The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency 2.2 Hz . The fundamental frequency of the closed organ pipe is 110 Hz, find the lengths of the pipes . Take velocity of sound= 330 m//s.

The first overtone of an open organ pipe beats with the first overtone

1.	The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency 2.2 Hz . The fundamental frequency of the closed organ pipe is 110 Hz, find the lengths of the pipes . Take velocity of sound= 330 m//s.
Answer» SOLUTION :Let `l_(1) and l_(2)` be lengths of open organ pipe and closed organ pipe respectively . FIRST overtone of open organ pipe ` = 2n_(1) = 2 xx (v)/( 2 l_(1)) = (v)/( l_(1))` First overtone of closed organ pipe ` = 3 n_(2) = 3 xx (v)/( 4 l_(2))` According to question , `(v)/(l_(1)) - ( 3v)/( 4 l_(2) = +- 2.2`(i) As `n_(2)` is the FUNDAMENTAL frequency of closed organ pipe `n_(2) = (v)/( 4 l_(2))` `l_(2) = (v)/( 4 n_(2)) = (330)/( 4 xx 110) = 0.75 m` From Eq.(i). `(v)/(l_(1)) = ( 3v)/( 4 l_(2)) +- 2.2` `(v)/(l_(1)) = 3 n_(2) +- 2.2` Taking positive SIGN `(v)/(l_(1)) = 3 xx 110 + 2.2 = 332.2` `:. l_(1) = (v)/( 332.2) = (330)/(332.2) m = 0.993 m = 99.3 cm` Taking negative sign `(v)/( l_(1)) = 3 n_(2) - 2.2 = 3 xx 110 - 2.2 = 327.8` `:. l_(1) = (v)/( 327.8) = (330)/(327.8) = 1.006 m = 100.6 cm`