Home Page Textbooks Physics: Principles and Problems Page 416: Practice Problems

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

Textbook solutions

All Solutions

Page 416: Practice Problems

Exercise 18

Step 1

1 of 2

Information given in the text are:

$f = 440, mathrm{Hz}$

$T = 20, mathrm{^oC}$

$dfrac{lambda}{2} = ?$

Since we have to determine resonance spacing, following equation:

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

becomes:

$$
dfrac{lambda}{2} = dfrac{v}{2f}
$$

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

$$
dfrac{lambda}{2} = dfrac{343, mathrm{m/s}}{2 cdot 440, mathrm{Hz}}
$$

$$
boxed{dfrac{lambda}{2} = 0.39, mathrm{m}}
$$

Result

2 of 2

$$
dfrac{lambda}{2} = 0.39, mathrm{m}
$$

Exercise 19

Step 1

1 of 2

Information given in the text are:

$f = 440, mathrm{Hz}$

$dfrac{lambda}{2} = 110, mathrm{cm} = 1.1, mathrm{m} rightarrow lambda = 2.2, mathrm{m}$

$v = ?$

Speed is given by:

$$
v = lambda f
$$

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

$$
v = 2.2, mathrm{m} cdot 440, mathrm{Hz}
$$

$$
boxed{v = 970, mathrm{m/s}}
$$

Result

2 of 2

$$
v = 970, mathrm{m/s}
$$

Exercise 20

Step 1

1 of 2

Information given in the text are:

$T = 27, mathrm{^oC}$

$dfrac{lambda}{2} = 20.2, mathrm{cm} = 0.202, mathrm{m} rightarrow lambda = 0.404, mathrm{m}$

$v = 347, mathrm{m/s}$

$f = ?$

Frequency is given by:

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

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

$$
f = dfrac{347, mathrm{m/s}}{0.404, mathrm{m}}
$$

$$
boxed{f = 859, mathrm{Hz}}
$$

Result

2 of 2

$$
f = 859, mathrm{Hz}
$$

Exercise 21

Solution 1

Solution 2

Step 1

1 of 2

The longest wavelength is:

$lambda_0 = (2) * (2.65) = 5.3 m$

The lowest frequency is:

$f_0 = dfrac{v}{lambda_0} = dfrac{343}{5.3} = 64.717 = 64.7 Hz$

$f_1 = 2 f_0 = (2)*(64.717) = 129 Hz$

$$
f_2 = 3 f_0 = (3)*(64.717) = 194 Hz
$$

Result

2 of 2

a) $64.7 Hz$

b) $129 Hz$ , $194 Hz$

Step 1

1 of 4

This is a basic problem in which we need to find the highest wavelength so we could calculate the lowest frequency because the frequency is inversely proportional to the wavelength.
$$lambda=2cdot Ltag1$$
From that we get the value for $lambda$. The value is $5.3~mathrm{m}$. Now we will insert that value in equation for frequency which is:
$$f=frac{v}{lambda}tag2$$

Step 2

2 of 4

a) Let’s now calculate the value for lowest frequency using equation (2):
$$begin{align*}
f&=frac{v}{lambda}\
&=frac{343~mathrm{frac{cancel{m}}{s}}}{5.3~mathrm{cancel{m}}}\\
&=boxed{64.7~mathrm{frac{m}{s}}}
end{align*}$$
That is the lowest frequency.

Step 3

3 of 4

b) Next two resonant frequencies $(f_{1}$ and $f_{2})$ are for wavelength $lambda_{1}$ and $lambda_{2}$ respectively. Keep in mind that:
$$begin{align*}
lambda_{1}&=L\
lambda_{2}&=frac{2}{3}L
end{align*}$$
Using the equation (2) we get:
$$begin{align*}
f_{1}&=frac{v}{lambda_{1}}\\
&=frac{343~mathrm{frac{cancel{m}}{s}}}{2.65~mathrm{cancel{m}}}\\
&=boxed{129.4~mathrm{frac{m}{s}}}end{align*}$$
Now the same for the second wavelength:
$$begin{align*}
f_{2}&=frac{v}{lambda_{2}}\
&frac{3cdot 343~mathrm{frac{cancel{m}}{s}}}{5.3~mathrm{cancel{m}}}\\
&=boxed{194.1~mathrm{frac{m}{s}}}
end{align*}$$
Those are the values for two resonant frequencies.

Result

4 of 4

$$begin{align*}
&text{a)}~ f= 64.7~mathrm{frac{m}{s}}\
&text{b)}~f_{1}=129.4~mathrm{frac{m}{s}} ~text{and}~f_{1}=194.1~mathrm{frac{m}{s}}
end{align*}
$$

unlock

All Solutions

Page 416: Practice Problems

Haven't found what you were looking for?

Search for samples, answers to your questions and flashcards