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

  • question_answer
    A mass M is suspended from a light spring. An additional mass m added displaces the spring further by a distance x. Now the combined mass will oscillate on the spring with period [CPMT 1989, 1998 ; UPSEAT 2000]

    A)            \[T=2\pi \sqrt{\left( mg/x(M+m) \right)}\]                

    B)            \[T=2\pi \sqrt{\left( (M+m)x/mg \right)}\]

    C)            \[T=(\pi /2)\sqrt{\left( mg/x(M+m) \right)}\]

    D)            \[T=2\pi \sqrt{\left( (M+m)/mgx \right)}\]

    Correct Answer: B

    Solution :

                       As mg produces extension x, hence \[k=\frac{mg}{x}\] \ \[T=2\pi \sqrt{\frac{(M+m)}{k}}=2\pi \sqrt{\frac{(M+m)x}{mg}}\]


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