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Vladen's answer is the one I get. <br>
<p class="MsoNormal">f is the frequency of the source<br>
f<sub>o</sub> is the frequency heard by the observer<br>
c is the difference in cents (10)<br>
Vs is speed of sound in air (1100 ft/s)<br>
V is speed of observer (speed of the guy on bike -- the answer)</p>
<p class="MsoNormal">Doppler shift equation for stationary sound
source; observer
moving directly to or from the source:</p>
<p class="MsoNormal"><o:p> </o:p> f/f<sub>o</sub> – 1 = V/Vs</p>
<p class="MsoNormal"><o:p></o:p>You have to get f/f<sub>o</sub> from
the 10 cents
difference.</p>
<p class="MsoNormal"><o:p> </o:p>f/f<sub>o</sub> = 2^(c/1200) = 1.005793</p>
<p class="MsoNormal"><o:p></o:p>Plug it in and get</p>
<p class="MsoNormal"><o:p></o:p>1.005793 – 1 = V/Vs</p>
<p class="MsoNormal"><o:p> </o:p>0.005793 = V/Vs</p>
<p class="MsoNormal"><o:p> </o:p>0.005793 * Vs = V</p>
<p class="MsoNormal"><o:p> </o:p>V = 0.005793 * (1100 ft/s) = <span
style=""> </span>6.3723 ft/s <br>
</p>
<p class="MsoNormal">6.3723 ft/s (3600s/hr) / (5280 ft/mi ) = 4.3
mph<br>
</p>
<p class="MsoNormal">az<br>
</p>
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<br>
<br>
Mark Schecter wrote:
<blockquote cite="mid44939A42.4000904@pacbell.net" type="cite">Hi,
Vladan.
<br>
<br>
Well, your number and mine don't agree, and I'm not at all sure of
mine. So I'm going to show how I got to my result, and if I'm wrong,
I'd be delighted to know how. So here goes.
<br>
<br>
The fact that the tone goes flat 10 cents when going away merely
confirms that the difference between the stopped truck and the moving
cycle produces a 10 cent differential in pitch. So I considered the
pitch coming from the stopped truck to be 1, and the sound to be
travelling at 1100 feet per second. In order to reach a pitch of 2, the
cycle would have to be moving at the speed of sound toward the truck,
to achieve a total of 2200 feet per second closing speed. With that
thought in mind, I just calculated that a 10 cent increase in pitch
equalled 10/1200 of the speed of sound, so:
<br>
<br>
10 cents higher than nominal pitch =
<br>
10/1200 * (speed of sound in air)
<br>
or 1/120 * (1100 ft/sec) = 9.1666 ft/sec (speed of bicycle)
<br>
9.166 ft/sec * 3600 secs/hour = 33,000 ft/hour
<br>
33,000 / 5280 (ft/mi) = 6.25 mph
<br>
<br>
However, you arrived at 2 meters/second, which equals 7200 meters/hour,
which translates to 4.47 miles per hour. So would you tell me how you
got there? Thanks!
<br>
<br>
-Mark Schecter
<br>
<br>
V T wrote:
<br>
<blockquote type="cite">2 meters/second; I would have stopped for
some ice
<br>
cream. <br>
Vladan
<br>
<br>
=====================
<br>
I was out riding my bicycle this calm quiet evening
<br>
when I happened upon an ice cream truck playing music
<br>
to attract customers. The truck had stopped to
<br>
dispense ice cream, but the music continued. Since I
<br>
always carry my ETD when I ride my bike, I quickly
<br>
measured the pitch of a recurring note in the music
<br>
and found it to be 10 cents sharp as I was riding
<br>
straight towards the truck. Then after I passed the
<br>
truck, I measured the pitch again and found it to be
<br>
10 cents flat as I was riding directly away from it. How fast was I
riding my bicycle?
<br>
<br>
Robert Scott
<br>
Ypsilanti, MI
<br>
<br>
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</blockquote>
<br>
</blockquote>
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