2. Solved Papers
  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2012

  • question_answer
    The angular frequency and the amplitude of a simple pendulum are\[\omega \]and a respectively. If the pendulum is displaced through a distance\[x\] from the mean position then the ratio of its kinetic energy (K) and potential energy (U) will be

    A)  \[\frac{K}{U}=\frac{{{a}^{2}}-{{x}^{2}}{{\omega }^{2}}}{{{x}^{2}}{{\omega }^{2}}}\]

    B)  \[\frac{K}{U}=\frac{{{a}^{2}}-{{x}^{2}}}{{{x}^{2}}}\]

    C)  \[\frac{K}{U}=\frac{{{x}^{2}}}{{{a}^{2}}-{{x}^{2}}}\]

    D)  \[\frac{K}{U}=\frac{{{x}^{2}}{{\omega }^{2}}}{{{a}^{2}}-{{\omega }^{2}}{{x}^{2}}}\]

    Correct Answer: B

    Solution :

     \[\frac{K}{U}=\frac{\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{\omega }^{2}})}{\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}}\] \[=\frac{{{a}^{2}}-{{x}^{2}}}{{{x}^{2}}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner