A) \[\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-2}} \right]\]
B) \[\frac{3}{2}m{{r}^{2}}\]
C) \[\frac{m{{r}^{2}}}{2}\]
D) \[\frac{3}{8}m{{r}^{2}}\]
Correct Answer: A
Solution :
Let \[{{\phi }_{1}}\] and \[{{\phi }_{2}}\] represent angles of the first and second waves, then \[{{\phi }_{2}}=\frac{2\pi }{\lambda }[(vt-x)+{{x}_{0}}]\] and \[{{\phi }_{1}}=\frac{2\pi }{\lambda }(vt-x)\] But \[{{x}_{0}}=\frac{\lambda }{2}\] \[\therefore \] \[{{\phi }_{2}}-{{\phi }_{1}}=\pi \] Hence, phase difference, \[\phi =\pi .\]So, amplitude of the resultant wave \[R=\sqrt{{{a}^{2}}+{{b}^{2}}+2ab\,\cos \phi }\] \[=\sqrt{{{a}^{2}}+{{b}^{2}}+2ab\,\cos \pi }=\sqrt{{{(a-b)}^{2}}}=a-b\] or \[R|a-b|\]
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