A) \[\frac{\omega }{k}:1\]
B) \[\frac{1}{ka}:1\]
C) \[ka:1\]
D) \[\omega a:1\]
Correct Answer: C
Solution :
As, \[\operatorname{y}(x,\,t)=a\,sin\,\,\left( kx-\omega t \right)\] \[\operatorname{Particle} velocity \frac{dy}{dt} = - a\omega cos \left( kx - \omega t \right)\] Maximum particle velocity \[{{\left( \frac{dy}{dt} \right)}_{\max }}\,\,=\,\,a\omega \] ?(i) Wave velocity \[v=\frac{\omega }{k}\] ? (ii) From Eqs. (i) and (ii), we get \[{{\left( \frac{dy}{dt} \right)}_{\max }}\,\,\,:\,\,\,v=a\omega :\frac{\omega }{k}=ka:1\]
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