A) \[\lambda =2\pi {{y}_{0}}\]
B) \[\lambda =\frac{\pi {{y}_{0}}}{3}\]
C) \[\lambda =\frac{\pi {{y}_{0}}}{2}\]
D) \[\lambda =\pi {{y}_{0}}\]
Correct Answer: D
Solution :
| \[y={{y}_{0}}\sin \frac{2\pi }{\lambda }\,(\text{v}t-x)\] |
| Particle velocity\[\frac{dy}{dt}={{y}_{0}}\times \frac{2\pi }{\lambda }\text{v cos}\frac{2\pi }{\lambda }\,(\text{vt}\,\text{-}\,\text{x)}\]. |
| Maximum particle velocity \[={{y}_{0}}\times \frac{2\pi \text{v}}{\lambda }\] |
| Wave velocity = v [given] |
| So, \[{{y}_{0}}\times \frac{2\pi \text{v}}{\lambda }=2\text{v}\] |
| \[\lambda =\pi {{y}_{0}}\] |
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