InterviewSolution
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InterviewSolution
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A longitudinal standing wave ` y = a cos kx cos omega t` is maintained in a homogeneious medium of density `rho`. Here `omega` is the angular speed and `k` , the wave number and `a` is the amplitude of the standing wave . This standing wave exists all over a given region of space. The space density of the potential energy `PE = E_(p)(x , t)` at a point `(x , t)` in this space isA. `E_(p) = ( rho a^(2) omega^(2))/(2)`B. `E_(p) = ( rho a^(2) omega^(2))/(2) cos^(2) kx sin^(2) omega t`C. `E_(p) = ( rho a^(2) omega^(2))/(2) sin^(2) kx cos^(2) omega t`D. `E_(p) = ( rho a^(2) omega^(2))/(2) sin^(2) kx sin^(2) omega t` |
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Answer» Correct Answer - C The given longitudinal standing wave is `y = a cos kx cos omega t` `(i)` The nodes of this wave are located where `cos kx = 0 ` (i.e., ) at the values `x = (lambda)/( 4) , ( 3 lambda)/(4) ,…` and the antinodes are located where `cos kx = +- (i.e.,)` at the values `x = 0 , (lambda)/(2) ,...` At the nodes , the space density of kinetic energy ( kinetic energy per unit vanishes for the nodes , i.e., `x = (lambda)/(4) , ( 3 lambda)/(4) `etc. Also , `y` is maximum at `t = 0` , as we see from Eq.(i). Hence potential energy must be maximum at `t = 0` . Hence the time factor in potential energy density must enter as ` cos^(2) omega t`. Also , the sum of kinetic and potential energy densities must always be constant for a given `x` as it represents total energy at that point. Hence the potential energy density is `E_(p) = ( rho a^(2) omega^(2))/(2) sin^(2) kx cos^(2) omega t` `(ii)` and the kinetic energy density is `E_(k) = ( rho a^(2) omega^(2))/(2) cos^(2) kx sin^(2) omega t` `(iii)` |
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