NEET Sample Paper NEET Sample Test Paper-39

  • question_answer A simple pendulum performs simple harmonic motion about \[x=0\]with an amplitude a and time period T. The speed of the pendulum at \[x=a/2\]will be

    A) \[\frac{\pi a\sqrt{3}}{2T}\]

    B)  \[\frac{\pi a}{T}\]

    C) \[\frac{3{{\pi }^{2}}a}{T}\]

    D) \[\frac{\pi a\sqrt{3}}{T}\] 

    Correct Answer: D

    Solution :

    \[v=\frac{dy}{dt}=A\omega \cos \omega t=A\omega \sqrt{1-{{\sin }^{2}}\omega t}\] Here, \[y=\frac{a}{2}\] \[\therefore \] \[v=\omega \sqrt{{{a}^{2}}-\frac{{{a}^{2}}}{4}}=\omega \sqrt{\frac{3{{a}^{2}}}{4}}=\frac{2\pi }{T}\frac{a\sqrt{3}}{2}=\frac{\pi a\sqrt{3}}{T}\]

adversite


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

Free
Videos