JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • question_answer
    If a spring has time period T, and is cut into n equal parts, then the time period of each part will be [AIEEE 2002]

    A) \[T\sqrt{n}\]

    B) \[\frac{T}{\sqrt{n}}\]

    C) nT

    D) T

    Correct Answer: B

    Solution :

    [b] As we know that spring constant of spring is inversely proportional to length of spring, so new spring constant for each part is given by \[k'=nk\] where, k is the spring constant of whole spring. From the theory of spring pendulum, we know that time period of spring pendulum is inversely proportional to square root of spring constant.
                \[i.e.\],   \[T\propto \frac{1}{\sqrt{k}}\]     \[\left( \because T=2\pi \sqrt{\frac{m}{k}} \right)\]
                            \[T'=\propto \frac{1}{\sqrt{nk}}\]      (T'= new time period)
                So,       \[T'=\frac{1}{\sqrt{n}}\]


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