Question

A sonometer wire having cross-sectional area 0.85 ´ 10-6 m2 is stretched between two rigid supports 1.2 m apart. A tension of 20 N is applied at its free end. If the temperature is reduced by 12°C, calculate the final tension in the wire. Take coefficient of linear expansion (a) and isothermal Young’s modulus (g) to be 1.5 ´ 10-5 K-1 and 2.0 ´ 1011 Nm-2, respectively.

Answer 1

Answer on Question #66035, Physics / Mechanics | Relativity

A sonometer wire having cross-sectional area $0.85 \cdot 10^{-6} \, \mathrm{m}^2$ is stretched between two rigid supports $1.2 \, \mathrm{m}$ apart. A tension of $20 \, \mathrm{N}$ is applied at its free end. If the temperature is reduced by $12{}^{\circ}\mathrm{C}$, calculate the final tension in the wire. Take coefficient of linear expansion (a) and isothermal Young's modulus (g) to be $1.5 \cdot 10^{-5} \, \mathrm{K}^{-1}$ and $2.0 \cdot 10^{11} \, \mathrm{Nm}^{-2}$, respectively.

Solution:

When $\mathrm{dL} = 0$, we using the next equation

$$dF = -A\gamma\alpha dT$$

Integrating this equation

$$\int_{F_1}^{F_2} dF = -A\gamma\alpha \int_{T_1}^{T_2} dT$$

We get

$$F_2 - F_1 = A\gamma\alpha (T_1 - T_2)$$

Let $T_1 = 20{}^{\circ}\mathrm{C}$

$$F_2 - F_1 = 0.85 \cdot 10^{-6} \, \mathrm{m}^2 \times 1.5 \cdot 10^{-5} \, \mathrm{K}^{-1} \times 2.0 \cdot 10^{11} \, \mathrm{Nm}^{-2} \times 8 \, \mathrm{K}$$ $$F_2 - F_1 = 20.4 \, \mathrm{N}$$

So that

$$F_2 = 20.4 \, \mathrm{N} + 20 \, \mathrm{N} = 40.4 \, \mathrm{N}$$

Answer: 40.4 N

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