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The equation of a travelling plane sound wave has the form `y=60 cos (1800t-5.3x)`, where y is in micrometres, t in seconds and x in metres. Find (a). The ratio of the displacement amplitude with which the particle of the medium oscillate to the wavelength, (b).the velocity oscillation ammplitude of particles of the medium and its ratio to the wave propagation velocity ,(c). the particle acceleration amplitude. |
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Answer» The given wave equation `xi=60 cos (1800t-5.3x)` is of the type `xi= a cos ( omegat-kx), `where` a=60xx10^(-6)m` `omega=1800 ` per sec and `k=5.3 `per metre As `k=(2pi)/(lambda), `so` lambda=(2pi)/( k)` and also `k=(omega)/(v), `so` v=(omega)/(k)=340m//s` `(a)` Sought ratio `=(a)/(lambda)=(ak)/(2pi)=5.1xx10^(-5)` `(b)` Since `xi=a cos ( omegat-kx)` `(deltaxi)/(deltat)=-a omegasin (omegat-kx)` So velocity oscillation amplitude `(deltaxi)/(deltat)_(m)` or `v_(m)=a omega=0.11m//s ...(1)` and sought ratio of velocity oscillation amplitude to the wave propagation velocity `((v_(m))/(v))=(0.11)/(340)=3.2xx10^(-4)` `(c)` Relative deformation `=( delta_(m))/(v)=a k sini (omegat-kx)` So, relative deformation amplitude `=((delta xi)/(deltax))=ak=(60xx10^(-6)xx5.3)m=3.2xx10^(-4)m ...(2)` From Eqns `(1)` and `(2)` `((deltaxi)/(deltax))_(x)=ak=(aomega)/(v)=(1)/(v)((deltaxi)/(deltat))_(m)` Thus` ((deltaxi)/( deltax))_(m)=(1)/(v)((deltaxi)/(deltat)), `where` v= 340 m//s` is the wave velocity. |
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