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, \[{{\omega }_{1}}=500\,\pi \] \[\Rightarrow \] \[{{f}_{1}}=\frac{500\pi }{2\pi }=250\,beats\text{/}s\] and \[{{\omega }_{2}}=506\pi \] \[\Rightarrow \] \[{{f}_{2}}=\frac{506\pi }{2\pi }=253\,beats\text{/}s\] Thus, number of beats produced \[={{f}_{2}}-{{f}_{1}}=253-250\] \[=3\,beats\text{/}s\] \[=3\times 60\,beats\text{/}\min \] \[=180\,beats\text{/}\min \]
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