A) -ve\[x-\]direction with frequency 1 Hz
B) +ve\[x-\]direction with frequency n Hz and wavelength\[\lambda =0.2\,m\]
C) +ve\[x-\]direction with frequency 1 Hz and wavelength\[\lambda =0.2\,m\]
D) -ve \[x-\]direction with amplitude 0.25 m and Wavelength\[\lambda =0.2\,m\]
Correct Answer: C
Solution :
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 \[a=0.25\,m,\,\omega =2\pi ,k=10\pi \,m\] \[\therefore \]\[\frac{2\pi }{T}=2\pi \Rightarrow f=1\,Hz\] Also, \[\lambda =\frac{2\pi }{k}=\frac{2\pi }{10\pi }0.2\,m\] Therefore, the wave is travelling along +ve x-direction with frequency 1 Hz and wavelength 0.2 m.
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