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NEET Physics Simple Harmonic Motion NEET PYQ-Simple Harmonic Motion

  • question_answer
    The time period of a mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be:                                            [AIPMT 2003]

    A)  \[\frac{T}{2}\] 

    B)                   2T

    C)  \[\frac{T}{4}\] 

    D)                   T

    Correct Answer: A

    Solution :

    Key Idea: Force constant of spring is inversely proportional to length of spring.
    Time period of mass suspended from spring.
                \[T=2\pi \sqrt{\frac{m}{k}}\]                  …(i)
    Now we know that,
    spring constant \[\propto \frac{1}{\text{length}}\]
    or         \[k\propto \frac{1}{x}\]                           ...(ii)
    Since, spring is cut into four equal parts, hence force constant of each part becomes four times the previous. So,
                \[k'=4k\]
    So, new time period of same mass suspended from one of the parts,
    \[T'=2\pi \sqrt{\frac{m}{4k}}=\frac{1}{2}.2\pi \sqrt{\frac{m}{k}}=\frac{T}{2}\] 


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