It has been given that, Young modulus of copper $Y=1.2 \times 10^{11} N / m ^2$, Coefficient of linear expansion of copper $\alpha=1.6 \times 10^{-5} /{ }^{\circ} C$, Density of copper $\rho=9.2 \times 10^3 kg / m ^3$.

Young's modulus is also known as modulus of elasticity and is defined as, the mechanical property of a material to withstand the compression or the elongation with respect to its length.

It is denoted as $E$ or $Y$. Young's Modulus (also referred to as the Elastic Modulus or Tensile Modulus), is a measure of mechanical properties of linear elastic solids like rods, wires, and such.

Coefficient of Linear Expansion is the rate of change of unit length per unit degree change in temperature.

We know, the speed of the transverse wave is given by, $v=\sqrt{\frac{F}{m}}$ where $m$ is linear mass density and $F$ is force.

From thermal expansion, $\Delta l=l \alpha \Delta \theta$.

The thermal stress in the wire corresponds to change in temperature.

Hence, $\Delta T$ is $F=Y \alpha \Delta T$

If $A$ is the cross-sectional area of the wire, then tension produced in the wire, $F=f A=Y \alpha \Delta T A$

Hence, force $F$ is given by, $F=Y \alpha A \Delta \theta$.

Linear mass density $=$ mass $/$ length

$=\text { volume } \times \text { density } / \text { length }$

$=\text { area } \times \text { length } \times \text { density } / \text { length }$

$=\text { area } \times \text { density }=A \rho$

Putting equation $(2)$ and $(3)$ in equation $(1)$,

$v=\sqrt{\frac{F}{\mu}}=\sqrt{\frac{Y \alpha A \Delta T}{(A \times l) \rho}}=\sqrt{\frac{Y \alpha \Delta T}{\rho}}$

Here, $Y=1.2 \times 10^{11} N / m ^2, \alpha=1.6 \times 10^{-5} /{ }^{\circ} C , \rho=9.2 \times 10^3\,kg / m ^3$.

The change in angle is $\Delta \theta=50^{\circ} C -30^{\circ} C =20^{\circ} C$.

Here, the velocity of the transverse waves,

$v=\sqrt{\frac{1.2 \times 10^{11} \times 1.6 \times 10^{-5} \times 20}{9.2 \times 10^3}}$

$=64.60\,m / s$