JEE Main & Advanced Sample Paper JEE Main Sample Paper-39

  • question_answer
    A block is placed on a frictionless horizontal table. The mass of the block is m and springs are attached on either side with force constants \[{{K}_{1}}\]and \[{{K}_{2}}.\]If the block is displaced a little and left to oscillate, then the angular frequency of oscillation will be

    A)  \[{{\left( \frac{{{K}_{1}}+{{K}_{2}}}{m} \right)}^{\frac{1}{2}}}\]               

    B)  \[{{\left[ \frac{{{K}_{1}}{{K}_{2}}}{m({{K}_{1}}+{{K}_{2}})} \right]}^{\frac{1}{2}}}\]

    C)  \[{{\left[ \frac{{{K}_{1}}{{K}_{2}}}{({{K}_{1}}-{{K}_{2}})m} \right]}^{\frac{1}{2}}}\]  

    D)  \[{{\left[ \frac{K_{1}^{2}K_{2}^{2}}{({{K}_{1}}+{{K}_{2}})m} \right]}^{\frac{1}{2}}}\]

    Correct Answer: A

    Solution :

     When springs are in parallel, then\[T=2\pi \sqrt{\frac{m}{{{K}_{1}}+{{K}_{2}}}}\Rightarrow \frac{2\pi }{T}=\omega =\sqrt{\frac{{{K}_{1}}+{{K}_{2}}}{m}}\]


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