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Give explanation of stationary waves produced in closed pipe and obtain equations of natural frequency (normal modes). |
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Answer» Solution :Air column (glass tube) partially filled with water is example of closed pipe. The end in contanct with water is node and OPEN end is antinode. At nodes, pressure changes are maximum but displacement is minimum (ZERO). At antinodes, pressure changes are minimum but displacement is maximum If we take `X =0` at node and `x =L` at antinode, for ends, then `|sin kx| =1` `therefore |sinkL|=1 [because x =L]` `therefore kL = (n + (1)/(2) ) PI` `therefore (2pi )/(lamda ) L = (n + (1)/(2)) pi` `thereforeL = ( n + (1)/(2) ) (lamda )/(2)` `therefore L = (2n +1) (lamda)/(4)` where `n = 0,1,2,...,n` and possible wavelengths, `lamda = (2L)/((n + (1)/(2)) )= ( 4L)/( 2n +1 )` where `n = 0,1,2,3,...` Normal modes (Natural frequencies of system), `v _(n) = (n + (1)/(2)) (v)/(2L) = (2n + 1) (v)/(4L)` where ` n = 0,1,2,3..` By taking `n =0` for fundamental frequency, `v _(1) = (v)/(4L)` and odd number harmonic gives HIGHER frequencies. For example, `3 (v)/(4L), 5 (v)/(4L),...,(2n +1) (v)/(2L) ` are obtained. Thus only odd harmonic are obtained in closed pipe.   The above six figures shows the first 6 harmoniccs of air column in closed pipe. |
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