A) \[\frac{k}{\omega }\]
B) \[k\omega \]
C) \[\omega \]
D) \[k\]
E) \[{{k}^{2}}\omega \]
Correct Answer: A
Solution :
Given, \[E={{E}_{0}}\sin (kx-\omega t)\] Comparing with standard equation, we get \[\frac{k}{\omega }=\frac{2\pi /\lambda }{2\pi f}=\frac{1}{c}.\]
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