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The standing waves are set up by the superimposition of two waves: `y_(1)=0.05 sin (15pit-x)`and `y_(2)=0.05 sin (15pit+x)`, where x and y are in metre and t in seconds. Find the displacement of particle at `x=0.5m.` |
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Answer» According to superposition principle, resultant displacement of stationary wave is given by `y=y_(1)+y_(2)=0.05sin(15pit-x)` `+0.05sin(15pit+x)` `=0.05xx2sin((15pit-x+15pit+x)/(2))` `cos((15pit+x-15pit+x)/(2))=0.1cosxsin15pit` Amplitude of stationary wave, `2r=0.1cosx` At `x=0.5m,` `2r=0.1cos0.5=0.1cos((180^(@))/(pi))xx0.5` `=0.1 cos 28.65^(@)=0.1xx0.88=0.088m` |
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