All Solutions
Page 523: Assessment
$245-240=5;Hz$
$145-140=5$ Hz
The beat frequency for both cases are equal.
Beat frequency is defined as the absolute value of the difference of the two frequencies. Hence, the valid explanation is option A
Calculate the time by dividing the distance by the speed of sound.
$t=dfrac{2d}{v}=dfrac{2(75)}{1530;m/s}=0.098;s$
Now use the formula $v=lambda f$
$$
lambda=dfrac{v}{f}=dfrac{1530;m/s}{55times10^3;Hz}=28times10^{-3};m=28;mm
$$
28;mm
$$
$$
f_{beat}=|f_1-f_2|=|268-272|=4;Hz
$$
4;Hz
$$
$$
f_{beat}=f pm f_{beat}=261pm 3 =264;Hz;or258;Hz
$$
textbf{(a)}
$$
We know that sound has its speed in the air given as
$$
begin{align*}
v_s & = 343 mathrm{m/s}
end{align*}
$$
It is given that sound wave in air has a frequency as
$$
begin{align*}
f_s & = 425 mathrm{Hz}
end{align*}
$$
Now, the wavelength of the sound wave can be estimated as
$$
begin{align*}
v_s & = f_s cdot lambda_s \
lambda_s & = dfrac{v_s}{f_s} \
& = left(dfrac{343 }{425}right) mathrm{m} \
& = 0.807 mathrm{m} \
& hspace*{-6mm}boxed{ lambda_s = 0.807 mathrm{m}}
end{align*}
$$
textbf{(b)}
$$
In a medium the speed of the wave is constant. Changing the frequency of the wave will result in change of the wavelength and vice -versa.
$$
begin{align*}
v_s & = f_s cdot lambda_s = text{constant} \
implies f_s & propto dfrac{1}{lambda_s}
end{align*}
$$
Hence, we see that the frequency is inversely proportional to the wvaelength.
Therefore, if the frequency of the sound is increased then the wavelength of the wave will decrease.
textbf{(c)}
$$
It is given that sound wave in air has a frequency as
$$
begin{align*}
f_s & = 475 mathrm{Hz}
end{align*}
$$
Now, the wavelength of the sound wave can be estimated as
$$
begin{align*}
v_s & = f_s cdot lambda_s \
lambda_s & = dfrac{v_s}{f_s} \
& = left(dfrac{343 }{475}right) mathrm{m} \
& = 0.722 mathrm{m} \
& hspace*{-6mm}boxed{ lambda_s = 0.722 mathrm{m}}
end{align*}
$$
(b) Decrease.
(c) $displaystyle { lambda_s = 0.722 mathrm{m}}$
$d=v_0+dfrac{1}{2}gt^2$
$t_1=sqrt{dfrac{2d}{g}}$
Calculate the time required for the sound to travel up
$t_2=dfrac{d}{v}$
Add the two times
$t=t_1+t_2=sqrt{dfrac{2d}{g}}+dfrac{d}{v}$
Plug-in given values
1.5 s= $dfrac{d}{343;m/s}+sqrt{dfrac{2}{9.81;m/s^2}}sqrt{d}$
Solve for d
$$
d=10.6;m
$$
$f_2=f_1+Delta f_0$
$=320.0 Hz-(7.5;Hz-4.5;Hz)=317.0;Hz$
$$
f_{unknown}=f_1+f_{beat,1}=320.0Hz+4.50;Hz=324.5;Hz
$$
$f_n=dfrac{nv}{4L}$. Smaller animals produce higher pitch because L is small.
Using the fact that $f_n=dfrac{nv}{4L}$
Removing some water makes the vibrating column longer resulting to a sound with a Lower Frequency in comparison with the first tone.
$f_n=dfrac{nv}{2L}$ (for open pipe at both ends)
Fundamental frequency corresponds to n = 1.
