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 :
| [d] \[a=A\cos \omega {{t}_{o}}\] ............(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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