91Ó°ÊÓ

From the given mass and length, the linear mass density \(\mu\) of the string can be calculated as \(\mu = \frac{mass}{length}= \frac{2.5 g}{0.4 m} = 6.25 \times 10^{-3} \mathrm{~kg/m}\)

We can rearrange the wave speed equation to solve for Tension as \(T = \mu v^{2}\). The speed of wave \(v\) is given as \(v = f \lambda\). The frequency \(f\) is given as 440 Hz. For a string fixed at both ends, the wavelength of the fundamental frequency is twice the length of the string, i.e. \(\lambda = 2L = 2 \times 0.400 m = 0.800 m\). So, \(v = 440 Hz \times 0.800 m = 352 m/s\). Substituting the values, we get \(T = 6.25 \times 10^{-3} \mathrm{~kg/m} \times (352 m/s)^{2} = 780 N\) for the tension.

Using Young’s modulus \(Y = \frac{\sigma}{\epsilon}\), we can calculate the strain \(\epsilon = \frac{\sigma}{Y}\). The stress \(\sigma\) is the force per unit area. Assuming the cross section area A is uniform along the wire's length, the force on it is just the tension T. So, \(\sigma = \frac{F}{A} = \frac{T}{A}\). Substituting the values, \(\epsilon = \frac{780 N}{\pi r^{2}} / 20 \times 10^{10} Pa\). This gives the fractional change in length.

The change in length from original length is equal to the fractional length change \( \Delta L/ L = \alpha \Delta T\). The coefficient of linear expansion is given as \( \alpha = 1.2 \times 10^{-5} K^{-1}\). Solving for the change in temperature \( \Delta T = \Delta L / L \alpha\). The new length of the wire \(L_{new} = L + \Delta L = L (1 + \epsilon) = 0.400 m (1 + \epsilon)\). From step 3, we can find the value of \(\epsilon\). The tension does not change when the temperature changes, so the string's fundamental frequency with new length is given by \(f_{new} = \frac{v}{\lambda_{new}} = \frac{v}{2L_{new}}\). We can solve for \(L_{new}\) and thereafter \(\Delta L/L\). Hence, we can solve for \( \Delta T = \Delta L / L \alpha \). The final temperature is \(T_{final} = T_{initial} + \Delta T = 20.0 °C + \Delta T\). The absolute temperature is calculated in Celsius so the final answer should be transformed back into Celsius.