A) \[50m,50m{{s}^{-1}}\]
B) \[100m,100m{{s}^{-1}}\]
C) \[50m,100m{{s}^{-1}}\]
D) \[100m,50m{{s}^{-1}}\]
Correct Answer: B
Solution :
: The given equation of a transverse wave is \[y=0.05\sin \pi (2t-0.02x)=0.05\sin (2\pi t-0.02\pi x)\] Compare it with the standard equation, \[y=A\sin (\omega t-kx)\] We get \[A=0.05m,\] \[\omega =2\pi rad\,{{s}^{-1}}\] \[k=0.02\pi \,\,rad\,{{m}^{-1}}\] The minimum distance between points having the same phase is known as wavelength of the wave. \[\lambda =\frac{2\pi }{k}=\frac{2\pi }{0.02\pi }=100m\] Wave velocity, \[\upsilon =\frac{\omega }{k}=\frac{2\pi \,rad\,{{s}^{-1}}}{0.02\pi \,rad\,{{m}^{-1}}}=100m{{s}^{-1}}\]
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