Answered

A flutist assembles her flute in a room where the speed of sound is 340 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 343 m/s.

How many beats per second will she hear if she now plays the note A as the tuning fork is sounded?

Answer :

Answer:

Explanation:

Beat results due to difference in the frequency of tuning fork and sound produced by the flute at higher temperature.

frequency of tuning fork = 440 Hz

apparent frequency of flute at higher temperature = 440 x (343 / 340)

= 444 Hz (approx )

No of beats heard = 444 - 440 = 4 beats per second.  

Answer:

we will get 440 beats per second

Explanation:

initial speed of sound in the flute, [tex]v_i=340\ m.s^{-1}[/tex]

frequency of the note played, [tex]f_i=440\ Hz[/tex]

speed of the sound after the air gets warmer, [tex]v_f=343\ m.s^{-1}[/tex]

Initial wavelength:

[tex]\lambda_i=\frac{v_i}{f_i}[/tex]

[tex]\lambda_i=\frac{340}{440}[/tex]

[tex]\lambda_i=0.773\ m[/tex]

We know that the wavelength of sound inside a tube with both the ends open is twice the length of the tube.

So,the length of the tube:

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

[tex]2\times L=0.773[/tex]

[tex]L\approx0.3864\ m[/tex]

So the new frequency:

[tex]f_f=\frac{v_f}{\lambda_i}[/tex]

[tex]f_f=\frac{343}{0.773}[/tex]

[tex]f_f=443.88\ Hz[/tex]

we will get 440 beats per second