JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Simple Pendulum

  • question_answer
    The time period of a simple pendulum of length L as measured in an elevator descending with acceleration \[\frac{g}{3}\] is                    [CPMT 2000]

    A)            \[2\pi \sqrt{\frac{3L}{g}}\]     

    B)            \[\pi \sqrt{\left( \frac{3L}{g} \right)}\]

    C)            \[2\pi \sqrt{\left( \frac{3L}{2g} \right)}\]                    

    D)            \[2\pi \sqrt{\frac{2L}{3g}}\]

    Correct Answer: C

    Solution :

                       The effective acceleration in a lift descending with acceleration \[\frac{g}{3}\] is \[{{g}_{eff}}=g-\frac{g}{3}=\frac{2g}{3}\]            \ \[T=2\pi \sqrt{\left( \frac{L}{{{g}_{eff}}} \right)}\]\[=2\pi \sqrt{\left( \frac{L}{2g/3} \right)}\]\[=2\pi \sqrt{\left( \frac{3L}{2g} \right)}\]


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