question_answer 63)

                  For the harmonic travelling wave \[y=2\,\cos \,2\pi (10t-0.0080\,x+3.5)\]where \[x\] and \[y\] are in cm and \[t\] is second. What is the phase difference between the oscillatory motion at two points separated by a distance of                 (a) 4 m                                  (b) 0.5 m                 (c) \[\frac{\lambda }{2}\]                   (d) \[\frac{3\lambda }{4}\] (at a given instant of time)                    (e) What is the phase difference between the oscillation of a particle located at\[x=100\,cm,\]at\[t=Ts\] and\[t=5s\]

Answer:

                  \[y=2\cos \,2\pi (10t-0.0080\,x+3.5)\]   ?.. (i)                 Compare this equation with a standard equation                 \[y=a\,\cos (\omega t-kx+\phi )\]                 \[\therefore \] \[k=0.0080\,\times \,2\pi \]                 or \[\frac{2\pi }{\lambda }\,=\,0.0080\,\,\times \,2\pi \] or \[\lambda =\frac{1}{0.0080}\]                 \[=125\,cm=1.25\,m\]                 Now phase difference, \[\Delta \phi \,=\frac{2\pi }{\lambda }\times \,\text{path}\] difference (a) What path difference = 4 m,                 \[\Delta \phi =\,\frac{2\pi }{\lambda }\times \,4=\,\frac{2\pi }{1.25}\,\times \,4=\,6.4\] radian                 (b) When path difference = 0.5                 \[\Delta \phi =\frac{2\pi }{1.25}\,\times \,0.5\,=0.8\,\pi \] radian                 (c) When path distance \[=\frac{\lambda }{2}\].                 \[\Delta \phi =\,\frac{2\lambda }{\lambda }\times \frac{\lambda }{2}\,=\pi \] radian                 (c) When path difference \[=\frac{3\pi }{4},\]                 \[\Delta \phi \,=\frac{2\pi }{\lambda }\,\times \,\frac{3\lambda }{4}\,=\,1.5\,\pi \] radian                 (d) From given equation,                 \[\omega =\,20\pi \] or \[\frac{2\pi }{T}=20\,\pi \]                 or \[T=\,\frac{1}{10}\,=0.1\,s.\]                 At time \[t\], phase \[\phi =2\pi (10t-0.0080\,x+3.5)\]                 \[\therefore \] \[\Delta \phi =\,980\,\pi \] radians.