InterviewSolution
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InterviewSolution
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In an organ pipe ( may be closed or open of `99 cm` length standing wave is setup , whose equation is given by longitudinal displacement `xi = (0.1 mm) cos ( 2pi)/( 0.8) ( y + 1 cm) cos 2 pi (400) t` where `y` is measured from the top of the tube in metres and `t` in second. Here `1 cm` is th end correction. Equation of the standing wave in terms of excess pressure is - (Bulk modulus of air `B = 5 xx 10^(5) N//m^(2)`)A. `P_(ex) = (125 pN//m^(2)) sin ( 2pi)/(0.8) ( y + 1 cm) cos 2 pi (400 t)`B. `P_(ex) = (125 pN//m^(2)) cos ( 2pi)/(0.8) ( y + 1 cm) sin 2 pi (400 t)`C. `P_(ex) = (225 pN//m^(2)) sin ( 2pi)/(0.8) ( y + 1 cm) cos 2 pi (200 t)`D. `P_(ex) = (225 pN//m^(2)) cos ( 2pi)/(0.8) ( y + 1 cm) sin 2 pi (200 t)` |
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Answer» Correct Answer - A `P_(ex) = -B( d xi)/( d x)` `= ( 5 xx 10^(5)) xx (0.1 xx 10^(-3)) ( 2pi)/(0.8) sin ( 2pi)/( 0.8)` `(y + 1 cm) cos 2 p(400) t` `= (125 pi N//m^(2)) sin ( 2pi)/(0.8) ( y + 1 cm) cos 2 pi (400 t)` |
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