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You are right that the bicyclist slows down after he passes the ice
cream truck, but not because ten cents flat isn't the same as ten cents
sharp. It's because the bicyclist is traveling the same direction as
the sound waves away from the truck and in the opposite direction of
the sound waves toward the truck. Ten cents flat is the same frequency
ratio as ten cents sharp. <br>
az<br>
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V T wrote:
<blockquote
cite="mid20060622202909.96009.qmail@web30414.mail.mud.yahoo.com"
type="cite">
<pre wrap="">Hi Mark,
It is a good puzzler because it "tests" for three
pieces of knowledge:
1. The definition of the "cent"
2. The understanding of the Doppler effect, and
3. A subtle hitch that happens when using cents
instead of Hertz with the Doppler formula
Item three may be what Robert was trying to bring up
for discussion. When we use the Doppler formula with
frequencies given in Hz, for constant bicycle velocity
the pitch is sharp by the same number of Hertz when
approaching as it is flat when retreating from the ice
cream truck.
However - as you mentioned in a previous post - when
working in cents there is a slight difference in
velocity between approaching and retreating, since we
have the rule that the sharp/flat amount is +/- 10
cents.
In fact, in our puzzler there is a 0.579% difference
in speed between approach and retreat. That's about
41.7 m/hr difference. So he did actually experience a
sort of a sudden but small "ice cream pull" as he
passed. I jokingly said "he didn't even slow down" -
not true, he did!
Here is all the math needed for the puzzler:
f'=frequency heard by moving Robert (Hz)
f=frequency emitted by ice cream truck (Hz)
v=bicycle speed (positive for approaching, negative
for retreating)
c=speed of sound, about 345 m/sec
f'=f(1+v/c) (Doppler's formula)
You can rearrange this:
f'/f=1+v/c
Note now that f'/f is equal to our ratio of 1.00579
(10 cents), or 2^(1/120)
Vladan
========================================
Mark wrote:
Thanks for your thorough explanation. This puzzler
continues to
fascinate, in part because I feel close, but no cigar,
to having enough
math to figure it out myself. I do understand your
point about 2^1/12
and 2^1/1200. Out of context, I knew that. Where I get
into trouble is
connecting a mathematical representation to a
"real-world" problem. I
appreciate your (and Amy Z's and John D's) abilities
in that regard.
Sorry it took me so long to reply.
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</pre>
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