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\begin{center}
{\Large {Solution Derivations for Capa \#13\\[20pt]} }
\end{center}
1) Identify the following waves as T-Transverse, or L-Longitudinal. If the
first is T and the rets L, enter TLLL).
\begin{quote}
QUESTION:\\
A) The 'WAVE' made by fans at sports events.\\
B) Wave on a slinky toy, snugly confined in a long narrow tube.\\
C) Waves in a guitar string that is plucked.\\
D) Waves on the surface of a bass drum.\\
E) Band music at a football game.\\
ANSWER:\\
A) T, the fans are only moving up and down, not with the wave.\\
B) L, the slinky cannot move up and down due to its confinement.\\
C) T, the guitar string is vibrating back and forth perpendicular to the
axis.\\
D) T, on the surface, the material covering the drum is moving
perpendicular to the drum surface. This action compresses the air which
makes the sound.\\
E) L, the music we hear is from the air being compressed.
\end{quote}
2) The transverse displacement of a stretched string from equilibrium as a
function of time and position is given by:
\[
y = 0.13 \cos(9x - 99t).
\]
$x$ and $y$ are in $m$; $t$ is in $s$. (Select T-True, F-False, G-Greater
than, L-Less than, E-Equal to, If the first is T and the rest F, enter TFFF).
\begin{quote}
The general equation of the wave:
\[
y\left( x,t\right) =A\sin \left( kx-\omega t\right)
\]
QUESTION:\\
A) The wave travels in the positive x direction.\\
B) The speed of the wave is ..... \#\# m/s.\\
C) The period is ..... \#\# seconds.\\
D) The wavelength is ..... \#\# m.\\
ANSWER:\\
A) The wave travels in the positive x direction (to the right) if its
formula is the form $f\left( x-vt\right) $. It will travel in the negative
x direction (to the left) if the formula is $f\left( x+vt\right) $.\\
B) The speed of the wave can be calculated from $v=\frac{\omega }{k}$.\\
C) $\omega =\frac{2\pi }{T}$, so $T=\frac{2\pi }{\omega }$.\\
D) $k=\frac{2\pi }{\lambda }$, so $\lambda =\frac{2\pi }{k}$.
\end{quote}
3) A typical steel B-string in a guitar resonates in its fundamental
frequency at $240\ Hertz$. The length of the string is $0.660\ m$. What is
the wave velocity along the string?
\begin{quote}
$f$ = Given\\
$L$ = Given\\
To find the wavelength associated with the fundamental frequency, use the
equation
\[
\lambda _{n}=\frac{2L}{n}
\]
where $n=1$ for the fundamental frequency. This equation is what determines a
standing wave. Any integer number of half wavelengths determines a standing
wave since every half wavelength is a node in the wave. Whatever integer $n$
is used determines the harmonic level, where $n=1$ is the fundamental
frequency (1st harmonic). The velocity is then calculated from $v=\lambda f$.
\end{quote}
4) John is listening to a horn. He knows the frequency of the horn is
$300\ Hz$ when both he and the horn are at rest. If he hears a pitch of
$330\ Hz$, there are clearly several possibilities. (Give ALL correct
answers, i.e., B, AC, BCD...)
\begin{quote}
QUESTION:\\
A) John is moving towards the horn at rest.\\
B) Both can be moving and have the same speed.\\
C) Both can be moving and have different speeds.\\
D) Both can be moving, in opposite directions.\\
E) The distance between John and the horn is increasing with time.\\
F) Both cannot be moving in the same direction.\\
ANSWER:\\
A) True, when John moves towards the horn, it is the same as if he stood
still and the horn moved towards him.\\
B) True, if they are moving toward each other at the right speed, John
would hear more waves per unit time, thus a higher frequency.\\
C) True, the horn could be getting closer to John.\\
D) True, same as B, they can approach each other to achieve this effect.\\
E) False, it must be decreasing to hear a higher frequency.\\
F) False, they could be. If both are moving and the horn is catching
up to John, this effect could be achieved.\\
Be careful, this problem is looking for the TRUE answers... so answer ABCD
(or whatever answers are true).
\end{quote}
5) A 60 year old person has a threshold of hearing at $f = 10,000\ Hz$ of
$75.0\ dB$. By what factor must the intensity of a sound wave of that
frequency, audible to a typical young adult (sound level = $43.0\ dB$), be
increased so that it is heard by the older person.
\begin{quote}
$f$ = Given\\
$\beta _{old}$ = Given\\
$\beta _{young}$ = Given\\
In order for the older person to hear this sound, the intensity must be
increased above the threshold of hearing. That is, $\beta $ must be
increased. The problem is asking for what factor intensity must increase.
\begin{eqnarray*}
\beta _{young} & = & 10\log \left( \frac{I_{i}}{I_{0}}\right) \\
\beta _{old} & = & 10\log \left( \frac{I_{f}}{I_{0}}\right)
\end{eqnarray*}
where $I_{i}$ and $I_{f}$ are the initial and final intensities. We can
solve each equation for $I_{i}$ or $I_{f}$.
