A wire of length 40 cm and mass 3.2 g is stretched between two...

The fundamental frequency of a stretched string is given by:

$f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}$

where $L$ is the vibrating length and $\mu$ is the linear mass density. Since $\mu = \dfrac{m}{L}$, we have:

$f = \dfrac{1}{2L} \sqrt{\dfrac{T L}{m}} = \dfrac{1}{2} \sqrt{\dfrac{T}{mL}}$

Squaring both sides, the tension $T$ in the wire is:

$T = 4 f^2 mL$

Approximating $L \approx L_0 = 0.4 \text{ m}$ for the tension calculation:

$T = 4 \times (220)^2 \times (3.2 \times 10^{-3}) \times 0.4 = 247.8 \text{ N}$

The Young's modulus $Y$ is given by:

$Y = \dfrac{T L_0}{A \Delta L}$

where $L_0 = 0.4 \text{ m}$, $A = 1.0 \times 10^{-6} \text{ m}^2$, and $\Delta L = 40.05 - 40 = 0.05 \text{ cm} = 5 \times 10^{-4} \text{ m}$.

$Y = \dfrac{247.8 \times 0.4}{1.0 \times 10^{-6} \times 5 \times 10^{-4}} = \dfrac{99.12}{5 \times 10^{-10}} = 1.98 \times 10^{11} \text{ N/m}^2$

Answer: $1.98 \times 10^{11} \text{ N/m}^2$