Medium MCQ +4 / -1 PYQ · JEE Mains 2020

An object of mass m is suspended at the end of a massless wire of length 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. A $f = {1 \over {2\pi }}\sqrt {{{YA} \over {mL}}}$ Correct answer
  2. B $f = {1 \over {2\pi }}\sqrt {{{mL} \over {YA}}}$
  3. C $f = {1 \over {2\pi }}\sqrt {{{YL} \over {mA}}}$
  4. D $f = {1 \over {2\pi }}\sqrt {{{mA} \over {YL}}}$

Solution

An elastic wire can be treated as a spring with <br><br>k = ${{YA} \over l}$ <br><br>T = $2\pi \sqrt {{m \over k}}$ <br><br>$\Rightarrow$ f = ${1 \over {2\pi }}\sqrt {{k \over m}}$ = ${1 \over {2\pi }}\sqrt {{{YA} \over {ml}}}$

About this question

Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity

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