$f_1=dfrac{v}{2L}=dfrac{(1)(343)}{2(3.5)}=49;Hz$
49;Hz
$$
$f_n=dfrac{nv}{4L}$ (for closed at one end)
The fundamental frequency corresponds to n = 1
$$
f_1=dfrac{(1)(343m/s)}{4(3.5m)}=25;Hz
$$
25;Hz
$$
$f_n=dfrac{nv}{2L}$ (for open at both ends)
Use the formula for n =1,2,3
$f_1=dfrac{nv}{2L}=dfrac{(1)(343m/s)}{(2)(1.5;m)}=114;Hz$
$f_2=nf_1=(2)(114.3;Hz)=229;Hz$
$f_3=nf_1=(3)(114.3;Hz)=343;Hz$
114 ;Hz, 229;Hz, 343; Hz
$$
$f_n=dfrac{nv}{4L}$ (for open at one end)
Use the formula for n =1,3,5…
$f_1=dfrac{(1)(343)}{4(1.5)}=57;Hz$
$f_3=nf_1=3(57.2)=172;Hz$
$f_3=nf_1=5(57.2)=286;Hz$
57;Hz, 172;Hz,286;Hz
$$
$f_n=dfrac{nv}{4L}$ (for open at one end)
Solve for L
$$
L=dfrac{v}{4f_1}
$$
Plug-in the given values
$L=dfrac{343m/s}{4(275;Hz)}=0.312;m$
b.) Calculate the third harmonic frequency (n=3) and fifth harmonic (n=5)
$f_3=dfrac{3v}{4L}=dfrac{3(343;m/s)}{4(0.312;m)}=825;Hz$
$$
f_5=dfrac{5v}{4L}=dfrac{5(343m/s)}{4(0.312m)}=1.38;kHz
$$
0.312;m,825;Hz,1.38;kHz
$$
We know that when a standing wave is formed in a pipe open at both ends, the length of the pipe is equal to an integer number of wavelength halves, that is:
$$
begin{equation}
L = frac{n cdot lambda}{2} , text{ for } , n = 1,2,3…
end{equation}
$$
For such a pipe the frequency of the harmonics are given by the formula:
$$
begin{equation}
f_{text{n}} = n cdot f_1 = frac{v cdot V}{2L} , text{ for } , n = 1,2,3
end{equation}
$$
where $f_1$ is the fundamental frequency.
begin{enumerate}[a)]
item
We know that the following frequencies form standing waves in a pipe open at both ends:
begin{align*}
100text{ Hz} quad , quad 200text{ Hz} quad , quad 250text{ Hz} quad , quad 300text{ Hz}
end{align*}
Each of these frequencies is larger than the fundamental frequency by an integer number, as governed by formula (2). \
So we must find the smallest frequency $f_1$ that can be used to form the numbers 100, 200, 250, 300. \It is obvious that this frequency is not larger than 50, since we would then come to an impossibility. \\
We also know from the problem that there are no standing wave frequencies between $250text{ Hz}$ and $300text{ Hz}$, which tells us that the fundamental frequency can not be smaller than $50text{ Hz}$. \\
Now we conclude that it is $50text{ Hz}$ exactly. \
begin{align*}
f_1 = 50text{ Hz}
end{align*}
item
The third harmonic can easily be found using formula (2), the found fundamental frequency $f_1$, and the knowledge that for it the integer $n = 3$.
begin{align*}
f_3 = n cdot f_1 = 3 cdot 50text{ Hz} = 150text{ Hz}
end{align*}
end{enumerate}
item
The fundamental frequency is $f_1 = 50text{ Hz}$
item
The frequency of the third harmonic is $f_3 = 150text{ Hz}$
end{enumerate}
$f_n=dfrac{nv}{4L}$
Calculate the fundamental frequency (n=1)
$f_1=dfrac{nv}{4L}=dfrac{(1)(343;m/s)}{4(0.024;m)}=3.6;kHz$
Calculate the wavelength
$lambda _1=dfrac{4L}{n}=4(2.4;cm)/1=9.6;cm$
b.) Calculate the third harmonic
$f_3=dfrac{nv}{4L}=dfrac{3(343)}{4(0.024;m)}=11;kHz$
$lambda_3=dfrac{4L}{n}=dfrac{4(2.4;cm/s)}{3}=3.2;cm$
c.) The fundamental frequency varies inversely to the length of the tube. Hence, if the ear canal is shorter, the fundamental frequency must be greater than that calculated in part a.
$f_n=dfrac{nv}{4L}$
Solving for n=3 (third harmonic)
$f_3=dfrac{3v}{4L}=dfrac{3(343;m/s}{4(2.75;m)}=93.5;Hz$
b.) Solving for the fundamental frequency
$$
f_1=dfrac{f_3}{3}=dfrac{93.5;Hz}{3}=31.2;Hz
$$
$f_n=dfrac{nv}{2L}$
a.) Solving for $f_1$
$f_1=dfrac{f_n}{n}=dfrac{202;Hz}{2}=101;Hz$
b.) Solving for L
$$
L=dfrac{v}{2;f_1}=dfrac{343m/s}{2(101;Hz)}=1.70;m
$$
$f_{listener}=f_{source}dfrac{v_{sound}pm v_{listener}}{v_{sound}pm v_{source}}$
The rules are:
If the listener is travelling towards the source, the sign of $v_{listener}$ is positive. If it’s moving away, then it is negative.
