The speed of a longitudinal wave in a continuous medium is given by: \[ v = \sqrt{\frac{Y}{\rho}} \] where \(Y\) is the Young’s Modulus (Elasticity) and \(\rho\) is the volume density.

Young’s Modulus is defined as Stress over Strain: \(Y = \frac{\text{Stress}}{\text{Strain}}\).
Consider the spring has an effective cross-sectional area \(A\) and length \(l\).
For a small extension \(\Delta l\), the restoring force is \(F = k \Delta l\).

• Stress: \(\sigma = \frac{F}{A} = \frac{k \Delta l}{A}\)
• Strain: \(\varepsilon = \frac{\Delta l}{l}\)

Dividing Stress by Strain: \[ Y = \frac{k \Delta l / A}{\Delta l / l} = \frac{k l}{A} \]

Volume density is Mass per unit Volume. \[ \rho = \frac{m}{V} = \frac{m}{A \cdot l} \]

Substitute \(Y\) and \(\rho\) into the velocity equation: \[ v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{k l / A}{m / (A l)}} \] The area \(A\) cancels out: \[ v = \sqrt{\frac{k l^2}{m}} = l \sqrt{\frac{k}{m}} \]

The time \(t\) taken to travel the length \(l\) is: \[ t = \frac{\text{Distance}}{v} = \frac{l}{l \sqrt{k/m}} = \sqrt{\frac{m}{k}} \]

Answer: \(\sqrt{\frac{m}{k}}\)