All Solutions
Page 527: Standardized Test Prep
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:
$$
begin{equation}
f_{text{beat}} = abs{f_1 – f_2}
end{equation}
$$
When the open plastic bottles are struck they both release a frequency of $f = 260text{ Hz}$.
When one of them is shortened its frequency will change. Since the wavelength of the sound a pipe produces is proportional to the pipe’s length, shortening a pipe will decrease the wavelength of its sound wave. This corresponds to an increase in frequency. So we conclude:
$$
begin{align*}
f_{text{shorter}} > f_{text{longer}} tag{2}
end{align*}
$$
Now when they are both struck together the different frequencies interfere and a beat is heard. The beat frequency is $f_{beat} = 4text{ Hz}$.
Using formula (1) we can write:
$$
begin{align*}
f_{text{beat}} = abs (f_{text{shorter}} – f_{text{longer}})
end{align*}
$$
Conclusion (2) allows us to simply remove the absolute bracket, so we get:
$$
begin{align*}
f_{text{beat}} = f_{text{shorter}} – f_{text{longer}}
end{align*}
$$
Since the longer bottle is unchanged from the beginning we can write:
$$
begin{align*}
f_{text{longer}} = f = 260text{ Hz}
end{align*}
$$
Now rearranging for $f_{text{shorter}}$:
$$
begin{align*}
f_{text{shorter}} = f_{text{beat}} + f_{text{longer}}
end{align*}
$$
Plugging in the values we get:
$$
begin{align*}
f_{text{shorter}} = 4text{ Hz} + 260text{ Hz} = 264text{ Hz}
end{align*}
$$
The correct answer is C
We know that the wavelength and frequency determine the speed of a wave. They are related by the following formula:
$$
begin{equation}
v = lambda cdot f
end{equation}
$$
The fundamental wavelength of a tube-like wind instrument is directly proportional to it’s length.
As the length of the clarinet doesn’t changed noticeably during the heating we can say that the wavelength is constant:
$$
begin{align*}
lambda_1 = lambda_2
end{align*}
$$
We know that the wave speed in a medium that is in gas state is dependant on the temperature of the medium. As the temperature increases, so does the wave speed.
This can be understood on the molecular level, with an increase of the speed of all molecules, the speed of any directed motion also increases.
We conclude
$$
begin{align*}
v_1 > v_1
end{align*}
$$
Finally using formula (1) we can write:
$$
begin{align*}
f_1 = frac{v_1}{lambda_1} \
f_2 = frac{v_2}{lambda_2}
end{align*}
$$
Now using the previous conclusions for the wavelength and speed we can conclude:
$$
begin{align*}
f_2 > f_1
end{align*}
$$
The only correct solution is C
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:
$$
begin{align*}
L = frac{n cdot lambda}{4} , text{ for } , n = 1,2,3… tag{2}
end{align*}
$$
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 pitch of a note produced in the straw can be analyzed by analyzing the fundamental frequency of the standing wave formed in the straw.
Let us denote the initial length of the straw $L_{text{i}}$, and the final length $L_{text{f}}$.
The initial fundamental frequency $f_{1text{i}}$ and the final fundamental frequency $f_{1text{f}}$.
Even though we could have started from the formulas (4) and (5) we will derive them first.
Firstly for the initial fundamental frequency of the straw.
By inserting formula (1) into formula (3) we remove the wavelength and get the more usable formula:
$$
begin{align*}
L = frac{n cdot v}{2 , f} , text{ for } , n=1,2,3…
end{align*}
$$
For the fundamental frequency $f = f_{1text{i}}$, the integer $n = 1$ so we can rearrange for the initial fundamental frequency:
$$
begin{align*}
f_{1text{i}} = frac{v}{(2 , L_{text{i}})} tag{4}
end{align*}
$$
Now for the final fundamental frequency of the straw.
By inserting formula (1) into formula (2) we remove the wavelength and get the more usable formula:
$$
begin{align*}
L_{text{f}} = frac{n cdot v}{4 , f} , text{ for } , n=1,3,5…
end{align*}
$$
For the fundamental frequency $f = f_{1text{f}}$, the integer $n = 1$ so we rearrange for the new fundamental frequency:
$$
begin{align*}
f_{1text{f}} = frac{v}{4 , L_{text{f}}} tag{5}
end{align*}
$$
Now using formulas (4) and (5) we can compare the initial and final fundamental frequencies.
$$
begin{align*}
& f_{1text{i}} = frac{v}{(2 , L_{text{i}})} \
& f_{1text{f}} = frac{v}{(4 , L_{text{f}})}
end{align*}
$$
We also know that the straw is halved so:
$$
begin{align*}
L_{text{i}} = 2 , L_{text{f}}
end{align*}
$$
Using this we can se that the following line of equalities holds:
$$
begin{align*}
f_{1text{i}} = frac{v}{(2 , L_{text{i}})} = frac{v}{(2 cdot 2 , L_{text{f}})} = frac{v}{(4 , L_{text{f}})} = f_{1text{f}}
end{align*}
$$
The fundamental frequency remains the same, hence so does the pitch!