Furthermore, if the source is moving towards from the listener, the sign of $v_{source}$ is negative. If the source is moving away from the listener, it is positive.
$$
f_{listener}=136left(dfrac{343}{343-31.8}right)=150;Hz
$$
150;Hz
$$
$f_{listener}=f_{source}dfrac{v_{sound}pm v_{listener}}{v_{sound}pm v_{source}}$
The rules are:
If the listener is travelling towards the source, the sign of $v_{listener}$ is positive. If it’s moving away, then it is negative.
Furthermore, if the source is moving towards from the listener, the sign of $v_{source}$ is negative. If the source is moving away from the listener, it is positive.
$$
f_{listener}=136left(dfrac{343+31.8}{343}right)=149;Hz
$$
149;Hz
$$
$f_{listener}=f_{source}dfrac{v_{sound}pm v_{listener}}{v_{sound}pm v_{source}}$
The rules are:
If the listener is travelling towards the source, the sign of $v_{listener}$ is positive. If it’s moving away, then it is negative.
Furthermore, if the source is moving towards from the listener, the sign of $v_{source}$ is negative. If the source is moving away from the listener, it is positive.
$f_{listener}=(35.0)left(dfrac{343}{343-3.25}right)=35.3;kHz$
b.) Since the bat is going towards the moth, Doppler shift results to higher frequencies. Hence, the moth will hear a HIGHER frequency.
c.) Use the equation of Doppler effect and plug-in the given values
$$
f_{observer}=dfrac{35.0}{35.0-4.25}=35.4;kHz
$$
35.4;kHz
$$
$f_{listener}=f_{source}dfrac{v_{sound}pm v_{listener}}{v_{sound}pm v_{source}}$
The rules are:
If the listener is travelling towards the source, the sign of $v_{listener}$ is positive. If it’s moving away, then it is negative.
Furthermore, if the source is moving towards from the listener, the sign of $v_{source}$ is negative. If the source is moving away from the listener, it is positive.
$$
f_{listener}=502left(dfrac{343+13}{343}right)=521;Hz
$$
521;Hz
$$
$f_{listener}=f_{source}dfrac{v_{sound}pm v_{listener}}{v_{sound}pm v_{source}}$
The rules are:
If the listener is travelling towards the source, the sign of $v_{listener}$ is positive. If it’s moving away, then it is negative.
Furthermore, if the source is moving towards from the listener, the sign of $v_{source}$ is negative. If the source is moving away from the listener, it is positive.
$f_{listener}=495left(dfrac{v+0.825v}{v}right)=903;Hz$
b.) For the second case, the sign of $v_{source}$ is negative since it is moving towards the observer.
$$
f_{listener}=495left(dfrac{v}{v-0.825v}right)=2.83;kHz
$$
903 Hz, 2.83;kHz
$$
Our perception of sound is such that doubling the loudness corresponds to increasing the intensity by a factor of 10, which corresponds to an increase of $10text{ dB}$.
Keeping that in mind, it is easily found that for doubling of loudness we need a source of
$$
begin{align*}
45text{ dB} + 10text{ dB} = 55text{ dB}
end{align*}
$$
The amount of energy carried by a sound wave through a given area in a given time is the intensity $I$. It can be calculated via power $P$ and area $A$ as follows:
$$
begin{align*}
I = frac{P}{A}
end{align*}
$$
For spherical wave expansion the area becomes:
$$
begin{align*}
A = 4 , r^2 , pi
end{align*}
$$
So we get finally:
$$
begin{equation}
I = frac{P}{4 , r^2 , pi}
end{equation}
$$
where $r$ is the distance from the sound source to the wave front.
We know the power output of the sparrow song is $P =3.15 cdot 10^{-5}text{ W}$, and the minimum sound intensity that the human ear can detect is $I = 10^{-12} ; frac{text{W}}{text{m}^2}$
Now using formula (1) we find the maximum distance from the observer to the sound source (the bird) $r$ as follows:
$$
begin{align*}
I = frac{P}{4 , r^2 , pi}
end{align*}
$$
Rearranging for $r$ we get:
$$
begin{align*}
r = sqrt{frac{P}{4 , pi , I}}
end{align*}
$$
Plugging in the values we get:
$$
begin{align*}
r = sqrt{frac{3.15 cdot 10^{-5}text{ W}}{4 , pi cdot 10^{-12} ; frac{text{W}}{text{m}^2}}} approx 1,584text{ m}
end{align*}
$$
The amount of energy carried by a sound wave through a given area in a given time is the intensity $I$. It can be calculated via power $P$ and area $A$ as follows:
$$
begin{align*}
I = frac{P}{A}
end{align*}
$$
For spherical wave expansion the area becomes:
$$
begin{align*}
A = 4 , r^2 , pi
end{align*}
$$
So we get finally:
$$
begin{equation}
I = frac{P}{4 , r^2 , pi}
end{equation}
$$
where $r$ is the distance from the sound source to the wave front.
