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 with 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)

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boffeemadrid boffeemadrid · Answer 1 · 2019-12-12T06:47:35+01:00

Answer:

5.15348 Beats/s

4.55 mm

Explanation:

[tex]v_1[/tex] = Velocity of sound = 342 m/s

[tex]v_2[/tex] = Velocity of sound = 346 m/s

[tex]f_1[/tex] = First frequency = 440 Hz

Frequency is given by

[tex]f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz[/tex]

Beat frequency is given by

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

Beat frequency is 5.15348 Hz

Wavelength is given by

[tex]\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m[/tex]

Relation between length of the flute and wavelength is

[tex]\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m[/tex]

At v = 346 m/s

[tex]\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m[/tex]

[tex]L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m[/tex]

Difference in length is

[tex]\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm[/tex]

It extends to 4.55 mm