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:

\[m = \frac{0 . 0032}{0 . 4} = 8 \times  {10}^{- 3}   kg/m\] 

\[\text{ Change  in  length, }   ∆ L = 40 . 05 - 40 = 0 . 05  cm\] 

\[= 0 . 05 \times  {10}^{- 2}   m\] 

\[Strain = \frac{∆ L}{L} = \frac{0 . 05 \times {10}^{- 2}}{0 . 4}\] 

\[ = 0 . 125 \times  {10}^{- 2} \] 

\[ f_0  = \frac{1}{2L}\sqrt{\frac{T}{m}}\] 

\[  = \frac{1}{2 \times \left( 0 . 4005 \right)}  \sqrt{\frac{T}{8 \times {10}^{- 3}}}\]

\[\Rightarrow 220 \times 220 = \left[ \frac{1}{\left( 0 . 801 \right)^2} \right] \times T \times \left( \frac{{10}^3}{8} \right)\] 

\[ \Rightarrow T \times 1000 = 220 \times 220 \times 0 . 641 \times 0 . 8\] 

\[ \Rightarrow T = 248 . 19  N\] 

\[Stress = \frac{Tension}{Area} = \frac{248 . 19}{1  {mm}^2} = \frac{248 . 19}{{10}^{- 6}}\] 

\[ \Rightarrow Stress = 248 . 19 \times  {10}^6 \] 

\[\text{ Young's  modulus, }   Y = \frac{stress}{strain}\] 

\[  = \frac{248 . 19 \times {10}^6}{0 . 125 \times {10}^{- 2}}\] 

\[ \Rightarrow Y = 19852 \times  {10}^8 \] 

\[ = 1 . 985 \times  {10}^{11}   N/ m^2\]
Hence, the required Young's modulus of the wire is
\[1 . 985 \times {10}^{11} N/ m^2\]