We know that the warning siren produces a sound intensity $I_0 = 0.5 ; frac{text{W}}{text{m}^2}$ at a distance of $r_0 = 2text{ m}$.
Now, different observers will experience different sound intensities since they are located at different distances from the sound source (warning siren). But since it is the same source its power is constant, so we can write formula (1) as follows:
$$
begin{align*}
I_0 = frac{P}{4 , r_0^2 , pi} \
I_x = frac{P}{4 , r_x^2 , pi}
end{align*}
$$
Rearranging into a more useful form we get:
$$
begin{equation*}
frac{P}{4 , pi} = I_0 cdot r_0^2 = I_x cdot r_x^2 tag{2}
end{equation*}
$$
We will use this formula to solve the rest of the problem.
begin{enumerate}[a)]
item
Using formula (2) we simply find the intensity $I_x=I_A$ at distance $r_x = r_A = 12text{ m}$ as follows:
begin{align*}
I_0 cdot r_0^2 = I_A cdot r_A^2
end{align*}
Rearranging for $I_A$ we get:
begin{align*}
I_A = frac{I_0 cdot r_0^2}{r_A^2}
end{align*}
Plugging in the values we get:
begin{align*}
I_A = frac{0.5 ; frac{text{W}}{text{m}^2} cdot (2text{m})^2 }{(12text{ m})^2 } approx 1.4 cdot 10^{-2} ; frac{text{W}}{text{m}^2}
end{align*}
item
Once again, using formula (2) we simply find the intensity $I_x = I_B$ at distance $r_x = r_B = 21text{ m}$ as follows:
begin{align*}
I_0 cdot r_0^2 = I_B cdot r_B^2
end{align*}
Rearranging for $I_B$ we get:
begin{align*}
I_B = frac{I_0 cdot r_0^2}{r_B^2 }
end{align*}
Plugging in the values we get:
begin{align*}
I_B = frac{0.5 ; frac{text{W}}{text{m}^2} cdot (2text{ m})^2 }{(21text{ m})^2 } approx 0.45 cdot 10^{-2} ; frac{text{W}}{text{m}^2}
end{align*}
end{enumerate}
item
We can find how far the siren can be heard by setting the sound intensity $I_x$ to be equal to $I_C = 10^{-12} ; frac{text{W}}{text{m2}^2}$ (which is the minimum sound the human ear can hear) and using formula (2)
begin{align*}
I_0 cdot r_0^2 = I_C cdot r_C^2
end{align*}
Now rearranging our formula for $r_C$ we get:
begin{align*}
& r_c^2 = frac{I_0 cdot r_0^2}{I_C} \
& r_c = sqrt{frac{I_0 cdot r_0^2}{I_C} }
end{align*}
Plugging in the values we get:
begin{align*}
r_C = 1.41 cdot 10^{6}text{ m} approx 1410text{ km}
end{align*}
end{enumerate}
item
The intensity of the siren at distance $r_A = 12text{ m}$ is $I_A = 1.4 cdot 10^{-2} ; frac{text{W}}{text{m}^2}$
item
The intensity of the siren at distance $r_B = 21text{ m}$ is $I_B = 0.45 cdot 10^{-2} ; frac{text{W}}{text{m}^2}$
item
The maximum distance the siren can be heard is $r_C = 1.41 cdot 10^{6}text{ m} = 1410text{ km}$
end{enumerate}
$A=pi r^2=pi (4.0times10^{-3})^2=5.0times10^{-5};m^2$
Since intensity is power over area, you can solve for the power as
$P=I;A=(1times10^{-12}W/m^2)(5.0times10^{-5})=5.0times10^{-17};W$
$$
P=I;A=(1;W/m^2)(5.0times10^{-5}m^2)=5.0times10^{-5}W
$$
$70-10=60;dB$
b.) If the number of violins is increased to 100, the sound level will become 20 dB higher than each violin. Thus, the intensity level produced in simultaneously playing 100 violin is equal to 80 dB.
