A) \[{{C}_{2}}\]
B) \[\frac{{{C}_{1}}}{{{C}_{2}}}\]
C) \[\frac{a}{b}\]
D) \[\frac{2a}{b}\]
Correct Answer: C
Solution :
(c.)From the figure \[2{{l}_{0}}{{\cos }^{2}}\left( \frac{\pi y}{\beta } \right)\] \[{{l}_{0}}{{\cos }^{2}}\left( \frac{\pi y}{\beta } \right)\] \[\frac{{{l}_{0}}}{2}{{\cos }^{2}}\left( \frac{2\pi y}{\beta } \right)\] and \[4{{l}_{0}}{{\cos }^{2}}\left( \frac{\pi y}{\beta } \right)\] At \[{{C}_{1}}\] and \[{{C}_{2}}\] is \[+ve\] Velocity of particle \[vp=-v\left( \frac{dy}{dx} \right)=-ve\] Velocity of particle at \[x=0\] and \[t=0\] downwards. So, initial phase \[\phi =\pi \] So, \[y=A\sin (\omega t-kx+\pi )=A\sin (kx-\omega t)\] \[=A\sin (5\pi x-2000\pi t)=A\sin 5\pi (x-400t)\]
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