A) \[2\pi \]
B) \[4\pi \]
C) \[6\pi \]
D) \[8\pi \]
Correct Answer: C
Solution :
Given, \[y=2\cos (20\pi t-2\pi \times 0.008x+0.7\pi )\] Comparing with general equation \[y=a\cos (\omega t\pm kx\pm {{\phi }_{0}})\] We get \[k=2\pi \times 0.008\] \[\frac{2\pi }{\lambda }=2\pi \times 0.008\] \[\left[ As\,k=\frac{2\pi }{\lambda } \right]\] \[\Rightarrow \] \[\lambda =\frac{1}{0.008}\] \[\lambda =125\,cm\] \[=1.25\,m\] Now, phase difference\[(\Delta \phi )\] \[=\frac{2\pi }{\lambda }\]path difference\[(\Delta x\,)\] \[=\frac{2\pi }{1.25}\times 4=6\pi \]You need to login to perform this action.
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