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Haryana PMT Haryana PMT Solved Paper-2005

  • question_answer
    A particle is moving with S.H.M in a straight line. When the distance of the particle from equilibrium position has values \[{{x}_{1}}\]and \[{{x}_{2}}\] the corresponding values of velocities are \[{{u}_{1}}\] and \[{{u}_{2}}\] The time period of oscillation Is :

    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            \[\omega =\left( \frac{u_{1}^{2}-u_{2}^{2}}{x_{2}^{2}-x_{1}^{2}} \right)\] \[\therefore \]  \[T=\frac{2\pi }{\omega }=2\pi {{\left[ \frac{x_{2}^{2}-x_{1}^{2}}{u_{1}^{2}-u_{2}^{2}} \right]}^{1/2}}\]


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