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 0.2 m. |
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