• # question_answer                 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? A)                 $\lambda =2\,\pi {{y}_{0}}$      B)                 $\lambda =\frac{\pi {{v}_{0}}}{3}$ C)                 $\lambda =\frac{\pi {{y}_{0}}}{2}$         D)                 $\lambda =\pi {{y}_{0}}$

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 $or{{y}_{0}}\frac{2\pi v}{\lambda }=2v$ $or\lambda =\pi {{y}_{0}}$