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

T = 2 π \sqrt{\frac{E A}{M L}, k} = \frac{L}{E A}

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T = \frac{1}{2 π} \sqrt{\frac{E A}{M L}, k} = \frac{A}{E L}

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T = \frac{1}{2 π} \sqrt{\frac{E L}{M A}, k} = \frac{E A}{L}

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T = 2 π \sqrt{\frac{M L}{E A}, k} = \frac{E A}{L}

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Solution

The correct option is D $T = 2 π \sqrt{\frac{M L}{E A}, k} = \frac{E A}{L}$
According to the formula of Young's Modulus
$E = \frac{F L}{A . Δ L}$
Where $Δ L$ is the extension in the spring
$E = \frac{E A . Δ L}{L}$ ............ $(1)$
Now, according to Hooke's law
$f = k Δ L$ ........... $(2)$
where $k$ is the spring constant
By comparing $(1)$ and $(2)$
$k Δ L = \frac{E A Δ L}{L}$
$k = \frac{E A}{L}$
Time period, $T = 2 π \sqrt{\frac{M}{k}}$
$T = 2 π \sqrt{\frac{M L}{E A}}$

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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

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