1is it that the displaement is in y direction....so work done will b in y direction and hence the power transmitted will also be in y direction...
hey ppl./..wen we calculate the rate of propagation of energy(ie,power) across a moving wave.....we take only the sin component and say that
sin\Theta = tan \Theta \Theta is very small and is equal to ∂y/∂x.....
ie the force exerted by the left part on the section in considerration is F∂y/∂x
and the power is
(F∂y/∂x)∂y/∂t
so why isnt the F cos\Theta considered or why is it left out while calculating the power transmitted????
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3 Answers
11Its probably because you are calculating the rate of energy "across" the wave, in which case, the cos@ component is perpendicular to the direction of motion of the particles which are moving along the sin@ component. Thus force is perpendicular to the displacement, which makes the work along that direction zero.