BHU PMT BHU PMT (Mains) Solved Paper-2008

  • question_answer

    Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:

    Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by\[45{}^\circ ,\] then (1) the resultant amplitude is\[(1+\sqrt{2})a\] (2) the phase of the resultant motion relative to the first is\[90{}^\circ \] (3) the energy associated with the resulting motion is\[(3+2\sqrt{2})\]times the energy associated with any single motion (4) the resulting motion is not simple harmonic

    A)  1, 2 and 3 are correct.

    B)  1 and 2 are correct.

    C)   2 and 4 are correct.

    D)  1 and 3 are correct.

    Correct Answer: D

    Solution :

                     Let simple harmonic motion be represented by \[{{y}_{1}}=a\sin \left( \omega t-\frac{\pi }{4} \right);{{y}_{2}}=a\sin \,\omega t\]and \[{{y}_{3}}=a\sin \left( \omega t+\frac{\pi }{4} \right)\] On superimposing the resultant SHM will be \[y=a\left[ \sin \left( \omega t-\frac{\pi }{4} \right)+\sin \omega t+\sin \left( \omega t+\frac{\pi }{4} \right) \right]\] \[=a\left[ 2\sin \omega t\cos \frac{\pi }{4}+\sin \omega t \right]\] \[=a[\sqrt{2}\sin \omega t+\sin g\omega t]\] \[=a(1+\sqrt{2})\sin \omega t\] Resultant amplitude\[=(1+\sqrt{2})a\] Energy is\[SHM\propto (amplitude)\] \[\therefore \]\[\frac{{{E}_{\operatorname{Re}sul\tan t}}}{{{E}_{\sin gle}}}={{\left[ \frac{A}{a} \right]}^{2}}={{(\sqrt{2}+1)}^{2}}=(3+2\sqrt{2})\] \[\Rightarrow \]\[{{E}_{\operatorname{Re}sula\tan t}}=(3+2\sqrt{2}){{E}_{\sin gle}}\]


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