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, f

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 in N/ m 2 ).

TestHub · Accepted Answer

μ = 3.2 × 1 0 − 3 /0.40 = 8 × 1 0 − 3 kg/m. v = 2 × 0.4005 × 220 = 176.2 m/s. T = μ v 2 = 248.4 N. Strain = 0.05/40 = 1.25 × 1 0 − 3 . Y = A × strain T ​ = 1 0 − 6 × 1.25 × 1 0 − 3 248.4 ​ = 1.99 × 1 0 11 N/m² (rounding).

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