A) \[720m{{s}^{-1}}\]
B) \[384m{{s}^{-1}}\]
C) \[250m{{s}^{-1}}\]
D) \[1m{{s}^{-1}}\]
Correct Answer: B
Solution :
It is given that\[v=120Hz\], phase difference \[\Delta \phi =\frac{\pi }{2}\]and path difference\[\Delta x=0.8\,\,m\]. As \[\Delta \phi =\frac{2\pi }{\lambda }\cdot \Delta x\] \[\Rightarrow \] \[\lambda =\frac{2\pi }{\Delta \phi }\cdot \Delta x=\frac{2\pi \times 0.8}{\frac{\pi }{2}}=3.2\,\,m\] Hence, wave velocity \[v=v\lambda =120\times 3.2=384m{{s}^{-1}}\]
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