2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Simple Harmonic Motion

JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Spring Pendulum

  • 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                            [MP PMT 1994]

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

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

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

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

    Correct Answer: A

    Solution :

                       In this case springs are in parallel, so\[{{k}_{eq}}={{k}_{1}}+{{k}_{2}}\] and \[\omega =\sqrt{\frac{{{k}_{eq}}}{m}}=\sqrt{\frac{{{k}_{1}}+{{k}_{2}}}{m}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner