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Consider a sinusoidal travelling wave shown in figure. The wave velocity is `+ 40 cm//s`. Find (a) the frequency (b) the phase difference between points `2.5 cm` apart ( c ) how long it takes for the phase at a given position to (d) the velocity of a particle at `P` at the instant shown. |
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Answer» Correct Answer - A::B::C::D (a) `(lambda)/(2) = 2 cm` rArr `lambda=4 cm` `f =(v)/(lambda) = (40)/(4) =10 Hz` (b) `Deltaphi = (2pi)/(lambda)Deltax = ((2pi)/(4))(2.5) = (5pi)/(4)` ( c ) `omegat = theta` or `(2pi//t) = theta` `:. t=(theta)/(2pif) = (pi//3)/((2pi)(10))` `=(1)/(60)s` (d) At `P`, particle is at mean position. So, `v= "maximum velocity"` `= omegaA` `=2pifA` `=(2pi)(10)(2)` `= (40pi) cm//s` `=125.7 cm//s = 1.26 m//s` Further, `v_(P) = - (v)((dely)/(delx))` ...(i) Sign of `v_(1)`, the wave velocity is given positive. Sign of `(dely)/(delx)`, slope of `y - x` graph is also positive. Hence, from Eq. (i) particle velocity is negative. `:. v_(P) = - 1.26 m//s` |
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