Question

An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat frequency is generated by the first overtone of the closed pipe with the fundamental of the open pipe

Answer 1

fb = 240.35 Hz

Step-by-step explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

`f=(v_s)/(2L)` (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

`f=(343m/s)/(2(0.47228m))=362.36Hz`

Closed tube:

`f'=(v_s)/(4L')`

L': length of the closed tube = 0.702821m

`f'=(343m/s)/(4(0.702821m))=122.00Hz`

Next, you use the following formula for the beat frequency:

`f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz`

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz