A) \[\lambda =\frac{\pi {{Y}_{0}}}{4}\]
B) \[\lambda =\frac{\pi {{Y}_{0}}}{2}\]
C) \[\lambda =\pi {{Y}_{0}}\]
D) \[\lambda =2\pi {{Y}_{0}}\]
Correct Answer: B
Solution :
Comparing the given equation with \[y=a\sin (\omega t-kx)\], We get a = Y0, w = 2pf, \[k=\frac{2\pi }{\lambda }\]. Hence maximum particle velocity \[{{({{v}_{\max }})}_{particle}}=a\omega ={{Y}_{0}}\times 2\pi f\] and wave velocity \[{{(v)}_{wave}}=\frac{\omega }{k}=\frac{2\pi f}{2\pi /\lambda }=f\lambda \] \[\because \,\,\,{{({{v}_{\max }})}_{Particle}}=4{{v}_{Wave}}\] Þ \[{{Y}_{0}}\times 2\pi f=4f\lambda \] Þ \[\lambda =\frac{\pi {{Y}_{0}}}{2}\].
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