Doppler Cents Puzzler

[Amy Zilk]

Vladen's answer is the one I get.

f is the frequency of the source
f_o is the frequency heard by the observer
c is the difference in cents (10)
Vs is speed of sound in air (1100 ft/s)
V is speed of observer (speed of the guy on bike -- the answer)

Doppler shift equation for stationary sound source; observer moving 
directly to or from the source:

     f/f_o -- 1 = V/Vs

You have to get f/f_o from the 10 cents difference.

 f/f_o = 2^(c/1200) = 1.005793

Plug it in and get

1.005793 -- 1 = V/Vs

 0.005793 = V/Vs

 0.005793 * Vs = V

 V = 0.005793 * (1100 ft/s) =  6.3723 ft/s 

6.3723 ft/s (3600s/hr) / (5280 ft/mi ) = 4.3 mph

az




Mark Schecter wrote:
> 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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