| A transverse wave is represented by the equation |
| \[y={{y}_{0}}\sin \frac{2\pi }{\lambda }(vt-x)\] |
| For what value of \[\lambda \] is the maximum particle velocity equal to two times the wave velocity? [AIPMT 1998] |
A) \[\lambda =2\,\pi {{y}_{0}}\]
B) \[\lambda =\frac{\pi {{v}_{0}}}{3}\]
C) \[\lambda =\frac{\pi {{y}_{0}}}{2}\]
D) \[\lambda =\pi {{y}_{0}}\]
Correct Answer: D
Solution :
| The given equation is |
| \[y={{y}_{0}}\sin \,\frac{2\pi }{\lambda }(vt-x)\] .....(i) |
| In the wave equation v is the particle velocity. |
| Differentiating Eq. (i) with respect to time, |
| \[u=\frac{dy}{dx}={{y}_{0}}\frac{2\pi v}{\lambda }\cos \frac{2\pi }{\lambda }(vt-x)\] |
| Maximum particle velocity, \[{{u}_{\max }}={{y}_{0}}\frac{2\pi v}{\lambda }\] |
| Now it is given that, |
| maximum particle velocity \[=2\times \] wave velocity |
| or \[{{y}_{0}}\frac{2\pi v}{\lambda }=2v\] |
| or \[\lambda =\pi {{y}_{0}}\] |
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