Question

A 40 cm wire having a mass of 3⋅2 g is stretched between two fixed supports 40⋅05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of cross section of the wire is 1⋅0 mm², find its Young modulus.

Answer 1

Given:
Length of the wire (L) = 40 cm = 0.40 m
Mass of the wire = 3.2 g = 0.003 kg
Distance between the two fixed supports of the wire = 40.05 cm
Fundamental mode frequency = 220 Hz
Therefore, linear mass density of the wire (m) is given by:
$$\begin{gathered} m=\frac{0.0032}{0.4}=8\times {10}^{-3}\mathrm{kg}/\mathrm{m} \\ \mathrm{Change}\mathrm{in}\mathrm{length},∆\mathrm{L}=40.05-40=0.05\mathrm{cm} \\ =0.05\times {10}^{-2}\mathrm{m} \\ \mathrm{Strain}=\frac{∆L}{L}=\frac{0.05\times {10}^{-2}}{0.4} \\ =0.125\times {10}^{-2} \\ {f}_0=\frac{1}{2L}\sqrt{\frac{T}{m}} \\ =\frac{1}{2\times \left(0.4005\right)}\sqrt{\frac{T}{8\times {10}^{-3}}} \end{gathered}$$

$$\begin{gathered} \Rightarrow 220\times 220=\left[\frac{1}{{\left(0.801\right)}^2}\right]\times T\times \left(\frac{{10}^3}{8}\right) \\ \Rightarrow T\times 1000=220\times 220\times 0.641\times 0.8 \\ \Rightarrow \mathrm{T}=248.19\mathrm{N} \\ \mathrm{Stress}=\frac{\mathrm{Tension}}{\mathrm{Area}}=\frac{248.19}{1{\mathrm{mm}}^2}=\frac{248.19}{{10}^{-6}} \\ \Rightarrow \mathrm{Stress}=248.19\times {10}^6 \\ \mathrm{Young}\text{'}\mathrm{s}\mathrm{modulus},Y=\frac{\mathrm{stress}}{\mathrm{strain}} \\ =\frac{248.19\times {10}^6}{0.125\times {10}^{-2}} \\ \Rightarrow Y=19852\times {10}^8 \\ =1.985\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$
Hence, the required Young's modulus of the wire is $1.985\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2$.