A) \[\frac{\pi A}{2}\]
B) \[\pi A\]
C) \[2\pi A\]
D) \[A\]
Correct Answer: C
Solution :
[c] \[y=A\sin (\omega t-kx)\] Particle velocity, \[{{v}_{p}}=\frac{dy}{dt}=A\,\omega \cos (\omega t-kx)\] \[\therefore \,\,{{v}_{p\,\max }}=A\,\omega \] wave velocity \[=\frac{\omega }{k}\] \[\therefore \,\,\,\,\,A\,\omega =\frac{\omega }{k}\] \[i.e.,\,\,A=\frac{1}{k}\] But \[k=\frac{2\pi }{\lambda }\] \[\therefore \,\,\lambda =2\pi A\]
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