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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.
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Solution
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:
\[m = \frac{0 . 0032}{0 . 4} = 8 \times {10}^{- 3} kg/m\]
\[\text{ Change in length, } ∆ L = 40 . 05 - 40 = 0 . 05 cm\]
\[= 0 . 05 \times {10}^{- 2} m\]
\[Strain = \frac{∆ L}{L} = \frac{0 . 05 \times {10}^{- 2}}{0 . 4}\]
\[ = 0 . 125 \times {10}^{- 2} \]
\[ f_0 = \frac{1}{2L}\sqrt{\frac{T}{m}}\]
\[ = \frac{1}{2 \times \left( 0 . 4005 \right)} \sqrt{\frac{T}{8 \times {10}^{- 3}}}\]
\[\Rightarrow 220 \times 220 = \left[ \frac{1}{\left( 0 . 801 \right)^2} \right] \times T \times \left( \frac{{10}^3}{8} \right)\]
\[ \Rightarrow T \times 1000 = 220 \times 220 \times 0 . 641 \times 0 . 8\]
\[ \Rightarrow T = 248 . 19 N\]
\[Stress = \frac{Tension}{Area} = \frac{248 . 19}{1 {mm}^2} = \frac{248 . 19}{{10}^{- 6}}\]
\[ \Rightarrow Stress = 248 . 19 \times {10}^6 \]
\[\text{ Young's modulus, } Y = \frac{stress}{strain}\]
\[ = \frac{248 . 19 \times {10}^6}{0 . 125 \times {10}^{- 2}}\]
\[ \Rightarrow Y = 19852 \times {10}^8 \]
\[ = 1 . 985 \times {10}^{11} N/ m^2\]
Hence, the required Young's modulus of the wire is
\[1 . 985 \times {10}^{11} N/ m^2\]
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