JEE Main & Advanced JEE Main Paper (Held On 11 April 2014)

  • question_answer
    The angular frequency of the 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%, the change in time period compared to the undamped oscillator is approximately as follows:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

    A) increases by 1%               

    B) increases by 8%

    C) decreases by 1%             

    D) decreases by 8%

    Correct Answer: B

    Solution :

    The change in time period compared to the un damped oscillator increases by 8%.


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