begin{enumerate}[a)]
item
First a reminder: \\
The amount of energy carried by a sound wave through a given area in a given time is the intensity $I$. It can be calculated via power $P$ and area $A$ as follows:
begin{align*}
I = frac{P}{A}
end{align*}
For spherical wave expansion the area becomes:
begin{align*}
A = 4 , r^2 , pi
end{align*}
So we get finally:
begin{align*}
I = frac{P}{4 , r^2 , pi} tag{1}
end{align*}
where $r$ is the distance from the sound source to the wave front. \\
We can see from formula (1) that as the distance r decreases, the intensity of the sound increases!
end{enumerate}
item
First let us remember: \\
We know that for a moving sound source the frequency shift is given by the following Doppler Effect formula:
begin{equation*}
f_{text{observer}} = frac{f_{text{source}} }{left(1 pm frac{v_{text{source}}}{v_{text{sound}}} right)} tag{2}
end{equation*}
Where the plus sign holds when the source is moving away from the observer,
and the minus sign when the source is moving toward the observer. \\
Now looking at formula (2) we can see that when the sound source approaches the observer the observed frequency will increase.
end{enumerate}
item
Solving this part requires both parts b) and d) solved, so the reader is encouraged to look at them first. \
Then we will remember an important formula for wave speed:
We know that the wavelength and frequency determine the speed of a wave. They are related by the following formula:
begin{equation*}
v = lambda cdot f tag{3}
end{equation*}
Now, since we know that the speed of the sound wave is constant, and that the frequency increases, by looking at formula (3) it is obvious that the wavelength has to decrease!
end{enumerate}
item
The speed of a wave in a given substance depends on the properties of that substance, such as density, temperature, and so forth… \\
Since our wave never changes mediums (it always travels through air) it’s speed will not change!
end{enumerate}
item
The intensity of the sound increases
item
The frequency of the sound wave increases
item
The wavelength of the sound wave decreases
item
The speed of the sound wave remains the same
end{enumerate}
The blow on the nail creates a sound due the vibration of the nail. It works similar to a tuning fork.
$$
begin{align*}
text{frequency} & propto text{length of the nail}
end{align*}
$$
As the length of the nail gets shorter the frequency of the sound wave increases.
$$
begin{align*}
text{Higher frequency} & longrightarrow text{Higher pitch}
end{align*}
$$
Therefore, we hear the higher pitch with each hammer blow.
We know that the wavelength and frequency determine the speed of a wave. They are related by the following formula:
$$
begin{equation}
v = lambda cdot f
end{equation}
$$
We know that the speed of sound in air is $v =343 ; frac{text{m}}{text{s}}$, and we know the measured frequencies of the two baseball hits $f_1 = 510text{ Hz}$ and $f_2 = 170text{ Hz}$.
Now using formula (1) we separately calculate the wavelengths of the sound waves produced by the hit as follows:
$$
begin{align*}
v = lambda cdot f
end{align*}
$$
Rearranging for lambda we get:
$$
begin{align*}
lambda = frac{v}{f}
end{align*}
$$
Now plugging in the values we get:
$$
begin{align*}
& lambda_1 = frac{343 ; frac{text{m}}{text{s}}}{510text{ Hz}} approx 0.67text{ m} \
& lambda_2 = frac{343 ; frac{text{m}}{text{s}}}{170text{ Hz}} approx 2.02text{ m}
end{align*}
$$
Now the wavelength difference is simply found:
$$
begin{align*}
& Delta , lambda = lambda_2 – lambda_1 \
& Delta , lambda = 2.02text{ m} – 0.67text{ m} = 1.35text{ m}
end{align*}
$$
We know that the wavelength and frequency determine the speed of a wave. They are related by the following formula:
$$
begin{equation}
v = lambda cdot f
end{equation}
$$
We know that the speed of sound in human body is $v = 1500 ; frac{text{m}}{text{s}}$, and we know the used ultrasound frequency to equal $f = 3.5text{ MHz}$
Now using formula (1) we calculate the wavelength of the ultrasound waves in the human body:
$$
begin{align*}
v = lambda cdot f
end{align*}
$$
Rearranging for lambda we get:
$$
begin{align*}
lambda = frac{v}{f}
end{align*}
$$
Now plugging in the values we get:
$$
begin{align*}
lambda = frac{1500 ; frac{text{m}}{text{s}}}{3.5 cdot 10^{6}text{ Hz}} approx 4.29 cdot 10^{-4}text{ m}
end{align*}
$$
We know that the wavelength and frequency determine the speed of a wave. They are related by the following formula:
$$
begin{equation}
v = lambda cdot f
end{equation}
$$
We know that when a standing wave is formed in a bottle or pipe open at one end, the length of the pipe is equal to an odd number of wavelength fourths, that is:
$$
begin{align*}
L = frac{n cdot lambda}{4} , text{ for} , n = 1,3,5… tag{2}
end{align*}
$$
We know that when a standing wave is formed in a pipe open at both ends, the length of the pipe is equal to an integer number of wavelength halves, that is:
$$
begin{align*}
L = frac{n cdot lambda}{2} , text{ for } , n = 1,2,3… tag{3}
end{align*}
$$
The fundamental frequency of the pipe closed at one end is $f_1 = 261.6text{ Hz}$.
