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The equation of a transverse wave is given by y = 0.05 sin pi (2t - 0.02 x) , where x,y are in metre and t is in second. The minimum distance of separation between two particles which are in phase and the wave velocity are respectively |
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Answer» 50m 50 `ma^(-1)` y = 0.05 sin `(2pi t - 0.02 pi x)` COMPARING it with standard equation y = r sin `(omega t - KX) ` We have `omega= 2pi and K = 0.02 pi` and k = `(2pi)/(lambda) RARR -.01 pi= (2pi)/(lambda) ` `lambda = (2pi)/(0.02 pi) = 100 ` m so CORRECT choice is (b). |
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