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Question

(a)

It has been given that, Young modulus of copper $Y=1.2 	imes 10^{11} N / m ^2$, Coefficient of linear expansion of copper $\alpha=1.6 	imes 1

Vedclass · Accepted Answer

$64.6$

Vedclass · Answer

(a)

It has been given that, Young modulus of copper $Y=1.2 	imes 10^{11} N / m ^2$, Coefficient of linear expansion of copper $\alpha=1.6 	imes 10^{-5} /{ }^{\circ} C$, Density of copper $ho=9.2 	imes 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, $\Delta l=l \alpha \Delta 	heta$.

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, $F=Y \alpha A \Delta 	heta$.

Linear mass density $=$ mass $/$ length

$=	ext { volume } 	imes 	ext { density } / 	ext { length }$

$=	ext { area } 	imes 	ext { length } 	imes 	ext { density } / 	ext { length }$

$=	ext { area } 	imes 	ext { density }=A ho$

Putting equation $(2)$ and $(3)$ in equation $(1)$,

$v=\sqrt{\frac{F}{\mu}}=\sqrt{\frac{Y \alpha A \Delta T}{(A 	imes l) ho}}=\sqrt{\frac{Y \alpha \Delta T}{ho}}$

Here, $Y=1.2 	imes 10^{11} N / m ^2, \alpha=1.6 	imes 10^{-5} /{ }^{\circ} C , ho=9.2 	imes 10^3\,kg / m ^3$.

The change in angle is $\Delta 	heta=50^{\circ} C -30^{\circ} C =20^{\circ} C$.

Here, the velocity of the transverse waves,

$v=\sqrt{\frac{1.2 	imes 10^{11} 	imes 1.6 	imes 10^{-5} 	imes 20}{9.2 	imes 10^3}}$

$=64.60\,m / s$

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