Doppler Cents Puzzler #2

[Amy Zilk]

LOL!!! 
You would want to shove the piano directly at the conductor at 3.4 mph. 
Would that sufficiently flatten the conductor? 
Amy

ed440 at mindspring.com wrote:
> Amy-
> Excellent reply.
> Now, please, can you tell us: If the conductor wants the piano at 
> 442hz and I don't want to raise pitch at the pin block, how fast do I 
> have to push the piano across the stage?
> Mathematically challenged folks want to know.
> Ed Sutton
>
>
>     -----Original Message-----
>     From: Amy Zilk
>     Sent: Jun 17, 2006 9:18 AM
>     To: schecter at pacbell.net, Pianotech List
>     Subject: Re: Doppler Cents Puzzler
>
>     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
>>>
>>>     __________________________________________________
>>>     Do You Yahoo!?
>>>     Tired of spam?  Yahoo! Mail has the best spam protection around
>>>     http://mail.yahoo.com
>>
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