\begin{eqnarray*}
\frac{\beta _{young}}{10} & = & \log \left( \frac{I_{i}}{I_{0}}\right)\\
10^{\frac{\beta _{young}}{10}} & = & \frac{I_{i}}{I_{0}}\\
I_{0}10^{\frac{\beta _{young}}{10}} & = & I_{i}
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{\beta _{old}}{10} & = & \log \left( \frac{I_{f}}{I_{0}}\right) \\
10^{\frac{\beta _{old}}{10}} & = & \frac{I_{f}}{I_{0}} \\
I_{0}10^{\frac{\beta _{old}}{10}} & = & I_{f}
\end{eqnarray*}
where
\[
I_{0}=10^{-12}\frac{W}{m^{2}}
\]
is the threshold of human hearing. To find the factor the intensity must
increase, think that some number $x$ must be multiplied by $I_{i}$ to get
$I_{f}$. So,
\begin{eqnarray*}
x\ast I_{i} & = & I_{f}\\
x & = & \frac{I_{f}}{I_{i}}
\end{eqnarray*}
This ratio is unitless.
\end{quote}
6) A certain loudspeaker has a diameter of $26\ cm$. At what frequency will
the wavelength of the sound it emits in air be equal to its diameter?
({\em Note:} If the wavelength is large compared to the diameter of the
speaker, then the sound waves spread out almost uniformly in all directions
from the speaker; but when the wavelength is small compared to the diameter of
the speaker, the wave energy is propagated mostly in the forward direction.)
The SI unit of frequency is ``Hz''.
\begin{quote}
$d$ = Given in $cm$.\\
To find the frequency at which the wavelength equals the diameter, simply
note that $\lambda =d$. The general equation relating frequency and
wavelength is $v=\lambda f$. This is a constant for any medium. To find
the frequency,
\[
f=\frac{v}{\lambda }
\]
Since we are concerned with the sound in the air, the velocity in air is
$343\ m/s$. Remember to perform the conversion from $cm$ to $m$ for the
diameter in this problem. The answer will have units of $s^{-1}$ or $Hz$.
\end{quote}
7) The noise level in an empty examination hall is $41\ dB$. When 100 physics
students are writing an exam, the sounds of heavy breathing and pens traveling
rapidly over paper cause the noise level to rise to $57\ dB$ (not counting the
occasional groans). Assuming that each student contributes and equal amount
of noise power, find the noise level when half of the students have left the
examination hall.
\begin{quote}
$\beta _{i}$ = Given\\
$\beta _{f}$ = Given\\
100 students\\
We must find the sound intensity when the room is empty and when it is full.
Then we can determine how much each student contributes to that intensity.
\[
\beta _{i}=10\log \left( \frac{I_{i}}{I_{0}}\right)
\]
where
\[
I_{0}=10^{-12}.
\]
Looking back to problem 5, the intensity $I_{i}$ is
\[
I_{i}=I_{0}10^{\frac{\beta _{i}}{10}}
\]
Similarly, the final intensity $I_{f}$ is
\[
I_{f}=I_{0}10^{\frac{\beta _{f}}{10}}
\]
The change in intensity is $\Delta I=I_{f}-I_{i}$. Each student contributes
an amount of $\frac{\Delta I}{100}$ to the change in intensity. If half the
students leave, we can multiply the intensity each student creates by 50 to
find the intensity created by half the students. Then we can find the decibel
rating corresponding to that intensity. So,
\begin{eqnarray*}
I_{half} & = & 50\ast \frac{\Delta I}{100}\\
& = & \frac{1}{2}\left( I_{f}-I_{i}\right)\\
& = & \frac{1}{2}\left( I_{0}10^{\frac{\beta _{f}}{10}}-I_{0}10^{\frac{\beta _{i}}{10}}\right)\\
& = & \frac{1}{2}I_{0}\left( 10^{\frac{\beta _{f}}{10}}-10^{\frac{\beta _{i}}{10}}\right)\\
\beta _{half} & = & 10\log \left( \frac{I_{half}}{I_{0}}\right)\\
& = & 10\log \left(\frac{\frac{1}{2}I_{0}\left( 10^{\frac{\beta _{f}}{10}}-10^{\frac{\beta _{i}
}{10}}\right) }{I_{0}}\right)\\
& = & 10\log \left( \frac{1}{2}\left( 10^{\frac{\beta _{f}}{10}}-10^{\frac{\beta _{i}}{10}}\right) \right)
\end{eqnarray*}
\end{quote}
8) The pilot of a light aircraft wishes to reduce significantly engine noise.
