A) \[y=0.2\sin \,2\pi \,\left( 6\,t+\frac{x}{60} \right)\]
B) \[y=0.2\sin \,\pi \,\left( 6\,t+\frac{x}{60} \right)\]
C) \[y=0.2\sin \,2\pi \,\left( 6\,t-\frac{x}{60} \right)\]
D) \[y=0.2\sin \,\,\pi \,\left( 6\,t-\frac{x}{60} \right)\]
Correct Answer: C
Solution :
Key Idea: The expression of travelling wave is sine or cosine function of \[\omega t\pm kx\]. The general expression of travelling wave can be written as \[y=A\,\sin \,(\omega t\pm kx)...(i)\] For travelling wave along positive x-axis we should use minus (-) sign only. \[\therefore y=A\,\sin \,(\omega t-kx)\] but \[\omega =\frac{2\pi v}{\lambda }\,\,and\,\,k=\frac{2\pi }{\lambda }\] So, \[y=A\sin \frac{2\pi }{\lambda }(vt-x)\] Given, A = 0.2, m, v = 360 m/s, \[\lambda \]= 60 m, Substituting in Eq. (ii) we have \[y=0.2\sin \frac{2\pi }{60}(360\,t-x)\] \[ory=0.2\,\sin \,2\pi \,\left( 6\,t-\frac{x}{60} \right)\]
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