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InterviewSolution
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The equation to a transverse wave travelling in a rope is given by `y=A cos(pi)/(2)[kx-omega-alpha]` where `A=0.6 m, k=0.005 cm^(-1),omega=8.0 s^(-1) and `alpha` is a non-vanishing constant. Then for this wave,A. the wavelength of the wave is `lambda=8m`B. the maximum velocity `v_(m)` of a particle of the rope will be, `v_(m)=7.53 m//s`.C. the equation of a wave which, when superposed with the given wave can produce standing waves in the rope is `y=A cos(pi)/(2)(kx+omega-alpha)`D. The equation of the wave can be represented by `y 2a cos(v)/(a)[t+(x)/(V)]` |
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Answer» Correct Answer - a.,b.,d. Given wave is `y=A cos.(pi)/(2)[kx-omega-alpha]` here wave number, `kxx(pi)/(2)=(2pi)/(lambda)` giving `lambda=(4)/(k)` here `k=0.005 cm^(-1)`. Hence `lambda=(4)/(0.005)cm=8m` maximum velocity `V_(m)=Axx` angular velocity. Here angular velocity `=(piomega)/(2)=(3.14xx8)/(2)=12.56 rad//s` hence `V_(m)=0.6xx12.56 m//s=7.53 m//s` Also, to produce stationary waves, the two waves should travel in opposite direction and have same frequency. the wave given by `y=A cos(pi)/(2)(kx+omegat-alpha)` fulfils this condition. |
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