The correct answer is A
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.
The fact that the intensity of the sound at the speaker increases twice means that it’s power is doubled, so we write:
$$
begin{align*}
P_2 = 2 , P_1
end{align*}
$$
We also know that the distance from the observer to the speaker is doubled, so we write:
$$
begin{align*}
r_2 = 2 , r_1
end{align*}
$$
Now we simply find the new intensity with respect to the old one using formula (1) as follows:
$$
begin{align*}
I_2 = frac{P_2}{(4 cdot r_2^2 cdot pi) } = frac{2 , P_1}{(4 cdot (2 , r_1)^2 cdot pi) } = frac{2}{4 cdot I_1} = frac{1}{2} cdot I_1
end{align*}
$$
We have found $I_2 = frac{I_1}{2}$ , so we conclude that the final solution is B
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
The length of the tube is $L = 1.2text{ m}$
The frequencies of the first and second harmonics in the tube can be found using formula (2) by setting the integer $n = 1$ for the first harmonic and $n = 2$ for the second, which leads to
begin{align*}
& f_1 = frac{n cdot v}{(2 , L)} = frac{v}{(2 , L)} \
& f_2 = frac{n cdot v}{(2 , L)} = frac{2 cdot v}{(2 , L)}
end{align*}
Now using $v = 343 ; frac{text{m}}{text{s}}$ for the speed of sound in air, and plugging in the given length we get:
begin{align*}
& f_1 = frac{343 ; frac{text{m}}{text{s}}}{(2 cdot 1.2text{ m}) } approx 143text{ Hz} \
& f_2 = 2 cdot f_1 = 286text{ Hz}
end{align*}
We have found that the correct answer is C
end{enumerate}
We know that the wavelength and frequency determine the speed of a wave. They are related by the following formula:
$$
begin{equation}
v = lambda cdot f
end{equation}
$$
$$
begin{align*}
v sim sqrt{frac{F}{mu}} = sqrt{frac{F cdot L}{m}} tag{2}
end{align*}
$$
First let us say that all relevant wavelengths are proportional to the length of the guitar string.
This means that Formula (1) becomes:
$$
begin{align*}
v sim f cdot L
end{align*}
$$
Rearranging for frequency we get:
$$
begin{align*}
f sim frac{v}{L}
end{align*}
$$
Now using formula 2 we get
$$
begin{align*}
& f sim frac{sqrt{frac{F cdot L}{m}}}{L} \
& f sim sqrt{frac{F}{m cdot L}} tag{3}
end{align*}
$$
Now let us look at all the different changes that are offered as answers in the problem separately:
Tightening the string increases $F$, looking at formula (3) we see that will increase the frequency.
Option A is incorrect!
Shortening the string decreases $L$ , so looking at formula (3) we see that will increase the frequency.
Option B is incorrect!
Replacing the string with a thinner string of the same length will decrease the mass $m$ (supposing they are of the same material and hence share the same density)
Looking at formula (3) we see that decreasing the mass will increase the frequency.
Option C is incorrect!
Finally, plucking the string with more force doesn’t change the tension force $F$, nor the mass $m$, nor the length $L$, so we conclude that this change doesn’t affect the frequency (the pitch).
Option D is correct!
This leads us to answer C. The speed of sound is largest in a rigid solid, such as steel.
Side note: It is interesting to note that the speed of sound in a vacuum is 0 (or not defined) because sound can’t travel without a medium to carry it.
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 concluded that a change in intensity for a ratio lesser that 10 (4 in this problem) can only bring about a change in decibels smaller that $10text{ dB}$
The only option that fulfils this condition is A ( a change of $6text{ dB}$)
Precise calculation would require a logarithmic formula that exceeds the range of this course.
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.
Both the frequency heard by an observer moving toward a stationary source, and the frequency heard by a stationary observer approached by a moving sound source increase, and are higher that the original frequency of the source.
Let us name the original frequency $f_{text{source}}$ , and now we will find the new frequencies using the Doppler effect.
In the case of a moving sound source we will use formula (1) with the minus sign. The speed of the source is
$$
begin{align*}
v_{text{source}} = 0.1 cdot v_{text{sound}}
end{align*}
$$
$$
begin{align*}
& f_{text{observer1}} = frac{f_{text{source}}}{ left(1 – frac{0.1 cdot v_{text{sound}}}{v_{text{sound}}} right) } \
& f_{text{observer1}} = frac{f_{text{source}}}{ 0.9 }
end{align*}
$$
In the case of the moving observer we will use formula (2) with the plus sign, the speed of the observer is
$$
begin{align*}
v_{text{observer}} = 0.1 cdot v_{text{sound}}
end{align*}
$$
$$
begin{align*}
& f_{text{observer2}} = f_{text{source}} left(1 + frac{0.1 cdot v_{text{sound}}}{v_{text{sound}}} right) \
& f_{text{observer2}} = f_{text{source}} cdot 1.1
end{align*}
$$
In the case of the moving sound source we have $f_{text{observer1}} = frac{f_{text{source}}}{ left(1 – 0.1right) }$
