InterviewSolution
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InterviewSolution
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When a train of plane wave traverses a medium, individual particles execute periodic motion given by the equation `y 4sin(pi)/(2)(2t x/8)` Where the length are expressed in centimetres and time in seconds. Calculate the amplitute, wavelength, (a) the phase different for two positions of the same particle which are occupied at time interval 0.4 s apart and (b) the phase difference at any given instant of two particle 12 cm apart. |
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Answer» The equation of a wave motion is given by `y=A sin(2pi)/(lambda)(vt+x)`(i) Here, `y 4 sin(pi)/(2)(2t x/8)` This equation canbe written as `y 4sin(2pi)/(32)(16t x)` (ii) comparing `Eq`. (i)with `Eq`. (ii), we get Amplitude `A=4 cm,`wavelength `(lambda)=32 cm`, wave velocity `v=16 cm//s` here frequency is given as `f=(v)/(lambda)=(16)/(32)=(1)/(2)=0.5 Hz` (a). phase of a particle at instant `t_(1)` is given by `phi_(1) (pi)/(2)(2t_(1) x/8)` The phase at instant `t_(2)` is given by `phi_(2) (pi)/(2)(2t_(1) x/8)` The phase difference is given as `phi_(1)-phi_(2)=(pi)/(2)[(2t_(1)+x/8)-(2t_(2)+x/8)]` `pi(t_(1)-t_(2)) pi(0.4)` `(As t_(1)-t_(2)=0.4)` `=180xx0.4=72^@` `(pi rad=180^@)` (b). phase different at an instant between two particle with path different `(Delta)` is `phi=(2pi)/(lambda)xxDelta` `(2pi)/(32)xx12` `(As Delta=12 cm` `(3pi)(4)` |
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