Chapter 12, Problem 25P

(a)

To determine

To Calculate: The fundamental and first three audible overtones.

Answer to Problem 25P

The fundamental frequency is $76.6\text{Hz}$ and first three audible overtones are

  $230\text{Hz},383\text{Hz},\text{and}536\text{Hz}$

Explanation of Solution

Given:

Length of the organ pipe, $l=112\text{cm}$

Formula used:

The frequency is,

  $f=\frac{nv}{4L}$

Where L is the length of the string

M is the masse of the vibrating portion of the string

V is the sound’s velocity

Calculation:

Only odd harmonic are present if only one end of the pipe is closed. Then the frequency is given as,

  $f_n=\frac{nv}{4L}=n{f}_1$ for $n=1,3,\mathrm{5...}$

Here,

  $\begin{array}{l}{f}_1=\frac{v}{4L}=\frac{343\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}{4\times \left({\text{112×10}}^{\text{-2}}\text{m}\right)}\\ f_1=76.6\text{Hz}\end{array}$

Similarly for $n=3$

  $\begin{array}{l}{f}_3=\frac{3v}{4L}=\frac{3\times 343\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}{4\times \left(112\times {10}^{-2}\text{m}\right)}\\ f_3=230\text{Hz}\end{array}$

Similarly, for $n=5$ and $n=7$

  $f_5=5f_1=383\text{Hz}$ and $f_7=7f_1=536\text{Hz}$ respectively.

Conclusion:

The frequencies are $76.6\text{Hz},230\text{Hz},383\text{Hz},\text{and}536\text{Hz}$ .

(b)

To determine

To Calculate: The fundamental and first three audible overtones.

Answer to Problem 25P

The fundamental frequency is $153\text{Hz}$ and first three audible overtones are

  $306\text{Hz},459\text{Hz},\text{and}612\text{Hz}$

Explanation of Solution

Given:

Length of the organ pipe, $l=112\text{cm}$

Formula used:

The frequency is,

  $f=\frac{v}{2L}$

Where L is the length of the string

M is the masse of the vibrating portion of the string

V is the sound’s velocity

Calculation:

All harmonic frequencies are present if both ends of the pipe is open. So, the frequencies are

  $f_n=\frac{nv}{2L}=n{f}_1$ for $n=1,2,\mathrm{3...}$

  $f_1=\frac{1\times \left(343\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)}{2\times \left(1.12\text{m}\right)}=153\text{Hz}$ for, $n=1$

Similarly, for $n=2$

  $f_2=2f_1=306\text{Hz}$

For, $n=3$

  $f_3=3f_1=459\text{Hz}$

For, $n=4$

  $f_4=4f_1=612\text{Hz}$

Conclusion:

The frequencies are $153\text{Hz},306\text{Hz},459\text{Hz},\text{and}612\text{Hz}$