A wire of length 40 cm and mass 3.2 g is stretched between two fixed supports located 40.05 cm apart. The wire vibrates at 220 Hz in its fundamental mode. Determine the Young modulus of the wire, given that its cross-sectional area is 1.0 mm ^2 .

Step-by-step solution

The fundamental frequency of a stretched string is given by: f = 1 2L T where L is the vibrating length and is the linear mass density. Since = m L , we have: f = 1 2L T L m = 1 2 T mL Squaring both sides, the tension T in the wire is: T = 4 f^2 mL Approximating L L₀ = 0.4 m for the tension calculation: T = 4 (220)^2 (3.2 10⁻³) 0.4 = 247.8 N The Young's modulus Y is given by: Y = T L₀ A L where L₀ = 0.4 m , A = 1.0 10⁻⁶ m ^2 , and L = 40.05 - 40 = 0.05 cm = 5 10⁻⁴ m . Y = 247.8 0.4 1.0 10⁻⁶ 5 10⁻⁴ = 99.12 5 10⁻¹⁰ =