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For a travelling harmonic wave `y=2.0 cos(10t-0.0080x+0.35)`, where x and y are in centimetres and t in seconds. What is the phase difference between oscillatory motion of two points separated by a distance of (a) 4cm (b) 0.5 cm (c ) `lambda//2` (d) `3lambda//4` |
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Answer» Given `y=2.0 cos (10t-0.0080x+0.35)`. The standard equation of travelling harmonic wave can be writtien as `y =A xcos(omegat-kx+phi)`. on comparing two equation, we have `omega =10 red//s` and `k=0.0080 cm^(-1)=0.80 m^(-1)` `lambda=(2pi)/(k)=(2pi)/(0.80)m` phase dufference `Deltaphi=(2pi)/(lambda)xxDeltax` (i). when `DeltaX=4 M,Delta phi=(2pi)/((2pi//0.80))xx4=3.2 rad` (ii) when `Deltax=0.5 m, Delta `phi=(2pi)/((2pi/0.80))xx0.5=0.40` rad (iii) when `Deltax=(pi)/(2), Delta phi=(2pi)/(lambda)xx(lambda)/(2)=pi rad` (iv). when `Deltax=(3lambda)/(4), Delta phi=(2pi)/(lambda)xx(3lambda)/(4)=(3pi)/(2) rad` |
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