All Solutions
Page 410: Section Review
The ones that are not affected are: frequency and period.
$110 = 10 log((P_1)/(2 times 10^{-5}))$
$log((P_1)/(2 times 10^{-5})) = 11$
$(P_1)/(2 times 10^{-5}) = 10^{11}$
$P_1 = (2 times 10^{-5}) * (10^{11})$
$P_1 = 2 times 10^{6} Pa$
————————————-
$B_2 = 10 log(P_2/P_0)$
$50 = 10 log((P_2)/(2 times 10^{-5}))$
$log((P_2)/(2 times 10^{-5})) = 5$
$(P_2)/(2 times 10^{-5}) = 10^{5}$
$P_2 = (2 times 10^{-5}) * (10^{5})$
$P_2 = 2 Pa$
————————————
$$
Delta P = P_1 – P_2 = 2 times 10^{6} – 2 = 2 times 10^{6} Pa
$$
2 times 10^{6} Pa
$$
$$B=10cdot log bigg(frac{p1}{p_{0}}bigg)$$
$$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*}$$
$$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*}$$
$$begin{align*}
Delta P&= p_{2} – p_{1}\
&=2cdot 10^6~mathrm{pa} – 2~mathrm{pa}\
&= boxed{1 999 998 ~mathrm{pa}}
end{align*}$$
That is the end of this solution.
Therefore, sound of the train will travel faster through the rails then through the air, and in that way, they could hear it faster.
textbf{color{#c34632}(a)}
$$
Smaller Insect will reflect back less sound compared to larger Insect. Hence the bat will get echoes of smaller intensity from a smaller insect.
textbf{color{#c34632}(b)}
$$
If the Insect is flying towards the bat, the echo will have higher frequency than the emitted sound by bat
If the Insect is flying away from the bat, the echo will have lower frequency than the emitted sound by bat
This is due to doppler effect
Because at the Instant the car passes the trooper, the cars velocity has no component along the line joining the car and the trooper.
The radar detector uses doppler effect to measure the speed of the car
In this case $v_s = v_d=0$
Therefore the doppler equation reduces to
$$
f_d = f_sleft( dfrac{v-0}{v-0}right)implies f_d = f_s
$$
This cannot be used to find the speed of the car
text{color{#4257b2}Click to see
$ $
color{default}Hint : It says, at the instant the car passes the trooper }
$$
