A) \[v\]
B) \[v/2\]
C) 2
D) 1
Correct Answer: C
Solution :
If (v - 1), v, (v + 1) be the frequencies of the three waves and a be the amplitude of each then \[{{y}_{1}}=a\sin \,2\pi \,(v-1)\,t\], \[{{y}_{2}}=a\sin 2\pi \,\,vt\] and \[{{y}_{3}}=a\,\sin 2\pi \,(v+1)\,t\] Resultant displacement due to all three waves is \[y={{y}_{1}}+{{y}_{2}}+{{y}_{3}}\] \[=a\sin 2\pi vt+a\,[\sin \,2\pi \,(v-1)\,t\] \[+\sin \,2\pi \,(v+1)\,t]\] \[=a\sin 2\pi vt+a\,[2\sin \,2\pi \,vt\,\cos \,2\pi ]\] \[=a\,[2\cos 2\pi \,t+1]\,\sin \,2\pi vt\] \[=d\,\sin \,2\pi \,vt\] with \[d=a[1+2\cos 2\pi ]\] So, \[I\propto {{(d)}^{2}}\propto {{a}^{2}}{{(1+2\cos 2\pi t)}^{2}}\] For \[I\] to be max. or min. \[\frac{dI}{dt}=0\Rightarrow \frac{d}{dt}{{(1+2\cos 2\pi t)}^{2}}=0\] ie, \[2(1+2cos\,\,\pi \,t)=(2\sin 2\pi t)\times 2\pi =0\] \[\sin \,2\pi \,t=0\] or \[1+2\cos \,2\pi \,t=0\] So, if \[1+2\cos \,2\pi \,t=0\Rightarrow 2\pi \,t=2\pi \,n\pm \frac{2\pi }{3}\] with n = 0, 1, 2,... \[t=\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{5}{3}\], .... and for these value of t \[\cos 2\pi t=-\left( \frac{1}{2} \right),\,I=0\], ie, \[I\] is minimum and if\[\sin 2\pi \,t=0\] \[2\pi \,t=n\pi ,n=0,\,1,2,\,..\,\Rightarrow \,t=0,\frac{1}{2},1,\frac{3}{2},2....\] \[I\] is therefore \[9{{a}^{2}},{{a}^{2}},9{{a}^{2}},{{a}^{2}}\] ie, intensity is maximum (with two different values) ie, number of beats per sec is two.
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