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A flute is designed so that it plays a frequency of 261.6 Hz,middle C, when all the holes are covered and the temperature is20.0 C.

(a) consider the flute to be a pipe open at both ends, andfind its length, assuming that the middle-C frequency is thefundametal frequency.

(b) A second player, nearby in a colder roomalso attempts to play middle C on an identical flute. A beatfrequency of 3.00 beats/s is heard. What is the temperatureof the room?

Respuesta :

cjmejiab

Our values are given as,

[tex]f_1 = 261.6Hz[/tex]

[tex]T = 20.0\°C = 293K[/tex]

[tex]f_{beat} = 3.00Hz[/tex]

PART A) To get the speed of sound we use the equation

[tex]v = (331 m/s)\sqrt{\frac{T}{273K}}[/tex]

So the speed of sound at [tex]20.0\°C[/tex] will be,

[tex]v = (331m/s)\sqrt{\frac{293K}{273K}}[/tex]

[tex]v = 343m/s[/tex]

We know that for an open pipe the first harmonic or the fundamental mode will be

[tex]\lambda_1 = 2L[/tex]

[tex]\lambda_1 = \frac{v}{f_1}[/tex]

[tex]\lambda_1 = (\frac{343m/s}{261.6Hz})[/tex]

[tex]\lambda_1 = 1.31m[/tex]

So the length of the flute will be

[tex]L = \frac{\lambda_1}{2}[/tex]

[tex]L = \frac{1.31m}{2}[/tex]

[tex]L = 0.65m[/tex]

[tex]L = 65cm[/tex]

PART B) As the person in the colder room also attempts to play middle C the fundamental wavelength is unchanged so that,

[tex]\lambda_1' = \lambda_1[/tex]

[tex]\lambda_1' = 1.31m[/tex]

So from the second player in the colder room the frequency will be

[tex]f_1' = f_1-f_{beat}[/tex]

[tex]f_1' = 261.6Hz - 3.00 Hz[/tex]

[tex]f_1' = 258Hz[/tex]

So the speed of sound in this room will be

[tex]v' = \lambda_1' f_1'[/tex]

[tex]v' = (1.31m)(258.6Hz)[/tex]

[tex]v' = 339m/s[/tex]

So we get the temperature in the room from

[tex]v = (331m/s)\sqrt{\frac{T}{273K}}[/tex]

Rearranging to find the temperature

[tex]T = (273K)(\frac{339}{331})^2[/tex]

[tex]T = 286K[/tex]

[tex]T = 13.0\°C[/tex]

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