Home Page Textbooks Physics: Principles and Problems Page 429: Standardized Test Practice

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

Textbook solutions

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Page 429: Standardized Test Practice

Exercise 1

Step 1

1 of 2

Sound travels from its source to our ear by changes in air pressure.

Result

2 of 2

$textit{A}$, by changes in air pressure

Exercise 2

Step 1

1 of 2

Information given in the text are:

$f = 327, mathrm{Hz}$

$v = 1493, mathrm{m/s}$

$lambda = ?$

Wavelength is given by:

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

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

$$
lambda = dfrac{1493, mathrm{m/s}}{327, mathrm{Hz}}
$$

Finally:

$$
boxed{lambda = 4.57, mathrm{m}}
$$

Result

2 of 2

$$
lambda = 4.57, mathrm{m}
$$

Exercise 3

Step 1

1 of 2

Information given in the text are:

$v = 351, mathrm{m/s}$

$f = 298,mathrm{Hz}$

$lambda = ?$

Wavelength is given by:

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

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

$$
lambda = dfrac{351, mathrm{m/s}}{298,mathrm{Hz}}
$$

$$
boxed{lambda = 1.18, mathrm{m}}
$$

Result

2 of 2

$textit{C}$, $1.18, mathrm{m}$

Exercise 4

Solution 1

Solution 2

Step 1

1 of 2

Information given in the text are:

$v_{o} = 60, mathrm{km/h} = 16.6, mathrm{m/s}$

$f_{s} = 512, mathrm{Hz}$

$v= 343, mathrm{m/s}$

$f_{o} = ?$

We will be using Doppler effect which gives us the following formula:

$$
f_{o} = f_{s} dfrac{v – v_{o}}{v – v_{s}}
$$

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

$$
f_{o} = 512, mathrm{Hz} dfrac{343, mathrm{m/s} – 0}{343, mathrm{m/s} – 16.6, mathrm{m/s}}
$$

$$
boxed{f_{o} approx 538, mathrm{Hz}}
$$

Result

2 of 2

$textit{C}$, $538, mathrm{Hz}$

Step 1

1 of 2

$f_o = f_s (dfrac{v – v_o}{v – v_s})$

$f_o = (512) (dfrac{343 – 0.0}{343 – (60/3.6)})$

$$
f_o = 538 Hz
$$

Result

2 of 2

$$
538 Hz
$$

Exercise 5

Solution 1

Solution 2

Step 1

1 of 2

Information given in the text are:

$v_{o} = 72, mathrm{km/h} = 20, mathrm{m/s}$

$f_{s} = 657, mathrm{Hz}$

$f_{o} = ?$

We will be using Doppler’s effect which gives us the following formula:

$$
f_{o} = f_{s} dfrac{v – v_{o}}{v – v_{s}}
$$

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

$$
f_{o} = 657, mathrm{Hz} dfrac{343, mathrm{m/s} – 20, mathrm{m/s}}{343, mathrm{m/s} – 0}
$$

$$
boxed{f_{o} approx 620, mathrm{Hz}}
$$

Result

2 of 2

$textit{B}$ $620, mathrm{Hz}$

Step 1

1 of 2

$f_o = f_s (dfrac{v – v_o}{v – v_s})$

$f_o = (657) (dfrac{343 – (72/3.6)}{343 – 0.0})$

$$
f_o = 620 Hz
$$

Result

2 of 2

$$
620 Hz
$$

Exercise 6

Step 1

1 of 2

We have:

$f_{beat} = dfrac{20}{5.0} = 4.0 = |f_1 – f_2|$

Thus:

$f_1 – f_2 = 4.0 ===> f_2 = f_1 – 4.0 = 262 – 4.0 = 258 Hz$

or:

$f_1 – f_2 = -4.0 ===> f_2 = f_1 + 4.0 = 262 – 4.0 = 266 Hz$

Result

2 of 2

258 Hz or 266 Hz

Exercise 7

Step 1

1 of 2

A clamped string and an open pipe have resonant frequencies at each whole number multiple of the lowest frequency.

Result

2 of 2

$textit{B}$, a clamped string and an open pipe

Exercise 8

Solution 1

Solution 2

Step 1

1 of 2

$wavelength = lambda = 4 L = (4)*(0.168) = 0.672 m$

$$
speed = v = lambda f = (0.672)*(488) = 328 m/s
$$

Result

2 of 2

$$
328 m/s
$$

Step 1

1 of 3

As the figure is showing we can simply calculate the wavelength using the given information. Keep in mind that we need to convert the units of measurements first.
$$L=16.8~mathrm{cm}=0.168~mathrm{m}$$
Now we can calculate the wavelength using the equation:
$$lambda=4cdot Ltag1$$
When we calculate it we will insert the value in the equation for speed of sound:
$$v=fcdotlambdatag2$$

Step 2

2 of 3

We will now calculate the wavelength:
$$begin{align*}
lambda&=4cdot L\
&=4cdot 0.168~mathrm{m}\
&= 0.672~mathrm{m}
end{align*}$$
Let’s now insert that value in equation (2):
$$begin{align*}
v&=fcdotlambda\
&=488~mathrm{Hz} ~cdot 0.672~mathrm{m}\
&=boxed{328~mathrm{frac{m}{s}}}
end{align*}$$
And that is the speed of sound in this problem.

Result

3 of 3

$v=328~mathrm{frac{m}{s}}$

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