A) \[{{60}^{o}}\]
B) \[{{120}^{o}}\]
C) \[{{160}^{o}}\]
D) \[{{180}^{o}}\]
Correct Answer: B
Solution :
In the first case, \[\frac{a}{2}=a\sin \omega \,t\] or \[\sin \omega \,t=\frac{1}{2}\] In the second case, \[\frac{a}{2}=a\sin \,(\omega \,t+\phi )\] or \[\frac{1}{2}=\sin \,\omega \,t\cos \phi +\cos \,(\omega \,t)\sin \phi \] or \[\frac{1}{2}\cos \phi +\sqrt{1-\frac{1}{4}}\sin \phi =\frac{1}{2}\] or \[\cos \phi +\sqrt{3}\sin \phi =1\] or \[\sqrt{3}{{\sin }^{2}}\phi =1-\cos \phi \] or \[3{{\sin }^{2}}\phi =1+{{\cos }^{2}}\phi -2\cos \phi \] or \[3(1-{{\cos }^{2}}\phi )=1+{{\cos }^{2}}\phi -2\cos \phi \] or \[4{{\cos }^{2}}\phi -2\cos \phi -2=0\] or \[2{{\cos }^{2}}\phi -\cos \phi -1=0\] \[\cos \phi =\frac{-(-1)\pm \sqrt{1+8}}{4}\] or \[\cos \phi =\frac{1\pm 3}{4}=\frac{4}{4}\] or \[\frac{-2}{4}=1\] or \[-\frac{1}{2}\]when \[\cos \phi =1,\,\phi ={{0}^{o}}\] This value is not acceptable in the given problem. Again, \[\cos \phi =\frac{-1}{2}\] or \[\phi ={{120}^{o}}\]
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