A metal wire, with density p and Young’s modulus Y, is stretched between rigid supports. At temperature T, the speed of a transverse wave is found to be v₁. When the temperature increased to $\mathbf{T}\mathbf{=}\mathbf{∆}\mathbf{T}$, the speed decreases to v₂ < v₁. Determine the coefficient of linear expansion of the wire.

The speed of transverse wave is,

$\mathbf{v}\mathbf{=}\sqrt{\frac{\mathbf{F}}{\mathbf{\mu }}}$ ...(i)

And the expansion relation is given by,

$\mathbf{∆}\mathbf{L}\mathbf{=}{\mathbf{L}}_{\mathbf{0}}\mathbf{\alpha }\mathbf{∆}\mathbf{T}$ ...(ii)

Also, Young’s Modulus expression is given as,

$\mathbf{Y}\mathbf{=}\frac{{\displaystyle \frac{\mathbf{F}}{\mathbf{A}}}}{{\displaystyle \frac{\mathbf{∆}\mathbf{L}}{\mathbf{L}}}}$ ...(iii)

From these equations, relation for change in tension with respect to change in temperature is,

$\mathbf{∆}\mathbf{F}\mathbf{=}\mathbf{‐}\mathbf{Y\alpha A}\mathbf{∆}\mathbf{T}$ ...(iv)

Take the velocities to be v₁ and v₂.

$$\begin{gathered} v_1=\sqrt{\frac{F}{\mu }} \\ {v}_1^2=\frac{F}{\mu } \\ F=\mu {\nu }_1^2 \end{gathered}$$

For v₂, the tension decreases according to equation (iv),

$$\begin{gathered} v_2=\sqrt{\frac{\left(F+∆F\right)}{\mu }} \\ =\sqrt{\frac{F-Y\alpha A∆T}{\mu }} \\ {v}_2^2=\frac{F}{\mu }-Y\alpha A\frac{∆T}{\mu } \\ =v_1^2-\frac{Y\alpha A∆T}{\mu } \end{gathered}$$

Therefore,

$v_1^2-v_2^2=\frac{Y\alpha A∆T}{\mu }$

Substitute $\mu =\frac{1}{\rho }$in above equation and solve for $\alpha$,

$\alpha =\frac{v_1^2-v_2^2}{\left({\displaystyle \frac{Y}{\rho }}\right)∆T}$

This is the expression of the coefficient of thermal expansion.