A) \[{{\cos }^{-1}}\frac{7}{9}\]
B) \[{{\cos }^{-1}}\frac{5}{9}\]
C) \[{{\cos }^{-1}}\frac{4}{9}\]
D) \[{{\cos }^{-1}}\frac{1}{9}\]
Correct Answer: A
Solution :
[a] Let \[{{x}_{1}}=a\sin \omega t\]and \[{{x}_{2}}=a\sin (\omega t+\delta )\]to two SHMs. \[\frac{a}{3}=a\sin \omega t\] and \[-\frac{a}{3}=a\sin (\omega t+\delta )\] \[\sin \omega t=1/3\]and \[\sin (\omega t+\delta )=-1/3\] Eliminating \[t,\frac{1}{3}\cos \delta +\sqrt{1-\frac{1}{9}}\sin \delta =-\frac{1}{3}\] \[9{{\cos }^{2}}\delta +2\cos \delta -7=0\] \[\cos \delta =-1\] and \[\frac{7}{9}\] i.e., \[\delta =180{}^\circ \] or \[{{\cos }^{-1}}\left( \frac{7}{9} \right)\] If we put\[180{}^\circ \], we find that \[{{v}_{1}}\]and \[{{v}_{2}}\]are of opposite signs. Hence \[\delta =180{}^\circ \]is not applicable. \[\therefore \delta ={{\cos }^{-1}}\left( \frac{7}{9} \right)\]
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