InterviewSolution
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InterviewSolution
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A plane longitudinal harmonic wave propagates in a medium with density rho. The velocity of the wave propagation si v . Assuming that the density variations of the medium, induced by the propagating wave, Delta rho lt lt rho , demonstrate that ltbr. (a) the pressure incrementin the medium Deltap=- rho v^(2)( delta xi//delta x),where delta xi // delta xis the relative deformation, (b)the wave intensity is defined by Eq. (4.3i) |
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Answer» Solution :`(a)` Let us CONSIDER the motion of an element of the medium of thikness `dx` os unit area of cross `-` section. Let `xi=` displacement of the PARTICLES of the medium at location `x`. Then by the equation of motion. `rho dx dot(xi)=-dp` where `dp` is the pressure increment over the length `dx` Recalling the wave equation `ddot(xi)=v^(2)(delta^(2)xi)/( delta x^(2))` we can write the foregoing equationas `rho v^(2)(delta^(2)xi)/(deltax^(2))dx=-dp ` Integrating THIE equation, we get `Delta p=` surplus pressure `=- rho v^(2) ( deltaxi)/( deltax)+ Const.` In the absence of a deformation `(` a wave `),` the surplus pressure is `Delta p=0`. So' Const ' `=0` and LTBR. `Deltap=-rho v^(2)( deltaxi)/( deltax)`. `(b)` We have found earlier that `w=w_(k)+ w_(p)=` total energy density `w_(k)=(1)/(2)rho((deltaxi)/( delta))^(2), w_(p) =(1)/(2)E((deltaxi)/( deltax))^(2)=(1)/(2) rho v^(2)((deltaxi)/(deltax))^(2)` It is easy to see that the space `-` time average of both densities is the same and the space time average of total energy density is then `lt w gt = lt rho v^(2)((deltaxi)/(deltax))^(2)` The intensity of the wave is `I=vlt w GE lt ((Deltap)^(2))/(rhov) gt` Using `lt (Deltap)^(2) gt =(1)/(2) (Deltap)_(m)^(2)` we get `I=((Deltap)_(m)^(2))/(2 rho v)` |
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