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Found in: Page 513

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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# Pipe has only one open end; pipe B is four times as long and has two open ends. Of the lowest 10 harmonic numbers ${{\mathbit{n}}}_{{\mathbf{B}}}$ of pipe B , what are the (a) smallest, (b) second smallest, and (c) third smallest values at which a harmonic frequency of matches one of the harmonic frequencies of A ?

1. The smallest value of ${n}_{B}$ at which a harmonic frequency of B matches that of A is 2 .
2. The second smallest value of ${n}_{B}$ at which a harmonic frequency of matches that of A is 6 .
3. The third smallest value of ${n}_{B}$ at which a harmonic frequency of B matches that of A is 10.
See the step by step solution

## Step 1: Write the given data

The length of pipe B is ${L}_{B}=4{L}_{A}$

## Step 2: Determine the concept of the Doppler Effect

Use the concept of two open-ended and one open-ended pipe. With both ends of the pipe open, any harmonic can be set up in the pipe but with only one end open, only odd harmonic can be set up.

Formulae:

1. Resonant frequency for pipe with both ends open is $f=\frac{nv}{2L}$
2. Resonant frequency for pipe with both ends open is$f=\frac{nv}{4L}$

## Step 3: Calculate the smallest value of  nB

(a)

The resonant frequency for a pipe of length ${L}_{B}$ with two open ends using equation (i) is

$f=\frac{{n}_{B}v}{2{L}_{B}}$

For, $n=1,2,3\dots$

${L}_{B}=\frac{{n}_{B}v}{2f}$ …… (1)

The resonant frequency for a pipe of length with only one open end is

$f=\frac{{n}_{A}v}{4{L}_{A}}$

For, $n=1,3,5\dots$

${L}_{A}=\frac{{n}_{A}v}{4f}$ ….. (2)

The frequency of matches that of and the length of pipe B is ${L}_{B}=4{L}_{A}$. Hence, from equation (1) and equation (2), we get

$\begin{array}{l}\frac{{n}_{B}v}{2f}=4\left(\frac{{n}_{A}v}{4f}\right)\\ \frac{{n}_{B}}{2}={\left({n}_{A}\right)}_{odd}\\ {n}_{B}=2{\left({n}_{A}\right)}_{odd}\end{array}$ …… (a)

Using equation (a), the smallest value of ${n}_{B}$ at which a harmonic frequency of B matches that of A is given as:

$\begin{array}{c}{n}_{B}=2\left(1\right)\\ =2\end{array}$

Hence, the smallest value of ${n}_{B}$is 2.

## Step 4: b) Calculate the second smallest value of  nB

Using equation (a), the second smallest value of ${n}_{B}$ at which a harmonic frequency of B matches that of A is given as:

$\begin{array}{c}{n}_{B}=2\left(3\right)\\ =6\end{array}$

Hence, the second smallest value of ${n}_{B}$ is 6.

## Step 5: c) Calculate the third smallest value of  nB

Using equation (a), the third smallest value of ${n}_{B}$ at which a harmonic frequency of B matches that of A is given as:

$\begin{array}{c}{n}_{B}=2\left(5\right)\\ =10\end{array}$

Hence, the third smallest value of ${n}_{B}$ is 10.

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