Two successive resonant frequencies in an open organ pipe are 1944 Hz

1.	Two successive resonant frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air =324 m/s.
Answer» The resonant frequencies in an open organ pipe is given as νₙ = nV/2L Let the given frequencies are for n and n+1 harmonics, thus nV/2L = 1944 and (n+1)V/2L = 2592 →nV/2L + V/2L =2592 →1944+V/2L = 2592 →V/2L = 2592 - 1944 = 648 →L = V/(2648) = 324/1296 m = 0.25 m →L = 25 cm*

Two successive resonant frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air =324 m/s.

Answer»

The resonant frequencies in an open organ pipe is given as

νₙ = nV/2L

Let the given frequencies are for n and n+1 harmonics, thus

nV/2L = 1944 and

(n+1)V/2L = 2592

→nV/2L + V/2L =2592

→1944+V/2L = 2592

→V/2L = 2592 - 1944 = 648

→L = V/(2*648) = 324/1296 m = 0.25 m

→L = 25 cm

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Two successive resonant frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air =324 m/s.

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