Solve this following


An object of mass $m$ is suspended at the end of a massless wire of length $\mathrm{L}$ and area of crosssection, A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is:


  1. $\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{YA}}{\mathrm{mL}}}$

  2. $f=\frac{1}{2 \pi} \sqrt{\frac{Y L}{m A}}$

  3. $f=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{mA}}{\mathrm{YL}}}$

  4. $\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{mL}}{\mathrm{YA}}}$

Correct Option: 1


An elastic wire can be treated as a spring with


$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

$\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{YA}}{\mathrm{m} \ell}}$

Ans. (1)


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