A) \[b-ac\]
B) \[b=\frac{1}{ac}\]
C) \[c=\pi a\]
D) \[c=\frac{1}{\pi a}\]
Correct Answer: D
Solution :
[d] \[y=a\,\,sin\text{ (2}\pi \text{bt-2}\pi \text{cx)}\,.....\text{(1)}\] General equation \[y=a\,\,sin\text{ }\frac{\text{2}\pi t}{T}-\frac{\text{2}\pi \text{x}}{\lambda }.....\text{(2)}\] \[by\,(1)\,\And (2)\] \[\omega =\frac{2\pi }{T}=2\pi b2\pi c=\frac{2\pi }{\lambda }\] \[T=\frac{1}{b}c=\frac{1}{\lambda }\] partial velocity \[=2\pi ba\] \[\therefore \,{{V}_{p}}=\frac{dy}{dt}=2\pi ba\,\cos \,[2\pi bt-2\pi cx]\] wave velocity \[\therefore \,{{V}_{w}}=\frac{\lambda }{T}=\frac{b}{c}\] \[{{V}_{p}}=2{{V}_{w}}\] \[2\pi ba\,=2b\] \[c=\frac{1}{\pi a}\]You need to login to perform this action.
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