All Solutions
Page 405: Practice Problems
$f=18 text{Hz}$
$upsilon=343 dfrac{text{m}}{text{s}} quad text{at} quad 20^circtext{C}$
We have to find values of the wavelength $(lambda$) for sound wave which frequency is $18 text{Hz}$. As we know, the speed of wave (sound) in the air has value of $343 text{m}/text{s}$.
To find required value of wavelength we can start from the speed of wave relation which is given by following formula;
$$
upsilon=f cdot lambda.
$$
From previous formula, we can see that the speed of wave is defined as the product of the wave frequency and wavelength of wave.
After simple derivation the speed of wave equation, we get required relationship for calculate the value of wavelength.
$$
lambda=dfrac{upsilon}{f}.
$$
Plug in values into above relation and solve for wavelength,
$$
begin{align*}
lambda &=dfrac{343 dfrac{text{m}}{text{s}} }{18 text{Hz} }\
lambda &=19.05 text{m} \
end{align*}
$$
From the question, we know that the speed of the wave at 20$text{textdegree}$C is 343 $dfrac{text{m}}{text{s}}$ and the frequency of the wave is 18 hz
Wavelength is related to both wave speed and wave frequency. Wavelength is the distance between two corresponding points on adjacent waves.
Wavelength is directly proportional to the velocity and inversely proportional to the frequency
You can calculate the wavelength ($lambda$) with the formula below:
$mathbf{lambda = dfrac{v}{f}}$
Where:
$lambda$ = wavelength ,
v = speed of the wave
f = frequency of the wave
1) Using the formula stated in cell 1.
$mathbf{lambda = dfrac{v}{f}}$
2) Substituting the values in the formula
$lambda$ = $dfrac{343}{18}$
$boxed{lambda = 19.05 text{m}}$
boxed{lambda = 19.05 text{m}}
$$
text{color{#4257b2}Recall that : $$v = fcdot lambda$$}
$$
Therefore
$$
lambda=dfrac{v}{f}
$$
Substitute : $v=1533$ m/s, and $f=18$ Hz
$$
lambda=dfrac{1533text{ m/s}}{18text{ Hz}}approx 85text{ m}
$$
text{color{#4257b2}
$$lambdaapprox 85text{ m}$$ }
$$
$$lambda=frac{v}{f}tag1$$
In which the $v$ is the velocity of sound wave, while $f$ is the frequency. The speed of sound wave is about $1530~mathrm{frac{m}{s}}$. As we can see from the equation we don’t need the temperature for the calculations.
$$begin{align*}
lambda&=frac{v}{f}\
&=frac{1530~mathrm{frac{m}{s}}}{18~mathrm{Hz}}\
&=boxed{85~mathrm{m}}
end{align*}$$
And with that we finished this solution.
text{color{#4257b2}Recall that : $$v = fcdot lambda$$}
$$
Therefore
$$
f=dfrac{v}{lambda}
$$
Substitute : $v=5130$ m/s, and $lambda=1.25$ m
$$
f=dfrac{5130text{ m/s}}{1.25text{ m}}=4104text{ Hz}
$$
text{color{#4257b2}$$f=4104text{ Hz}$$ }
$$
$$f=frac{v}{lambda}tag1$$
Where the $f$ is frequency, $v$ is speed of the sound in iron and $lambda$ is the wavelength. We won’t use the temperature in the calculations as we can see from equation (1). Keep in mind that the speed of sound in iron is $5120~mathrm{frac{m}{s}}$.
$$begin{align*}
f&=frac{v}{lambda}\
&=frac{5120~mathrm{frac{cancel{m}}{s}}}{1.25~mathrm{cancel{m}}}\
&=boxed{4096~mathrm{Hz}}
end{align*}$$
And that concludes this problem.
$t_{echo}=0.8 text{s}$
$upsilon=343 dfrac{text{m}}{text{s}}$
In this case we need to determine the width of the canyon. Note that the echo passes through the canyon and back, so the distance that the sound travels in the canyon doubles.
$$
2d = upsilon cdot t_{echo}.
$$
To find echo distance which is actually represents the width of the canyon, we need to simplify above relation.
$$
d=dfrac{upsilon cdot t_{echo}}{2}
$$
Plug in values into above relation and solve for width of canyon.
$$
begin{align*}
d &=dfrac{bigg(343 dfrac{text{m}}{text{s}}bigg) cdot (0.8 text{s})}{2} tag{Plug in values.}\
d &=dfrac{274.4 text{m}}{2} \
d &=137.2 text{m} \
end{align*}
$$
So, the answer is: $text{underline{width of canyon is $137.2 text{m}.$}}$
$f=2280 text{Hz}$
$lambda=0.655 text{m}$
Our task is to determine the unknown medium which sound wave passes through. To find unknown medium, the best way is using the speed of wave equation, which is given by following equation.
$$
begin{align*}
upsilon &=f cdot lambda tag{Equation 1.}\
end{align*}
$$
From equation 1, clearly we can see that the speed of the wave is defined as the product of the wave frequency and wavelength. Table 15.1 has values of the speed of the wave for a different medium. Plug in values into equation 1 and solve for speed of the wave which passes through unknown medium.
$$
begin{align*}
upsilon &=(2280 text{Hz}) cdot 0.655 text{m} tag{$1 text{Hz}= text{s}^{-1}$.}\
upsilon &=(2280 text{s}^{-1}) cdot 0.655 text{m} \
upsilon &=1493.4 dfrac{text{m}}{text{s}}
end{align*}
$$
Using Table 15.1, we can see that the obtained value of the speed of wave corresponds to the water medium.
