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e-lub [12.9K]
3 years ago
15

A flutist assembles her flute in a room where the speed of sound is 342 m/s. When she plays the note A, it is in perfect tune wi

th a 440 Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is 346 m/s.
-got A: How many beats per second will she hear if she now plays the note A as the tuning fork is sounded? 5.15 Beats/s

-now gotta find B:
How far does she need to extend the "tuning joint" of her flute to be in tune with the tuning fork? (answer is d=_____mm)
Physics
1 answer:
sertanlavr [38]3 years ago
7 0

Answer:

5.15348 Beats/s

4.55 mm

Explanation:

v_1 = Velocity of sound = 342 m/s

v_2 = Velocity of sound = 346 m/s

f_1 = First frequency = 440 Hz

Frequency is given by

f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz

Beat frequency is given by

|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s

Beat frequency is 5.15348 Hz

Wavelength is given by

\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m

Relation between length of the flute and wavelength is

\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m

At v = 346 m/s

\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m

L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m

Difference in length is

\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm

It extends to 4.55 mm

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