By inserting formula (1) into formula (2) we remove the wavelength and get the more usable formula:
$$
begin{align*}
L = frac{n cdot v}{4 , f} , text{ for } , n = 1,3,5… tag{4}
end{align*}
$$
For the fundamental frequency $f = f_1$, the integer $n = 1$ so we can find the length:
$$
begin{align*}
L_1 = frac{v}{4 , f_1} = frac{343 ; frac{text{m}}{text{s}}}{(4 cdot 261.6text{ Hz}) } approx 0.33text{ m}
end{align*}
$$
The second harmonic frequency of the pipe opened at both ends is $f_2 = 261.6text{ Hz}$.
By inserting formula (1) into formula (3) we remove the wavelength and get the more usable formula:
$$
begin{align*}
L = frac{n cdot v}{2 , f} , text{ for } , n = 1,2,3…tag{ 5}
end{align*}
$$
For the second harmonic frequency $f = f_2$, the integer $n = 2$ so we can find the length:
$$
begin{align*}
L_2 = frac{ 2 , v}{2 , f_2} = frac{2 cdot 343 ; frac{text{m}}{text{s}}}{2 cdot 261.6text{ Hz}} approx 1.31text{ m}
end{align*}
$$
$120-12=108;machines$
must be shut-down.
We know that for a moving sound source the frequency shift is given by the following Doppler Effect formula:
$$
begin{equation*}
f_{text{observer}} = frac{f_{text{source}} }{left(1 pm frac{v_{text{source}}}{v_{text{sound}}} right)} tag{1}
end{equation*}
$$
Where the plus sign holds when the source is moving away from the observer,
and the minus sign when the source is moving toward the observer.
Beats are formed by the interfering of two frequencies that are played simultaneously. They can be heard as a periodical change in loudness of the resulting sound. The frequency of a beat is the difference of the interfering frequencies:
$$
begin{align*}
f_{text{beat}} = abs{f_1 – f_2} tag{2}
end{align*}
$$
The horn of the trains has a frequency of $f_{text{source}} = 124text{ Hz}$.
The slower of the two trains has a speed $V_{text{source1}} = 26 ; frac{text{m}}{text{s}}$, and the frequency of the sound the observer hears from its horn fobserver1 can be calculated by using formula (1).
We will use the formula with the minus sign, since the source is moving toward the observer.
$$
begin{align*}
f_{text{observer1}} = frac{f_{text{source}}}{ left(1 – frac{v_{text{source1}}}{v_{text{sound}} } right) }
end{align*}
$$
Plugging in the values we get:
$$
begin{align*}
f_{text{observer1}} = frac{124text{ Hz} }{ left(1 – frac{26 ; frac{text{m}}{text{s}}}{343 ; frac{text{m}}{text{s}} } right) } approx 134.2text{ Hz}
end{align*}
$$
$$
begin{align*}
f_{text{observer2}} > f_{text{observer1}} > f_{text{source}} tag{3}
end{align*}
$$
The heard beat frequency is $f_{text{beat}} = 4.4text{ Hz}$ so by formula (2) we can write
$$
begin{align*}
f_{text{beat}} = abs (f_{text{observer1}} – f_{text{observer2}})
end{align*}
$$
Now using our conclusion (3) we rewrite:
$$
begin{align*}
f_{text{beat}} = f_{text{observer2}} – f_{text{observer1}}
end{align*}
$$
Rearranging we find $f_{text{observer2}}$ :
$$
begin{align*}
f_{text{observer2}} = f_{text{beat}} + f_{text{observer1}}
end{align*}
$$
Plugging in the numbers we have:
$$
begin{align*}
f_{text{observer2}} = 4.4text{ Hz} + 134text{ Hz} = 138.6text{ Hz}
end{align*}
$$
$$
begin{align*}
f_{text{observer2}} = frac{f_{text{source}}}{ left(1 – frac{v_{text{source2}}}{v_{text{sound}} } right) }
end{align*}
$$
Rearranging for $v_{text{source2}}$:
$$
begin{align*}
v_{text{source2}} = v_{text{sound}} cdot left(1 – frac{f_{text{source}}}{f_{text{observer2}}} right)
end{align*}
$$
Plugging in the numbers we get:
$$
begin{align*}
v_{text{source2}} = 343 ; frac{text{m}}{text{s}} cdot left(1 – frac{124text{ Hz}}{138.6text{ Hz}} right) approx 36 ; frac{text{m}}{text{s}}
end{align*}
$$
We know that for a moving sound source the frequency shift is given by the following Doppler Effect formula:
$$
begin{equation*}
f_{text{observer}} = frac{f_{text{source}} }{left(1 pm frac{v_{text{source}}}{v_{text{sound}}} right)} tag{1}
end{equation*}
$$
Where the plus sign holds when the source is moving away from the observer,
and the minus sign when the source is moving toward the observer.
