Question

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 1

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 2

Answer:

we will get 440 beats per second

Explanation:

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

frequency of the note played, $f_i=440\ Hz$

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

Initial wavelength:

$\lambda_i=\frac{v_i}{f_i}$

$\lambda_i=\frac{340}{440}$

$\lambda_i=0.773\ m$

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:

$2L=\lambda_i$

$2\times L=0.773$

$L\approx0.3864\ m$

So the new frequency:

$f_f=\frac{v_f}{\lambda_i}$

$f_f=\frac{343}{0.773}$

$f_f=443.88\ Hz$

we will get 440 beats per second