The air in a pipe closed at one end is made to vibrate in its second overtone by a tuning fork of frequency 440 Hz. The speed of sound in a 330 m//s. End correction may be neglected . Let P_(0) denotes the mean pressure of any point in the pipe and Delta P_(0) the maximum amplitudes of pressure variation. a. Find the length L of the air column. b. What is the amplitude of pressure variation at the middle of the column c. What are maximum and minimum pressures at the open end of the pipe ?

The air in a pipe closed at one end is made to vibrate in its second o

1.	The air in a pipe closed at one end is made to vibrate in its second overtone by a tuning fork of frequency 440 Hz. The speed of sound in a 330 m//s. End correction may be neglected . Let P_(0) denotes the mean pressure of any point in the pipe and Delta P_(0) the maximum amplitudes of pressure variation. a. Find the length L of the air column. b. What is the amplitude of pressure variation at the middle of the column c. What are maximum and minimum pressures at the open end of the pipe ?
Answer» Solution :a. The fundamental frequency of the closed organ pipe ` = v// 4l`. In closed organ pipe only odd harmonics are present . SECOND overtone of pipe ` = 5 v// 4L` Given ` 5v//4L = 440` On solving , we get `L = ( 5 v)/( 4 XX 440) = ( 5 xx 330)/( 4 xx 440) = (15)/( 16) m = 0.9375 m = 93.75 cm` b. The equation of variation of pressure amplitude at any distance `x` from the node is `Delta P = Delta P_(0) cos kx` Pressure variation is maximum at a node and minimum (zero) at antinode. Distance of centre `C from N_(2) is LAMBDA//8)` `:. DeltaP = Delta P_(0) cos ( 2pi)/( lambda) xx (lambda)/( 8) = DeltaP_(0) (pi)/(4) = ( Delta P_(0))/( sqrt(2))` c. At antinode , the pressure variation is minimum (zero) , therefore at antinode remains equal to `P_(0)` (always). Therefore , at antinode `P_(MAX) = P_(min) = P_(0)`.