If he wears a head set which reduces the noise level by
$\Delta \beta = -26.1\ dB$, by what factor $f$ is the intensity of the noise
attenuated? (The factor $f$ is defined this way: if $I_0$ is the original
noise level, and $I$ is the attenuated noise level, then $I = f\;I_0$.)
\begin{quote}
First of all, attenuate means to lessen in severity, value, intensity, etc.\\
$I=f\ast I_{0}\qquad$ Given\\
$\Delta \beta $ = Given\\
So
\[
f=\frac{I}{I_{0}}=\frac{I_{f}}{I_{i}}
\]
The change of $I$ to $I_{f}$ and of $I_{0}$ to $I_{i}$ was done to avoid
confusion with $I_{0}$, the threshold of human hearing which equals $%
10^{-12} $. We need to find the initial and the final intensity. From
problem 5, the initial intensity $I_{i}$ is
\[
I_{i}=I_{0}10^{\frac{\beta _{i}}{10}}
\]
Similarly, the final intensity $I_{f}$ is
\[
I_{f}=I_{0}10^{\frac{\beta _{f}}{10}}
\]
Since we are looking at the change in intensity, it doesn't matter for what
initial decibel intensity we started with as long as the change is the same.
So we can choose $\beta _{i}$ to be anything and $\beta _{f}-\beta
_{i}=\Delta \beta $, or $\beta _{f}=\beta _{i}+\Delta \beta $. \ If $\Delta
\beta $ is negative, use it in the formula as such. Pick 100 if you are at
a loss for $\beta _{i}$ (or 50 if your calculator cannot handle large
numbers). Then calculate $I_{i}$ and $I_{f}$ and then the ratio $\frac{I_{f}}{I_{i}}$
to get $f$. The answer is unitless.
\end{quote}
9) Hovering over the pit of hell, Satan observes that, although a stationary
physics student shrieks at frequency $f_0 = 905.4\ Hz$, when the student falls
aways into the nether regions, the pitch of his shriek is $f_1 = 883.4\ Hz$.
What is the speed $v$ of descent of the physics student? Assume that the
velocity of sound is $v_{sound} = 340.0\ m/s$.
\begin{quote}
$f_{0}$ = Given (frequency at rest)\\
$f_{1}$ = Given (frequency when moving)\\
$v_{sound}$ = Given\\
$v_{student}=u$ = ?\\
The Doppler effect says that the frequency of a moving object ($f_{1}$) is
\[
f_{1}=\frac{f_{0}}{1-\frac{u}{v}}
\]
where $f_{0}$ is the frequency of the stationary object, $v$ is the wave
speed (in this case, the speed of sound $v_{sound}$), $u$ is the speed of
the source with respect to the medium (in this case, the falling rate of the
student $v_{student}$), and the minus sign in the denominator indicates that
the source is moving away from the observer. To solve for $u$,
\begin{eqnarray*}
\left( 1-\frac{u}{v}\right) f_{1} & = & f_{0}\\
1-\frac{u}{v} & = & \frac{f_{0}}{f_{1}}\\
1-\frac{f_{0}}{f_{1}} & = & \frac{u}{v}\\
u & = & v\left( 1-\frac{f_{0}}{f_{1}}\right)
\end{eqnarray*}
Using more symbolic names,
\[
v_{student}=v_{sound}\left( 1-\frac{f_{0}}{f_{1}}\right)
\]
The units of velocity are $m/s$ and CAPA is looking for the speed, so take
the absolute value of your result.
\end{quote}
10) A listener is a distance $r = 20.0\ m$ from a small explosion and reports
a sound of intensity $72.0\ dB$. What is the distance $R$ from the explosion
at which listeners hear an intensity of $52.0\ dB$?
\begin{quote}
$r$ = Given\\
$\beta _{i}$ = Given\\
$\beta _{f}$ = Given\\
$R$ = ?\\
We are given the intensity in decibels, but we can find the intensity $I$ in
$W/m^{2}$ (power/area) and proceed from there. Since the power of the loud
speaker does not change, it will connect the first and second parts. Again,
from problem 5, the initial intensity $I_{i}$ is
\[
I_{i}=I_{0}10^{\frac{\beta _{i}}{10}}
\]
Similarly, the final intensity $I_{f}$ is
\begin{eqnarray*}
I_{f} & = & I_{0}10^{\frac{\beta _{f}}{10}}\\
I_{i} & = & \frac{P}{A}
\end{eqnarray*}
Sound waves are emitted spherically, so the area is $\frac{4}{3}\pi r^{2}$.
This allows us to solve for the power.
\begin{eqnarray*}
P & = & A\ast I_{i}\\
P & = & \frac{4}{3}\pi r^{2}I_{i}
\end{eqnarray*}
Similarly,
\begin{eqnarray*}
\frac{P}{A} & = & I_{f}\\
\frac{P}{\frac{4}{3}\pi R^{2}} & = & I_{f}
\end{eqnarray*}
Solving for $R$,
\begin{eqnarray*}
P & = & I_{f}\frac{4}{3}\pi R^{2}\\
\frac{3}{4}\frac{P}{I_{f}\pi } & = & R^{2}\\
R & = & \sqrt{\frac{3}{4}\frac{P}{I_{f}\pi }}
\end{eqnarray*}
But
\[
P=\frac{4}{3}\pi r^{2}\ast I_{i}
\]
so
\[
R=\sqrt{\frac{3}{4}\frac{\frac{4}{3}\pi r^{2}I_{i}}{I_{f}\ast \pi }}=\sqrt{
\frac{r^{2}I_{i}}{I_{f}}}
\]
\end{quote}
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