If $A$ is the area of cross-section of a spring $L$ is its length $E$ is the Young's modulus of the material of the spring then time period and force constant of the spring will be respectively
AIIMS 2010, Medium
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According to the formula of Young's Modulus
$E=\frac{F L}{A \Delta L}$
where $\Delta \mathrm{L}$ is the extension in the spring.
$\mathrm{F}=\frac{\mathrm{EA} \Delta \mathrm{L}}{\mathrm{L}}$ $...(1)$
Now, according to Hooke's law
$\mathrm{F}=\mathrm{k} \Delta \mathrm{L}$ $...(2)$
where $\mathrm{k}$ is the spring constant By comparing $( 1)\, and\, ( 2)$
$\mathrm{k} \Delta \mathrm{L}=\frac{\mathrm{EA} \Delta \mathrm{L}}{\mathrm{L}}$
$\quad \mathrm{k}=\frac{\mathrm{EA}}{\mathrm{L}}$
Time period, $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{k}}}$
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{ML}}{\mathrm{EA}}}$
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