A) - ve x direction with frequency 1 Hz
B) + ve x direction with frequency \[\pi \] Hz and wavelength \[\lambda =0.2m\]
C) + ve x direction with frequency 1 Hz and wavelength \[\lambda =0.2m\]
D) -ve x direction with amplitude 0.25 m and wavelength \[\lambda =0.2m\]
Correct Answer: C
Solution :
Key Idea: The sign between two terms in argument of sine will define its direction. Writing the given wave equation \[y=0.25\sin (10\pi x-2\pi t)\] ...(i) The minus (-) between \[(10\pi x)\] and \[(2\pi t)\]implies that the wave is travelling along positive x direction. Now comparing Eq. (i) with standard wave equation\[y=a\sin (kx-\omega t)\] ...(ii) We have\[a=0.25m,\omega =2\pi ,k=10\pi m\] \[\therefore \]\[\frac{2\pi }{T}=2\pi \]\[\Rightarrow \]\[f=1Hz\] Also,\[\lambda =\frac{2\pi }{k}=\frac{2\pi }{10\pi }=0.2m\] Therefore, the wave is travelling along +ve x direction with frequency 1 Hz and wavelength 0.2m.
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