$v_{s} = -24.6, mathrm{m/s}$
$f= 524, mathrm{Hz}$
$f_{1}=?$
Since car are moving away from us, frequency will increase. By using Doppler’s effect we get:
$$
f_{1} = f dfrac{v+v_{0}}{v-v_{s}}
$$
When we put known values into the previous equation we get:
$$
f_{1} = 524, mathrm{Hz} cdot dfrac{1}{1 – dfrac{-24.6, mathrm{m/s}}{343, mathrm{m/s}}}
$$
$$
boxed{f_{1} = 489, mathrm{Hz}}
$$
f_{1} = 489, mathrm{Hz}
$$
color{#c34632}f_d = f_sleft(dfrac{v-v_d}{v-v_s} right)
$$
Substitute $v = 343$, $v_s=0$, $v_d = -25$ and $f_s = 365$
$$
f_d = 365cdot left(dfrac{343-left( -25right)}{343-0} right)
$$
$$
f_d = 365cdot left(dfrac{343+25}{343} right)approx391.6
$$
text{color{#4257b2}391.6 Hz}
$$
color{#c34632}f_d = f_sleft(dfrac{v-v_d}{v-v_s} right)
$$
Substitute $v = 343$, $v_s=24.6$, $v_d = -24.6$ and $f_s = 475$
$$
f_d =475left(dfrac{343-left( -24.6right)}{343-24.6} right)
$$
$$
f_d =475left(dfrac{343+24.6}{343-24.6} right)approx548.4
$$
text{color{#4257b2}548.4 Hz}
$$
$$f_{d}=f_{s}cdot frac{(v-v_{d})}{(v-v_{s})}tag1$$
$$begin{align*}
f_{d}&=475~mathrm{Hz}cdot frac{(343~mathrm{frac{m}{s}}-(-24.6~mathrm{frac{m}{s}})}{(343~mathrm{frac{m}{s}}-24.6~mathrm{frac{m}{s}})}\\
&=475~mathrm{Hz}cdot frac{367.6~mathrm{cancel{frac{m}{s}}}}{318.4~mathrm{cancel{frac{m}{s}}}}\
&=boxed{548.4~mathrm{Hz}}
end{align*}$$
That is the frequency we would hear from our car.
color{#c34632}f_d = f_sleft(dfrac{v-v_d}{v-v_s} right)
$$
Substitute $v = 1482$, $v_s=9.2$, $v_d = 0$ and $f_s = 3.5$
$$
f_d = 3.5cdot left(dfrac{1482-0}{1482-9.2} right)approx3.522
$$
text{color{#4257b2}3.522 MHz}
$$
$$f_{d}=f_{s}cdot frac{(v-v_{d})}{(v-v_{s})}tag1$$
Keep in mind that speed of sound in water is equal $1482~mathrm{frac{m}{s}}$.
$$begin{align*}
f_{d}&=f_{s}cdot frac{(v-cancel{v_{d}})}{(v-v_{s})}\
&=3.5~mathrm{MHz}cdot frac{(1482~mathrm{cancel{frac{m}{s}}})}{(1482~mathrm{cancel{frac{m}{s}}}-9.2~mathrm{cancel{frac{m}{s}}})}\
&=boxed{3.52~mathrm{MHz}}
end{align*}$$
color{#c34632}f_d = f_sleft(dfrac{v-v_d}{v-v_s} right)
$$
Substitute $v = 343$, $v_s=text{Unknown}$, $v_d = 0$, $f_d = 271$ and $f_s = 262$
$$
271 = 262left(dfrac{343-0}{343-v_s} right)
$$
Divide both sides by 262
$$
dfrac{271}{262} = dfrac{343}{343-v_s}
$$
Take reciprocal of both sides
$$
dfrac{262}{271} = dfrac{343-v_s}{343}
$$
$$
dfrac{262}{271} = 1-dfrac{v_s}{343}
$$
$$
dfrac{v_s}{343}= 1-dfrac{262}{271}
$$
Multiply both sides by 343
$$
v_s= 343left[ 1-dfrac{262}{271}right]approx11.39
$$
text{color{#4257b2}11.39 m/s}
$$
$$f_{d}=f_{s}cdot frac{(v-v_{d})}{(v-v_{s})}tag1$$
Where the $v$ is the speed of sound. In this problem $f_{d}$ is frequency of C sharp, while $f_{s}$ is the frequency of middle C.
$$begin{align*}
f_{d}=f_{s}cdot frac{(v-cancel{v_{d}})}{(v-v_{s})}
end{align*}$$
We cancel the $v_{d}$ because it equals zero as we have already said. Let’s express the $v_{s}$ now from equation (1).
$$begin{align*}
f_{d}&=f_{s}cdot frac{v}{(v-v_{s})}\
f_{d}cdot(v-v_{s})&=f_{s}cdot v\
(v-v_{s})&=frac{f_{s}}{f_{d}}cdot v\
v_{s}&=v-frac{f_{s}}{f_{d}}cdot v\
end{align*}$$
Now we can simplify it a bit more and then finish the solution:
$$begin{align*}
v_{s}=vcdot(1-frac{f_{s}}{f_{d}})
end{align*}$$
$$begin{align*}
v_{s}&=vcdot(1-frac{f_{s}}{f_{d}})\
&=343~mathrm{frac{m}{s}}cdot (1-frac{262~mathrm{cancel{Hz}}}{271~mathrm{cancel{Hz}}})\
&=343~mathrm{frac{m}{s}}cdot 0.03321\
&=boxed{11.39~mathrm{frac{m}{s}}}
end{align*}$$
And that is the end of this solution. We have found the $v_{s}$.
