91影视

The speed of transverse wave is,

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

And the expansion relation is given by,

$\mathbf{鈭�}\mathbf{L}\mathbf{=}{\mathbf{L}}_{\mathbf{0}}\mathbf{伪}\mathbf{鈭�}\mathbf{T}$ ...(ii)

Also, Young鈥檚 Modulus expression is given as,

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

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

$\mathbf{鈭�}\mathbf{F}\mathbf{=}\mathbf{鈥�}\mathbf{驰伪础}\mathbf{鈭�}\mathbf{T}$ ...(iv)

Take the velocities to be v₁ and v₂.

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

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

$$\begin{gathered} v_2=\sqrt{\frac{\left(F+鈭�F\right)}{渭}} \\ =\sqrt{\frac{F-Y伪A鈭�T}{渭}} \\ {v}_2^2=\frac{F}{渭}-Y伪A\frac{鈭�T}{渭} \\ =v_1^2-\frac{Y伪A鈭�T}{渭} \end{gathered}$$

Therefore,

$v_1^2-v_2^2=\frac{Y伪A鈭�T}{渭}$

Substitute $渭=\frac{1}{蚁}$in above equation and solve for $伪$,

$伪=\frac{v_1^2-v_2^2}{\left({\displaystyle \frac{Y}{蚁}}\right)鈭�T}$

This is the expression of the coefficient of thermal expansion.