A) v
B) v/2
C) 2
D) 1
Correct Answer: C
Solution :
If \[(v-1),v,(v+1)\]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 \[y={{y}_{1}}+{{y}_{2}}+{{y}_{3}}\] \[=a\sin 2\pi vt+a[\sin 2\pi (v-1)t+\sin 2\pi (v+1)t]\] \[=a\sin 2\pi vt+a[2\sin 2\pi \,vt\,\cos 2\pi t]\] \[=a[2cos2\pi t+1]sin2\pi vt\] \[=asin2\pi \,vt\]with \[a=a[1+2cos2\pi t]\] So, \[I\propto {{(a)}^{2}}\propto {{a}^{2}}{{(1+2cos2\pi t)}^{2}}\] For \[I\]to be max, or min. \[\frac{dI}{dt}=0\Rightarrow \frac{d}{dt}{{(1+2cos2\pi t)}^{2}}=0\] ie, \[2(1+2cos\pi t)(2sin2\pi t)\times 2\pi =0\] \[\sin 2\pi \,t=0\]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 \[n=0,1,2,...\] \[t=\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{5}{2},...\]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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