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JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Critical Thinking

  • question_answer
    One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of cross-section and Young's modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to                                                     [IIT 1993]

    A)            \[2\pi \left( \frac{m}{K} \right)\]                                     

    B)            \[2\pi {{\left\{ \frac{(YA+KL)m}{YAK} \right\}}^{1/2}}\]

    C)            \[2\pi \frac{mYA}{KL}\]   

    D)            \[2\pi \frac{mL}{YA}\]

    Correct Answer: B

    Solution :

                       The wire may be treated as a string for which force constant \[{{k}_{1}}=\frac{Force}{Extension}=\frac{YA}{L}\ \ \ \left( \because Y=\frac{F}{A}\times \frac{L}{\Delta L} \right)\] Spring constant of the spring \[{{k}_{2}}=K\] Hence spring constant of the combination (series) \[{{k}_{eq}}=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}=\frac{(YA/L)K}{(YA/L)+K}=\frac{YAK}{YA+KL}\] \[\because \]Time period \[T=2\pi \sqrt{\frac{m}{k}}=2\pi {{\left[ \frac{(YA+KL)m}{YAK} \right]}^{1/2}}\]


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