A) \[\pi \]
B) \[2\pi \]
C) \[\frac{\pi }{2}\]
D) \[\frac{B}{2\pi }\]
Correct Answer: B
Solution :
| Comparing given equation with |
| \[y=A\sin 2\pi \left( \frac{t}{T}-\frac{x}{\lambda } \right)\] |
| We have, \[T=\frac{1}{B}and\lambda =c\] |
| So, \[f=frequency=\frac{1}{T}=B\] |
| Hence, wave speed, \[{{\upsilon }_{wave}}=f\lambda \] |
| \[\Rightarrow \,\,\,\,\,\,\,{{\upsilon }_{wave}}=BC\] |
| Also, particle's maximum speed, |
| \[{{\upsilon }_{\max }}=A\omega =A\left( \frac{2\pi }{T} \right)\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,{{\upsilon }_{\max }}=A\left( 2\pi B \right)\] |
| Now given, \[{{\upsilon }_{wave}}={{\upsilon }_{\max }}\] |
| \[\Rightarrow BC=2\pi AB\] |
| \[\Rightarrow \frac{C}{A}=2\pi \] |
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