A) Amplitude A/2, frequency \[2n\] and wavelength \[\lambda /2\]
B) Amplitude A/2, frequency \[2n\] and wavelength \[\lambda \]
C) Amplitude A, frequency \[2n\] and wavelength \[2\lambda \]
D) Amplitude A, frequency \[n\] and wavelength \[\lambda \]
Correct Answer: A
Solution :
The given equation can be x written as \[y=\frac{A}{2}\cos \,\left( 4\pi nt-\frac{4\pi x}{\lambda } \right)+\frac{A}{2}\] \[\left( \because {{\cos }^{2}}\theta =\frac{1+\cos 2\theta }{2} \right)\] Hence amplitude \[=\frac{A}{2}\] and frequency \[=\frac{\omega }{2\pi }=\frac{4\pi n}{2\pi }=2n\] and wave length \[=\frac{2\pi }{k}=\frac{2\pi }{4\pi /\lambda }=\frac{\lambda }{2}\].
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