We know that for a moving observer the frequency shift is given by the following Doppler Effect formula:
$$
begin{align*}
f_{text{observer}} = f_{text{source}} left(1 pm frac{v_{text{observer}}}{v_{text{sound}}} right) tag{2}
end{align*}
$$
Where the plus sign holds when the observer is moving towards the source,
and the minus sign when the observer is moving away from the source.
Jim’s boat is moving with a speed of $v = 24 ; frac{text{m}}{text{s}}$ and sounds a horn of frequency $f_{text{source}} = 330text{ Hz}$.
begin{enumerate}[a)]
item
The frequency of the sound that Betsy hears $f_{text{observer1}}$ can be calculated using formula (1). \\
We will use the formula with the minus sign, since the source is moving toward the observer.
begin{align*}
f_{text{observer1}} = frac{f_{text{source}}}{ left(1 – frac{v_{text{source}}}{v_{text{sound}}} right) }
end{align*}
Plugging in the values we get:
begin{align*}
f_{text{observer1}} = frac{330text{ Hz} }{ left(1 – frac{24 ; frac{text{m}}{text{s}}}{343 ; frac{text{m}}{text{s}} } right) } approx 355text{ Hz}
end{align*}
end{enumerate}
item
When the horn sound is reflected from the cliff the frequency fobserver1 is effectively the new source frequency $f_{text{source2}}$, and Jim is now an observer moving toward the new stationary source.\
Obviously this situation is now described by formula (2). \\
Since he is moving toward the echo the frequency will increase further, so the frequency he hears $f_{text{observer2}}$ will be greater then the one heard by Betsy
begin{align*}
f_{text{observer1}} = f_{text{source2}}
end{align*}
item
The frequency of the echo Jim hears can be calculated using formula (2) as was already mentioned in part b). \
Along with $f_{text{source2}} = f_{text{observer1}}$ , using the formula with the plus sign (since the observer moves towards the source) yields:
begin{align*}
f_{text{observer2}} = 355text{ Hz} left(1 + frac{24 ; frac{text{m}}{text{s}} }{343 ; frac{text{m}}{text{s}} } right) approx 380text{ Hz}
end{align*}
end{enumerate}
item
The frequency heard by Betsy is $f_{text{observer1}} = 355text{ Hz}$
item
The frequency of the echo Jim hears is greater than the one Betsy hears.
item
The frequency of the echo Jim hears is $f_{text{observer2}} = 380text{ Hz}$
end{enumerate}
$$
begin{align*}
L = frac{n cdot lambda}{4} , text{ for } , n = 1,3,5… tag{1}
end{align*}
$$
The length of our pipe is $L = 2.5text{ m}$.
For the third harmonic the integer $n = 3$.
Since the nodes are the places of maximum elongation of the wave and the antinodes the places of minimum elongation of the wave, we know that the distance between adjacent nodes and antinodes is actually one quarter of the wavelength, so we write:
$$
begin{align*}
d = frac{lambda}{4}
end{align*}
$$
To find $lambda$ we simply use formula (1), along with the values for $n$ and $L$.
$$
begin{align*}
L = frac{n cdot lambda}{4}
end{align*}
$$
Rearranging for $lambda$:
$$
begin{align*}
lambda = frac{4 , L}{n}
end{align*}
$$
Now inserting the values we get:
$$
begin{align*}
lambda = frac{4 cdot 2.5text{ m}}{3} = 3.33text{ m}
end{align*}
$$
Finally we find the distance between adjacent nodes and antinodes:
$$
begin{align*}
d = frac{3.33text{ m}}{4} = 0.83text{ m}
end{align*}
$$
In order to do that, bats send out sound waves (ultrasound) using their mouth or nose. When these ultrasound waves hit an object, they produce echoes. This echo bounces off the object (preys) and returns to the bats’ ears.
