A) \[\frac{1}{2\pi {{t}_{0}}}{{\cos }^{-1}}\left( \frac{a+b}{2c} \right)\]
B) \[\frac{1}{2\pi {{t}_{0}}}{{\cos }^{-1}}\left( \frac{a+b}{3c} \right)\]
C) \[\frac{1}{2\pi {{t}_{0}}}{{\cos }^{-1}}\left( \frac{2a+3c}{b} \right)\]
D) \[\frac{1}{2\pi {{t}_{0}}}{{\cos }^{-1}}\left( \frac{a+c}{2b} \right)\]
Correct Answer: D
Solution :
\[a=A\cos \omega {{t}_{o}}\] ............(1) \[b=A\cos 2\omega {{t}_{o}}\] .............(2) \[c=A\cos 3\omega {{t}_{o}}\] .............(3) On adding (1) and (3) \[a+c=A(\cos \omega {{t}_{o}}+\cos 3\omega {{t}_{o}})\] \[a+c=2A\cos (\frac{3\omega {{t}_{o}}+\omega {{t}_{o}}}{2})cos(\frac{3\omega {{t}_{o}}-\omega {{t}_{o}}}{2})\] \[a+c=2A\cos 2\omega {{t}_{o}}\cos \omega {{t}_{o}}\] from (2), \[b=A\cos 2\omega {{t}_{o}}\] \[a+c=2b\cos \omega {{t}_{o}}\] \[{{\cos }^{-1}}(\frac{a+c}{2b})=2\pi f{{t}_{o}}\] \[f=\frac{1}{2\pi {{t}_{o}}}{{\cos }^{-1}}(\frac{a+c}{2b})\]
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