{\mathrm{A}}}\mathrm{v}$

Here,${\mathrm{f}}_0$ is frequency, $L$ is length and $v$is the velocity.

(a) Determine the length of pipe A

Fundamental frequency is defined as,

${\mathrm{f}}_0=\frac{\mathrm{n}}{2{\mathrm{L}}_{\mathrm{A}}}\mathrm{v}$

For fundamental frequency, $\mathrm{n}=1$

${\mathrm{f}}_0=\frac{\mathrm{v}}{2{\mathrm{L}}_{\mathrm{A}}}$

$\begin{array}{c}{\mathrm{L}}_{\mathrm{A}}=\frac{\mathrm{v}}{2{\mathrm{f}}_0}\\ =\frac{(343\mathrm{m}/\mathrm{s})}{2\times 300\text{Hz}}\\ =0.572\text{m}\end{array}$

Hence,the length of pipe A is $0.572\text{\hspace{0.17em}m}$.

(b) Determine the length of pipe B

Frequency of harmonics for one end open pipe is defined as,

${\mathrm{f}}_{\mathrm{n}}=\left(\mathrm{n}-\frac{1}{2}\right)\left(\frac{\mathrm{v}}{2{\mathrm{L}}_{\mathrm{B}}}\right)$

$\text{Forthirdharmonicsmeanssecondoverton,}\text{n}=2$

$\begin{array}{c}{\mathrm{f}}_3=\left(2-\frac{1}{2}\right)\left(\frac{\mathrm{v}}{2{\mathrm{L}}_{\mathrm{B}}}\right)\\ =\frac{3\times 343\mathrm{m}/\mathrm{s}}{2\times 2\times {\mathrm{L}}_{\mathrm{B}}}\end{array}$

${\mathrm{f}}_3=\frac{257.25}{{\mathrm{L}}_{\mathrm{B}}}\text{Hz}$

Now, second harmonic of A = third harmonic of B

$2{\mathrm{f}}_0={\mathrm{f}}_3$

$2\left(300\text{Hz}\right)=\frac{257.25}{{\mathrm{L}}_{\mathrm{B}}}\text{Hz-m}$

$\begin{array}{c}{\mathrm{L}}_{\mathrm{B}}=\frac{\text{257.25m}}{\text{600}}\\ =0.429\text{m}\end{array}$

Hence,the length of pipe B is $0.429\text{m}$.