A) moving along y direction with frequency \[2\pi \times \,{{10}^{6}}Hz\] and wavelength 200 m.
B) moving along x direction with frequency 106 Hz and wavelength 100 m
C) moving along x direction with frequency \[{{10}^{6}}Hz\] and wavelength 200 m
D) moving along - x direction with frequency \[{{10}^{6}}Hz\] and wavelength 200 m
Correct Answer: C
Solution :
Comparing the given equation \[y=2\sin (7.85\times -1005t)\]\[=(0.02)m\sin (7.85x-1005t)\] With the standard equation \[v=\frac{dy}{dt}=A\omega \cos \omega t=A\omega \sqrt{1-{{\sin }^{2}}\omega t}\] we get \[=\omega \sqrt{{{A}^{2}}-{{y}^{2}}}\] \[y=\frac{a}{2}\]\[\therefore \] Moreover, we know that \[v=\omega \sqrt{{{a}^{2}}-\frac{{{a}^{2}}}{4}}=\omega \sqrt{\frac{3{{a}^{2}}}{4}}=\frac{2\pi }{T}\frac{a\sqrt{3}}{2}=\frac{\pi a\sqrt{3}}{T}\] \[E=hv=h\frac{c}{\lambda }\]\[\Rightarrow \] Hence, the wave is moving along positive x-direction with frequency 106 Hz and wavelength 200 m.
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