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A transverse wave in a medium is described by the equationy = A sin 2(omegat - kx) . The magntiude of the maximum velocity of particles in the medium is equalthat of theof the wave velocity . If the value of A is |
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Answer» ` lambda/(2pi)` Velocityof the particle= ` (dy)/(dt) = 2 A omegacos 2 (omegat - kx) ` Maximumvelocity` = 2 A OMEGA` Velocityof the wave = ` ("COEFFICIENT of t ")/( "Coefficient of x ") = (2 Omega)/(2kg) = Omega/k` As per QUESTION ` 2 A omega = omega/k or 2A = 1/k = lamda/(2pi) RightarrowA = lamda/(4pi)` |
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