JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • question_answer
    If a simple pendulum has significant amplitude (up to a factor of 1/e of original) only in the period between \[t=0s\] to \[t=\tau s\], then \[\tau \] may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with 'b' as the constant of proportionality, the averatge life time of the pendulum is (assuming damping is small) in seconds:                     [AIEEE 2012]

    A) \[\frac{0.693}{b}\]

    B) b

    C) \[\frac{1}{b}\]

    D) \[\frac{2}{b}\]

    Correct Answer: D

    Solution :

    [d]
    \[m\frac{{{d}^{2}}x}{d{{t}^{2}}}=-kx-b\frac{dx}{dt}\]           
    \[m\frac{{{d}^{2}}x}{d{{t}^{2}}}+b\frac{dx}{dt}+kx=0\] here b is demping coefficient
    This has solution of type
    \[x={{e}^{\lambda t}}\] substituting this
    \[m{{\lambda }^{2}}+b{{\lambda }^{2}}+k=0\]
    \[\lambda =\frac{-b\pm \sqrt{{{b}^{2}}-4mk}}{2m}\]
    on solving for \[x\], we get
    \[x={{e}^{-\frac{b}{2m}t}}\]               \[a\,\cos \,({{\omega }_{1}}\,t-\alpha )\]                    
    \[{{\omega }_{1}}=\sqrt{\omega _{0}^{2}-{{\lambda }^{2}}}\] where \[{{\omega }_{0}}=\sqrt{\frac{k}{m}}\]
    \[\lambda =+\frac{b}{2}\]
    So, average life \[=\frac{2}{b}\]


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