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BCECE Engineering BCECE Engineering Solved Paper-2015

  • question_answer
    If two simple pendulums first of bob mass \[{{M}_{1}}\] and length \[{{l}_{1}}\], second of bob mass \[{{M}_{2}}\] and length \[{{l}_{2}}\]. Given \[{{M}_{1}}={{M}_{2}}\] and. \[{{l}_{1}}=2{{l}_{2}}\] If the vibrational energies of both are same, then which of the following is correct?

    A) Amplitude of B is smaller than A

    B) Amplitude of B is greater than B

    C)  Amplitude will be same

    D) None of the above

    Correct Answer: A

    Solution :

    Frequency,       \[n=\frac{1}{2\pi }\sqrt{\frac{g}{l}}\] or                            \[n\propto \frac{1}{\sqrt{l}}\] \[\therefore \]                  \[\frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{l}_{2}}}{{{l}_{1}}}}=\sqrt{\frac{{{l}_{2}}}{2{{l}_{2}}}}\]                                 \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{1}{\sqrt{2}}\] \[\Rightarrow \]                               \[{{n}_{2}}=\sqrt{2{{n}_{1}}}\] \[\Rightarrow \]                               \[{{n}_{2}}>{{n}_{1}}\] Energy,                                 \[E=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] and                        \[{{a}^{2}}\propto \frac{1}{m{{n}^{2}}}\]              {\[\because E\] is same} \[\therefore \]                  \[\frac{a_{1}^{2}}{a_{2}^{2}}=\frac{{{m}_{2}}n_{2}^{2}}{{{m}_{1}}n_{1}^{2}}\] Given, \[{{n}_{2}}>{{n}_{1}}\] and \[{{m}_{1}}>{{m}_{2}}\] \[\Rightarrow \]                               \[{{a}_{1}}>{{a}_{2}}\] So, amplitude of B is smaller than A.


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