Now, bats listen and analyze these echoes to figure out where the object is. They also figure out its size, shape and its moving speed.
The ears’ external structure plays an important role in receiving echoes. There is large variation in sizes, shapes, folds and wrinkles in bats’ ears. These factors considered to aid in the reception and funneling of echoes and sounds emitted from prey.
Bats can use echolocation to find their way around very quickly and efficiently in the darkness. Bats can detect very tiny objects using echolocation. It allows bats to find small insects such as mosquitoes, flies, beetles, moths, etc.
$$
begin{align*}
f_{bat} & = 20, 000 mathrm{Hz} – 200, 000 mathrm{Hz} \
left(f_{bat}right)_{ins} & = 20, 000 mathrm{Hz} – 60, 000 mathrm{Hz}
end{align*}
$$
Therefore the range of the wavelengths are
$$
begin{align*}
lambda_{bat} & = (0.0017 mathrm{m} – 0.017 mathrm{m}) = (1.7 mathrm{mm} – 17 mathrm{mm}) \
left(lambda_{bat}right)_{ins} & = (0.0057 mathrm{m} – 0.017 mathrm{m}) = (5.7 mathrm{mm} – 17 mathrm{mm} )
end{align*}
$$
When these insects hear an echolocating bat, they turn and try to fly away. To avoid the their capture, some insects try flying in some random patterns such as zigzag, spiral, or some looping pattern.
Some crickets and beetles are known to make clicking sounds. This kind of sound may startle the bat and scare it off thus avoiding being eaten.
We know that when a standing wave is formed in a pipe open at both ends, the length of the pipe is equal to an integer number of wavelength halves, that is:
$$
begin{align*}
L = frac{n cdot lambda}{2} , text{ for } , n = 1,2,3… tag{1}
end{align*}
$$
For such a pipe the frequency of the harmonics are given by the formula:
$$
begin{align*}
f_{text{n}} = n cdot f_1 = frac{n cdot v}{2 , L} , text{ for } , n = 1,2,3…tag{2}
end{align*}
$$
Where $f_1$ is the fundamental frequency.
begin{enumerate}[a)]
item
We know that nasal passage of the dinosaur corresponded to a bent tube $2.7text{ m}$ long, so $L = 2.7text{ m}$ \
The fundamental frequency of the tube can be found using formula (2) by setting the integer $n = 1$:
begin{align*}
f_1 = frac{n cdot V}{(2 , L)} = frac{v}{(2 , L)}
end{align*}
Using $v = 343 ; frac{text{m}}{text{s}}$ for the speed of sound in air we get:
begin{align*}
f_1 = frac{343 ; frac{text{m}}{text{s}}}{2 cdot 2.7text{ m}} approx 64text{ Hz}
end{align*}
We have found that the correct answer is C
end{enumerate}
We know that when a standing wave is formed in a pipe open at both ends, the length of the pipe is equal to an integer number of wavelength halves, that is:
$$
begin{align*}
L = frac{n cdot lambda}{2} , text{ for } , n = 1,2,3…tag{1}
end{align*}
$$
For such a pipe the frequency of the harmonics are given by the formula:
$$
begin{align*}
f_{text{n}} = n cdot f_1 = frac{n cdot v}{2 , L} , text{ for } , n = 1,2,3… tag{2}
end{align*}
$$
Where $f_1$ is the fundamental frequency.
a) we know that the nasal passage of the particular female dinosaur corresponded to a bent tube of length $L$, in which forms a standing wave of fundamental frequency $f_1 = 74text{ Hz}$
The fundamental frequency of the tube can be expressed using formula (2) by setting the integer $n = 1$, which leads to a solution for $L$:
$$
begin{align*}
f_1 = frac{n cdot v}{(2 , L)} = frac{v}{(2 , L)}
end{align*}
$$
Rearranging for $L$ we get:
$$
begin{align*}
L = frac{v}{2 , f_1}
end{align*}
$$
Now using $v = 343 ; frac{text{m}}{text{s}}$ for the speed of sound in air, and plugging in the given frequency we get:
$$
begin{align*}
L =frac{343 ; frac{text{m}}{text{s}}}{2 cdot 74text{ Hz} } approx 2.3text{ m}
end{align*}
$$
We have found that the correct answer is B
