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JEE Main & Advanced Physics Simple Harmonic Motion JEE PYQ-Simple Harmonic Motion

  • question_answer
    A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of \[{{10}^{12}}/\sec \]. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number\[\text{=6}\text{.02}\times \text{1}{{\text{0}}^{\text{23}}}\text{ gm mol}{{\text{e}}^{\text{-1}}}\]).                                                                                         [JEE Main Online 08-04-2018]

    A)  \[2.2\,\,N/m\]                

    B)  \[5.5\,\,N/m\]

    C)  \[6.4\,\,N/m\]    

    D)       \[7.1\,\,N/m\]

    Correct Answer: D

    Solution :

    [d] \[kx=m{{\omega }^{2}}X\]
    \[\Rightarrow k=m{{\omega }^{2}}\]
    \[=\frac{108}{6.02\times {{10}^{23}}}\times {{(2\pi \times {{10}^{12}})}^{2}}\times {{10}^{-3}}\]
    \[k\approx 7.1N/m\]


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