Doppler Cents Puzzler

[Mark Schecter]

Hi, Vladan.

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.

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:

10 cents higher than nominal pitch =
10/1200 * (speed of sound in air)
or 1/120 * (1100 ft/sec) = 9.1666 ft/sec (speed of bicycle)
9.166 ft/sec * 3600 secs/hour = 33,000 ft/hour
33,000 / 5280 (ft/mi) = 6.25 mph

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!

-Mark Schecter

V T wrote:
> 2 meters/second; I would have stopped for some ice
> cream. 
> 
> Vladan
> 
> =====================
> I was out riding my bicycle this calm quiet evening
> when I happened upon an ice cream truck playing music
> to attract customers.  The truck had stopped to
> dispense ice cream, but the music continued.  Since I
> always carry my ETD when I ride my bike, I quickly
> measured the pitch of a recurring note in the music
> and found it to be 10 cents sharp as I was riding
> straight towards the truck.  Then after I passed the
> truck, I measured the pitch again and found it to be
> 10 cents flat as I was riding directly away from it. 
> How fast was I riding my bicycle?
> 
> Robert Scott
> Ypsilanti, MI
> 
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