Q64P

Question

The frequency of the note F4 is 349 Hz. (a) If an organ pipe is open at one end and closed at the other, what length must it have for its fundamental mode to produce this note at 20.0°C? (b) At what air temperature will the frequency be 370 Hz, corresponding to a rise in pitch from F to F-sharp? (Ignore the change in length of the pipe due to the temperature change.)

Step-by-Step Solution

Verified
Answer

The length of the pipe for the given fundamental note is 0.246 m b) The air temperature corresponding the pitch is 35.50C

1STEP 1 Concept of the length of the pipe and the air temperature

The length of the pipe is given as L=v4f1 were. f1 is the fundamental frequency of an organ pipe, V is the speed of the sound, L is length of the pipe.

The new speed of the sound for the original temperature is v'=4fL were, v' is the new speed of the sound for the original temperature, F is the frequency.

2STEP 2 Calculate the length of the pipe

The fundamental frequency of an organ pipe is 349 Hz and the speed of the sound is 344 Substitute 349 Hz for  f1 and 344ms-1 for v to find the L

L=344×4×349=0.246m

Thus, the length of the pipe for the given fundamental note is 0.246 m

3STEP 3 Calculate the air temperature

The frequency of fundamental note is 370 Hz, The new speed of the sound for the original temperature is v'=4fL Substitute 370 Hz for f and 0.246 m for L to find the v' 

v'=4(370)(0.246)=364.08ms-1

Formula to calculate air temperature corresponding to the pitch is 

Substitute 364.08 T=v'-v0.6ms-1  for   v' and 344 ms-1for v to find T

T=364.08-344   =35.50C

Therefore, the air temperature corresponding the pitch is 35.50C