Physics, asked by Pavansm, 1 year ago

consider 2 cylindrical pipes of equal length. one of these acts as a closed organ pipe and the other as open pipe . the frequency of the third harmonic in the closed pipe is 200 Hz higher than the first harmonic of the open pipe. calvulate the fundamental frequency of the closed pipe.

Answers

Answered by duragpalsingh
3
We know,
L_i = L_{ii} = L
v_3^i  = 200+v_1^{ii}
v_1^i = ?

Closed Pipe:

L =  \frac{n\lambda}{4} =  \frac{nc}{4v^{ii}_n}, v^{ii}_n =  \frac{nc}{4L}

∴ Open pipe:

 \frac{n\lambda}{2} = \frac{nc}{2v^{ii}_n}, v^{ii}_n = \frac{nc}{2L}

 \frac{3\cdotp340}{4L} = 200 +  \frac{340}{2L}

L = 0.425m

v_1^i =  \frac{340}{4L} = 200 Hz



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