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<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>Wow.<span style='mso-spacerun:yes'> </span>I see this
topic (and A442 conductors—LOL) have been batted around all weekend. <span
style='mso-spacerun:yes'> </span>Some thoughts:<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>We don’t have enough data to determine the velocity. <span
style='mso-spacerun:yes'> </span>However, if we make some assumptions, we
can make a fair guess. <span style='mso-spacerun:yes'> </span>So, let’s
assume that conditions are right to give a standard speed of sound of 343 m/s. <span
style='mso-spacerun:yes'> </span>Let us further assume that the pitch
measured is A440.<span style='mso-spacerun:yes'> </span>And, one more
assumption:<span style='mso-spacerun:yes'> </span>both measurements are
taken at a distance of 5 meters from the truck.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>Okay, the Doppler formula is <span class=SpellE>fsubd</span>=<span
class=SpellE><span class=GramE>fsube</span></span><span class=GramE>(</span>1+v/u),
where <span class=SpellE>fsubd</span> is the observed frequency, <span
class=SpellE>fsube</span> is the emitted frequency, v is the observers
velocity, and u is the speed of the sound waves, when the observer is moving
toward the source. <span style='mso-spacerun:yes'> </span>When the
observer is moving away from the source, the sign of the “v/u” term
is negative.<span style='mso-spacerun:yes'> </span>A bit of mathematical gymnastics
gives ((<span class=SpellE>fsubd/fsube</span>)-1)*343=v<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>So, when you’re cycling toward the truck, your
velocity is 2.081 m/s, or 4.65 mph.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>When you’re cycling away from the truck, your velocity
is 1.925 m/s, or 4.31 mph.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>Your velocity has decreased by 0.156 m/s over 10 meters, so
you are slowing at a rate of 0.0156 m/s.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>The reason you are slowing, under these conditions, is that
10 cents sharp does not equal 10 cents flat!!<span
style='mso-spacerun:yes'> </span><o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>G#4= 440/1.0595=415.29<span class=GramE>;<span
style='mso-spacerun:yes'> </span>440</span>-415.29=24.71; 24.71/10=2.471
which is the frequency difference 10 cents flat of A440.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>A#4= 440* 1.0595=466.18<span class=GramE>;<span
style='mso-spacerun:yes'> </span>466.18</span>-440=26.18;<span
style='mso-spacerun:yes'> </span>26.18/10= 2.618, frequency difference 10
cents sharp of A440.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>If you had measured A220, instead, your respective
velocities would be 2.042 m/s, 1.918 m/s, and the rate of decrease in velocity
would be 0.0124 m/s.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><span class=GramE><font size=2 face=Arial><span
style='font-size:10.0pt;font-family:Arial'>New puzzle?</span></font></span><font
size=2 face=Arial><span style='font-size:10.0pt;font-family:Arial'><span
style='mso-spacerun:yes'> </span>How far would our 95 pound,
pencil-necked wisp of an A442 conductor fly if you shoved the Steinway D at him
with an acceleration of 2 m/s/s from 2 meters away?<span
style='mso-spacerun:yes'> </span>Assume standard conditions.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>John “Doomed to think like a Chemist forever”
Delmore<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'><o:p> </o:p></span></font></p>
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