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 \[={{f}_{2}}-{{f}_{1}}=253-250\] \[=3\,beats/s\] \[==3\times 60\,\,beats/\min \] \[=180\,beats/\min \]
You need to login to perform this action.
You will be redirected in
3 sec
