Physics
Physics
1st Edition
Walker
ISBN: 9780133256925
Textbook solutions

All Solutions

Page 500: Practice Problems

Exercise 7
Step 1
1 of 3
The original beat frequency is 4 Hz, as a result, the initial frequency of the string can be 436 Hz or 444 Hz. The final beat frequency is given to be 6 Hz, as a result, the final frequency can be 434 Hz or 446 Hz.
Step 2
2 of 3
Notice that only 446 Hz can satisfy both conditions. Therefore, the frequency increases from 444 Hz to 446 Hz when tightened.
Result
3 of 3
444 Hz to 446 Hz
Exercise 8
Step 1
1 of 2
The frequency of the three tuning forks are given as

$$
begin{align*}
f_{1} & = 252 mathrm{Hz} \
f_{2} & = 256 mathrm{Hz} \
f_{3} & = 259 mathrm{Hz}
end{align*}
$$

$$
textbf{Definition of Beat Frequency:}
$$

$$
begin{align*}
text{beat frequency} & = text{absolute value of the difference in frequency} \
f_{beat} & = big|f_A – f_B big|
end{align*}
$$

Therefore, the beat frequency between $f_1$ and $f_2$ is

$$
begin{align*}
f_{12} & = big|f_1 – f_2 big| \
& = big|(252 – 256) big| \
& = 4 mathrm{Hz}
end{align*}
$$

Similarly, the beat frequency between $f_2$ and $f_3$ is

$$
begin{align*}
f_{23} & = big|f_2 – f_3 big| \
& = big|(256 – 259) big| \
& = 3 mathrm{Hz}
end{align*}
$$

Similarly, the beat frequency between $f_1$ and $f_3$ is

$$
begin{align*}
f_{13} & = big|f_1 – f_3 big| \
& = big|(252 – 259) big| \
& = 7 mathrm{Hz}
end{align*}
$$

Hence, the beat frequencies are

$$
begin{align*}
f_{beat} & = 3 mathrm{Hz} , 4 mathrm{Hz} text{and} 7 mathrm{Hz}
end{align*}
$$

Result
2 of 2
The beat frequencies,
$displaystyle f_{beat} = 3 mathrm{Hz} , 4 mathrm{Hz} text{and} 7 mathrm{Hz}$
Exercise 9
Step 1
1 of 2
The second clarinet is expected to have higher frequency. Hence, its frequency is the sum of the first frequency and the beat frequency.

$$
f_2=f_1+f_{beat}=441;Hz+dfrac{8.00;beats}{2.00;s}=445;Hz
$$

Result
2 of 2
445 Hz
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