A) \[2\pi {{\left( \frac{x_{1}^{2}-x_{2}^{2}}{u_{1}^{2}-u_{2}^{2}} \right)}^{1/2}}\]
B) \[2\pi {{\left( \frac{x_{2}^{2}-x_{1}^{2}}{u_{1}^{2}-u_{2}^{2}} \right)}^{1/2}}\]
C) \[2\pi {{\left( \frac{{{x}_{2}}-{{x}_{1}}}{{{u}_{2}}-{{u}_{1}}} \right)}^{{}}}\]
D) \[2\pi {{\left( \frac{{{x}_{2}}-{{x}_{1}}}{{{u}_{1}}-{{u}_{2}}} \right)}^{{}}}\]
Correct Answer: C
Solution :
\[u_{1}^{2}={{\omega }^{2}}({{r}^{2}}-x_{1}^{2})\] and \[u_{2}^{2}={{\omega }^{2}}({{r}^{2}}-x_{2}^{2})\] \[\therefore \] \[u_{1}^{2}-u_{2}^{2}={{\omega }^{2}}(x_{2}^{2}-x_{1}^{2})\] or \[\omega =\left( \frac{u_{1}^{2}-u_{2}^{2}}{x_{2}^{2}-x_{1}^{2}} \right)\] \[\therefore \] \[T=\frac{2\pi }{\omega }=2\pi {{\left[ \frac{x_{2}^{2}-x_{1}^{2}}{u_{1}^{2}-u_{2}^{2}} \right]}^{1/2}}\]You need to login to perform this action.
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