Question

Two pipes, equal in length, are each open at one end. Each has a fundamental frequency of 479 Hz at 299 K. In one pipe the air temperature is increased to 306 K. If the two pipes are sounded together, what beat frequency results?

Answer 1

The speed of the sound at 299 K or 26 ^oC is

v = 347 m/s

The pipe which open at one end and closed at other end has a fundamental frequency euqla to

f = v / 4L

Here, length of the pipe is L.

From the above equation,

L = v/4f

Substitute 347 m/s for v and 479 Hz for f in the above equation,

L = v/4f

   = 347 m/s / 4 (479 Hz)

   = 0.181 m

This gives the length of the pipes which open at one end.

Now, the speed of the sound at 306 K or 33 ^oC is

v₃₀₆ = 351 m/s

The fundamental frequency of the other pipe at 306 K is  

f ₃₀₆= v₃₀₆ / 4L

       = 351 m/s / 4 (0.181 m)

       = 485 Hz

The beat frequency is equal to the difference in the frequency of the two fundamental frequencies produced by the two pipes at different temperatures.

Thus,

Beat frequency = f₃₀₆ - f                                                      

                        = 485 Hz - 479 Hz

= 6 Hz

Therefore, the beat frequency result is 6 Hz.