A) 360
B) 180
C) 3
D) 60
Correct Answer: B
Solution :
Key Idea: To reach the solution the given wave equations must be compared with standard equation of progressive wave. |
So, |
\[{{y}_{1}}=4\sin 500\,\pi t\] (i) |
\[{{y}_{2}}=2\sin 506\,\pi \,t\] (ii) |
Comparing Eqs. (i) and (ii) with |
\[y=a\sin \omega t\] (iii) |
We have, |
\[{{\omega }_{1}}=500\,\pi \] |
\[\Rightarrow \] \[{{f}_{1}}=\frac{500\,\pi }{2\pi }=250\,beats/s\] |
and \[{{\omega }_{2}}=506\,\pi \] |
\[\Rightarrow \] \[{{f}_{2}}=\frac{506\,\pi }{2\pi }=253\,beats/s\] |
Thus, number of beats produced |
\[={{f}_{2}}-{{f}_{1}}=253-250\] |
\[=3\,\text{beats}/s\] |
\[=3\times 60\,\,beats/\min \] |
\[=180\,beats/\min \] |
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