JEE Main & Advanced Sample Paper JEE Main Sample Paper-29

  • question_answer
    A particle of mass \[2\,kg\] has potential energy\[U=8{{x}^{2}}-4{{x}^{4}}\]. Then its angular frequency for small oscillation about its equilibrium position will be:

    A) \[2\sqrt{2}\,rad/\sec \] 

    B) \[2\,rad/\sec \]

    C) \[1\,rad/\sec \]                

    D) \[4\,rad/\sec \]

    Correct Answer: A

    Solution :

    \[\therefore \,\,-{{e}^{-2x}}\,.{{y}^{2}}=2\ell n\,\,y-1\] \[\Rightarrow \,\,{{y}^{2}}={{e}^{2x}}\,-2{{e}^{2x}}\,\ell ny\] \[\omega ={{x}_{1}}+i{{y}_{1}}\,\]                 \[z={{x}_{2}}+i{{y}_{2}}\] at \[(1+2i)\,({{x}_{1}}+i{{y}_{1}})\,\] \[\Rightarrow \,2{{x}_{1}}+{{y}_{1}}=0\] Hence U is minimum at x = 0 Hence \[\Rightarrow \,2{{x}_{2}}\,+{{y}_{2}}=0\] is stable equilibrium position \[\therefore \,\,2{{x}_{1}}+{{y}_{1}}=2{{x}_{2}}+{{y}_{2}}\]         when x is very small x3 can be neglected \[\Rightarrow -2({{x}_{2}}-{{x}_{1}})\,={{y}_{2}}-{{y}_{1}}\] \[\Rightarrow \,\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\,=-2\]

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