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The equation of a transverse wave travelling along a spring is y = 4.0 sin pi(2.0t -0.010x) , where y and x are in cm and t in second |
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Answer» Amplitude , a=4 cm, wavelength `lambda = 200 cm` and frequency of the wire `v= 1s^(-1)` `y=a SIN[(2pi)/lambda(vt -x) + phi_(0)]`.......(i) where `phi_(0)` is the initial phase. Here, the given equation is, `y = 4.0 sin[(2pi)/200 (200t -x)]`............(ii) COMPARING equations (i) and (ii), we get Amplitude, a=4 cm Wavelength, `lambda = 200 cm//s` Wave velocity, `u =v/lambda = 200/200 = 1s^(-1)` Initial phase of the origin, i.e. at t=0 and x=0 `phi_(0)=0` Phase difference between a time interval `Deltat` is given by `Deltaphi = 2piv Deltat = 2pi xx 1 xx 0.5 = pi = 180^(@)` |
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