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
    If a spring extends by x on loading, then energy stored by the spring is (if T is the tension in the spring and K is the spring constant)                                                            [AFMC 2000]

    A)            \[\frac{{{T}^{2}}}{2x}\]    

    B)            \[\frac{{{T}^{2}}}{2K}\]

    C)            \[\frac{2K}{{{T}^{2}}}\]    

    D)            \[\frac{2{{T}^{2}}}{K}\]

    Correct Answer: B

    Solution :

                       \[U=\frac{1}{2}K{{x}^{2}}\]but\[T=Kx\] So energy stored \[=\frac{1}{2}\frac{{{(Kx)}^{2}}}{K}=\frac{1}{2}\frac{{{T}^{2}}}{K}\]


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