Home Page Textbooks Physics: Principles and Problems Page 419: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

Textbook solutions

All Solutions

Page 419: Section Review

Exercise 22

Step 1

1 of 2

$textit{a.}$, Human voice is produced by the vocal chords.

$textit{b.}$, Sound of clarinet is produced by reeds.

$textit{c.}$, Sound of tuba is produced by the lips of the player.

$textit{d.}$, Sound of violin is produced by string.

Result

2 of 2

$textit{a.}$, vocal chords

$textit{b.}$ reeds

$textit{c.}$ lips of the player

$textit{d.}$, string

Exercise 23

Step 1

1 of 2

This is because length of the tube and resonant frequency are inversely proportional which means:

The longer the tube is the lower resonant frequency will be.

Result

2 of 2

The longer the tube is the lower resonant frequency will be.

Exercise 24

Step 1

1 of 2

Length of an open tube has to be one-half the wavelength: $L = dfrac{lambda}{2}$

Result

2 of 2

$L = dfrac{lambda}{2}$

Exercise 25

Solution 1

Solution 2

Step 1

1 of 2

Information given in the text are:

$f_{1} = 370, mathrm{Hz}$

$f_{2}$ , $f_{3}$, $f_{4}$ $= ?$

Harmonics are given by the product of a whole number and fundamental ($f_{1}$):

$$
f_{2} = 2 f_{1}
$$

$$
f_{3} = 3 f_{1}
$$

$$
f_{4} = 4 f_{1}
$$

When we put known values into the previous equations:

$$
f_{2} = 2 cdot 370, mathrm{Hz}
$$

$$
boxed{f_{2} = 740, mathrm{Hz}}
$$

$$
f_{3} = 3 cdot 370, mathrm{Hz}
$$

$$
boxed{f_{3} = 1100, mathrm{Hz}}
$$

$$
f_{4} = 4 cdot 370, mathrm{Hz}
$$

$$
boxed{f_{4} = 1480, mathrm{Hz}}
$$

Result

2 of 2

$$
f_{2} = 740, mathrm{Hz}
$$

$$
f_{3} = 1110, mathrm{Hz}
$$

$$
f_{4} = 1480, mathrm{Hz}
$$

Step 1

1 of 4

For this problem we are given a frequency of a note, and then asked to compute for the next $textit{three}$ harmonics produced by the said note.

Step 2

2 of 4

A string’s harmonic are whole numbers that are multiples of the fundamental, thus the frequencies are as follows.

Given a frequency of the F sharp note on a violin:

$$
begin{align*}
f_1 = 370 ,mathrm{Hz}
end{align*}
$$

Step 3

3 of 4

We can know solve for the next three harmonics: $f_2$, $f_3$, and $f_4$.

$$
begin{align*}
f_2 &= 2f_1 \
&= 2(370 ,mathrm{Hz})\
&= boxed{740 ,mathrm{Hz}}\
f_3 &= 3f_1 \
&= 3(370 ,mathrm{Hz})\
&= boxed{1110 ,mathrm{Hz}}\
f_4 &= 4f_1 \
&= 4(370 ,mathrm{Hz})\
&= boxed{1480 ,mathrm{Hz}}
end{align*}
$$

Result

4 of 4

$f_2 = 740 ,mathrm{Hz}$

$f_3 = 1110 ,mathrm{Hz}$

$$
f_4 = 1480 ,mathrm{Hz}
$$

Exercise 26

Step 1

1 of 3

Information given in the text are:

$L = 2.40, mathrm{m}$

$textit{a.}$, $f = ?$

Frequency is given by:

$$
f = dfrac{v}{lambda}
$$

Therefore, we need to determine wavelength since the speed of sound is known $v = 343, mathrm{m/s}$:

$$
begin{align*}
lambda &= 4L\
&= 4 cdot 4.20, mathrm{m}\
&= 9.60, mathrm{m}\
end{align*}
$$

Frequency will, finally, be:

$$
f = dfrac{343, mathrm{m/s}}{9.60, mathrm{m}}
$$

$$
boxed{f = 35/7, mathrm{Hz}}
$$

Step 2

2 of 3

$textit{b.}$, Information given in the text are:

$f _{2} = 1.40, mathrm{Hz}$

First we need to determine frequency, and since both pipes are played at the same time, it will now be:

$$
f ‘= f – f_{2}
$$

When we put known values into the previous equation we get:

$$
f = 35.7, mathrm{Hz} – 1.40, mathrm{Hz} = 34.3, mathrm{Hz}
$$

Since length is given by:

$$
L’ = dfrac{lambda}{4}
$$

We have to determine wavelength first:

$$
lambda = dfrac{v}{f}
$$

When we put known values into the previous equation we get:

$$
lambda = dfrac{343, mathrm{m/s}}{34.3, mathrm{Hz}} = 10, mathrm{m}
$$

Finally:

$$
L = dfrac{10}{4}
$$

$$
boxed{L’ = 2.50, mathrm{m}}
$$

Therefore difference will be:

$$
L – L’ = 2.50, mathrm{m} – 2.40, mathrm{m} = boxed{0.10, mathrm{m}}
$$

Result

3 of 3

$textit{a.}$, $f = 35/7, mathrm{Hz}$

$textit{b.}$, $0.10, mathrm{m}$

Exercise 27

Step 1

1 of 2

Various instruments sound different even when playing same note since every instrument brings out its own set of harmonic and fundamental frequencies which leads to different timbers.

Result

2 of 2

Because every instrument brings out its own set of harmonic and fundamental frequencies, therefore, they have different timbres.

Exercise 28

Step 1

1 of 2

Beat frequency is the difference between the frequencies of the two forks.

Therefore the frequency of the other fork is either :
$$
392-3 = 389hspace{10mm}text{or}hspace{10mm}392+3 = 395
$$

Result

2 of 2

$$
text{color{#4257b2}Frequency of the other fork is either 389 Hz or 395 Hz}
$$

Exercise 29

Step 1

1 of 2

All the objects will sound different because they have different timbres since they resonate with different harmonics.

Result

2 of 2

Objects will sound different since they resonate with different harmonics, therefore, timbres will be different.

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Page 419: Section Review

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