The lowest frequecy note of an organ is 16.4 Hz. What is the length of the shortest open organ pipe that will resonate at this frequency?

Answer 1

Scroll down to related links and look for "Closed or Gedackt Organ Pipes - Wikipedia". Organ builders still tend to use feet rather than metres when dealing with organ pipe length and tone e.g. an 8 foot pitched pipe is the same pitch as a piano, and is called '8 foot' because the longest pipe of the rank at this pitch ( the C below the C below Middle C) is 8 feet long. A 4-foot pipe is an octave higher, a 2 foot 2 octaves higher, and a 16 foot rank an octave lower than piano pitch. The fundamental tone of 16.4 Hz represents a 32-foot organ pipe, which is found, except in the very largest organs, only in the pedal section of large instruments. Although the 32 foot pitch is found frequently on large instruments (like cathedral organs) it is not the deepest note of an organ as stated in the question. Some organs, such as the Atlantic City Auditorium organ, USA and Liverpool Cathedral Organ in the UK have 64-foot ranks giving the lowest note as 8.2 Hz. On most organs offering 64-foot ranks, the sound is either produced by a stopped 32-foot pipe, or acoustically, where two shorter pipes are tuned so that the beats between them produce a 64-foot tone. Many huge organs do have a 64' rank (usually called gravissima), but nearly all of them are produced by either a stopped 32' or acoustically. There are not more then 5 organs that has a true (not acoustic, stopped, nor digital), (only counting the rank(S) that accually goes down to the sub-sub-contra C) 64 foot rank, three of them are the organ at the Atlantic City Convention Hall Main Auditorium, the organ at Sydney City Hall, and the one at Worcester Cathedral, in UK. The shortest lengh that will produce 16.4Hz (CCCC) is 16 ft, although this note is the C of the sub-contra octave, this 16ft pipe is stopped on the top, so it produces the note that corresponds to twice that lengh(32ft) . Also should be mentioned, the lowest (true) note on an acual organ is CCCCC which is 8 Hz, that single note cannot be heard alone. THE lowest note is produced by either stopping the 64' or combining the 64' and the fifth(42 2/3) to produce a 128' CCCCCC which is 4 Hz. Ｔhe　lowest note is produced by combinding a stopped 64' ( 128') and stopped 42 2/3' (85 1/3') to produce a resultant 256' which is 2Hz on CCC CCC C, this note is not considered audible by the human ear

Answer 2

The formula to calculate the length of the shortest open organ pipe that will resonate at a specific frequency is: Length = (v / 4f), where v is the speed of sound in air (approximately 343 m/s). Plugging in the values, the length of the shortest pipe would be around 5.22 meters.