A) 3 beats/second with intensity ratio of maxima to minima equal to 9
B) 3 beats/second with intensity ratio of maxima to minima equal to 2
C) 6 beats/second with intensity ratio of maxima to minima to equal to 2
D) 6 beats/second with intensity ratio of maxima to minima equal to 9
Correct Answer: A
Solution :
Comparing given equation with standard equations \[y=A\,\sin \,\omega t\] We get \[{{A}_{1}}=4,\,\,{{\omega }_{1}}=500\,\pi \] and \[{{A}_{2}}=2,{{\omega }_{2}}=506\pi \] Frequency \[n=\frac{\omega }{2\pi }\] \[\therefore \] \[{{n}_{1}}=\frac{{{\omega }_{1}}}{2\pi }=\frac{500\pi }{2\pi }=250\] \[{{n}_{2}}=\frac{{{\omega }_{2}}}{2\pi }=\frac{506\pi }{2\pi }=253\] Number of beats \[={{n}_{2}}-{{n}_{1}}=253-250=3\] \[\frac{{{I}_{\max }}}{{{\operatorname{I}}_{\min }}}=\frac{{{({{A}_{1}}+{{A}_{2}})}^{_{2}}}}{({{A}_{1}}-{{A}_{2}})}\] \[=\frac{{{(A+2)}^{_{2}}}}{{{(A-2)}^{2}}}={{\left( \frac{6}{2} \right)}^{2}}=\frac{9}{1}\]
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