The first obertone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fudamental frequency of clos3ed organ pipe is 110 Hz. Find length of the open pipe. (Given, sound in air =330 m/s )

Explanation

Let the lengths of open and closed pipes be `l_(1)` and `l_(2)` respectively. We know that
First overtone of open organ pipe, `n_(1) = (v)/(l_(1))`
first overtone of closed organ pipe,
`n_(2) = (3v)/(4l_2)`
fundamental frequency ofclosed organ pipe `n = (v)/(4l_2)`
According to the given question, as n = 110 Hz, we have
`110 = (v)/(4l_(2)) = (330)/(4l_(2))`
`l_(2) = (330)/(4 xx 110)` = 0.75 m
Further, it is given that beat frequency is 2.2 Hz, we have
`f_(B) = 2.2 = (v)/(l_(1)) - (3v)/(4l_(2))`
`2.2 = (2xx330)/(2l_(1)) - (3 xx 330)/(4 xx 0.75)`
`2.2 = (330)/(l_1) - 330`
`332.2 = (330)/(l_1)`
`l_(1) = (330)/(332.2)` = 0.993m Again, Beat frequency can also be given as
`f_(B) = 2.2 = (3v)/(4l_(2)) - (v)/(l_(1))`
`2.2 = (3xx 330)/(4 xx 0.75) - (2xx 330)/(2l_(1))`
`2.2 = 330 - (330)/(l_(1))`
`l_(1) = 1.006m`