Doppler Cents Puzzler #2

[John M. Formsma]

ROTFLOL !!!

 

JF

 

  _____  

From: pianotech-bounces at ptg.org [mailto:pianotech-bounces at ptg.org] On Behalf
Of Amy Zilk
Sent: Saturday, June 17, 2006 10:00 AM
To: Pianotech List
Subject: Re: Doppler Cents Puzzler #2

 

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
fo 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/fo - 1 = V/Vs

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

 f/fo = 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


 
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Tech%5CAZP-email-sig.gif> 



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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