Home Page Textbooks Physics: Principles and Problems Page 410: Section Review

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

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Page 410: Section Review

Exercise 11

Step 1

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This is the graph of the displacement of the eardrum versus the time of two cycles of a $1.0,mathrm{kHZ}$ tone and a $2.0,mathrm{kHZ}$ tone. The $2$nd cycle of the $1.0,mathrm{kHZ}$ tone ends at $1.0 ,mathrm{ms}$, where as for the $2.0,mathrm{kHZ}$ tone, it ends at $2.0 ,mathrm{ms}$.

Step 2

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Legend: $text{color{#4257b2} rule{5mm}{1mm} }$ $1.0,mathrm{kHZ}$ and $text{color{#c34632} rule{5mm}{1mm} }$ $2.0,mathrm{kHZ}$

Exercise scan

Exercise 12

Step 1

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Two sound characteristics that are affected by the medium through which the sound passes are: speed of sound and wavelength.

The ones that are not affected are: frequency and period.

Exercise 13

Step 1

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In order to change pitch of sound we have to change frequency and in order to change the loudness we have to change amplitude.

Exercise 14

Solution 1

Solution 2

Step 1

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$B_1 = 10 log(P_1/P_0)$

$110 = 10 log((P_1)/(2 times 10^{-5}))$

$log((P_1)/(2 times 10^{-5})) = 11$

$(P_1)/(2 times 10^{-5}) = 10^{11}$

$P_1 = (2 times 10^{-5}) * (10^{11})$

$P_1 = 2 times 10^{6} Pa$

————————————-

$B_2 = 10 log(P_2/P_0)$

$50 = 10 log((P_2)/(2 times 10^{-5}))$

$log((P_2)/(2 times 10^{-5})) = 5$

$(P_2)/(2 times 10^{-5}) = 10^{5}$

$P_2 = (2 times 10^{-5}) * (10^{5})$

$P_2 = 2 Pa$

————————————

$$
Delta P = P_1 – P_2 = 2 times 10^{6} – 2 = 2 times 10^{6} Pa
$$

Result

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$$
2 times 10^{6} Pa
$$

Step 1

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Remember that the human ears are sensitive to pressure variations in sound waves. Those are the amplitudes of the wave. They are calculated using the sound level (logarithmic scale). Keep in mind that sound level of $0~mathrm{dB}$ is for example a sound with a pressure amplitude $2cdot 10^{-5}~mathrm{pa}$. Equation for this logarithmic scale is:
$$B=10cdot log bigg(frac{p1}{p_{0}}bigg)$$

Step 2

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Let’s calculate first the pressure amplitude for the casual conversation using equation (1):
$$begin{align*}
B&=10cdot log bigg(frac{p1}{p_{0}}bigg)\
50~mathrm{dB}&=10cdot log bigg(frac{p1}{2cdot 10^{-5}~mathrm{pa}}bigg)\
5~mathrm{dB}&= log bigg(frac{p1}{2cdot 10^{-5}~mathrm{pa}}bigg)\
&Downarrow\
bigg(frac{p1}{2cdot 10^{-5}~mathrm{pa}}bigg)&= 10^5 ~mathrm{dB}\
&Downarrow\
p_{1}&=2~mathrm{pa}
end{align*}$$

Step 3

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Now let’s calculate the pressure amplitude for the rock concert using equation (1):
$$begin{align*}
B&=10cdot log bigg(frac{p2}{p_{0}}bigg)\
110~mathrm{dB}&=10cdot log bigg(frac{p2}{2cdot 10^{-5}~mathrm{pa}}bigg)\
11~mathrm{dB}&= log bigg(frac{p2}{2cdot 10^{-5}~mathrm{pa}}bigg)\
&Downarrow\
bigg(frac{p2}{2cdot 10^{-5}~mathrm{pa}}bigg)&= 10^{11} ~mathrm{dB}\
&Downarrow\
p_{2}&=2cdot 10^6~mathrm{pa}
end{align*}$$

Step 4

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Now let’s find out how much larger is the pressure amplitude at the rock concert by dividing it’s value with the pressure amplitude of the casual conversation:
$$begin{align*}
Delta P&= p_{2} – p_{1}\
&=2cdot 10^6~mathrm{pa} – 2~mathrm{pa}\
&= boxed{1 999 998 ~mathrm{pa}}
end{align*}$$
That is the end of this solution.

Result

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$Delta P=1 999 998 ~mathrm{pa}$

Exercise 15

Step 1

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It worked because speed of sound in steel (solid) is greater than the speed of sound in air (gas).

Therefore, sound of the train will travel faster through the rails then through the air, and in that way, they could hear it faster.

Exercise 16

Step 1

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$$
textbf{color{#c34632}(a)}
$$

Smaller Insect will reflect back less sound compared to larger Insect. Hence the bat will get echoes of smaller intensity from a smaller insect.

Step 2

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$$
textbf{color{#c34632}(b)}
$$

If the Insect is flying towards the bat, the echo will have higher frequency than the emitted sound by bat

If the Insect is flying away from the bat, the echo will have lower frequency than the emitted sound by bat

This is due to doppler effect

Exercise 17

Step 1

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The key part is $text{color{#4257b2}at the instant the car passes the trooper}$

Because at the Instant the car passes the trooper, the cars velocity has no component along the line joining the car and the trooper.

The radar detector uses doppler effect to measure the speed of the car

In this case $v_s = v_d=0$

Therefore the doppler equation reduces to
$$
f_d = f_sleft( dfrac{v-0}{v-0}right)implies f_d = f_s
$$

This cannot be used to find the speed of the car

Exercise scan

Step 2

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Although the trooper can stand at the side of the road and measure the speed of car BEFORE and AFTER it has passed the trooper.

Result

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$$
text{color{#4257b2}Click to see

$ $

color{default}Hint : It says, at the instant the car passes the trooper }
$$

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