Question

For the wave shown in figure, if its position shown is at $ 4{{l}_{0}}{{\cos }^{2}}\left( \frac{\pi y}{\beta } \right) $ . Then the equation of the wave is [when speed of wave, $ {{C}_{1}} $ ]

• A. $ {{C}_{2}} $
• B. $ \frac{{{C}_{1}}}{{{C}_{2}}} $
• C. $ \frac{a}{b} $
• D. $ \frac{2a}{b} $

Answer 1

Answer: (C)

(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) $