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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 10

  • question_answer
    The angular frequency of damped oscillator is given by \[\omega =\sqrt{\left( \frac{k}{m}-\frac{{{r}^{2}}}{4{{m}^{2}}} \right)}\] where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio \[\frac{{{r}^{2}}}{mk}\] is \[8%,\] then the undamped oscillator is, approximately, as follows

    A) Decreases by 1%

    B) Increases by 8% 

    C) Decreases by 8%

    D) Increases by 1%

    Correct Answer: D

    Solution :

    [d] \[\omega =\sqrt{\frac{k}{m}-\frac{{{r}^{2}}}{4{{M}^{2}}}}=\sqrt{\frac{k}{m}}\,{{\left( 1-\frac{{{r}^{2}}}{4mk} \right)}^{\frac{1}{2}}}\] \[=\sqrt{\frac{k}{m}}{{\left( 1-\frac{0.08}{4} \right)}^{\frac{1}{2}}}=\sqrt{\frac{k}{m}}\left( 1-\frac{0.08}{2\times 4} \right)=\sqrt{\frac{k}{m}}(0.99)\] \[\omega \] decreases by\[1%\]. Hence T increases by \[1%\].


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