A 40 cm long wire having a mass 3.2 g and area of cross section 1 mm² is stretched between the supports 40.05 cm apart. In its fundamental mode, it vibrates with a frequency 1000/64 Hz. Find the Young

Question

A 40 cm long wire having a mass 3.2 g and area of cross section 1 mm² is stretched between the supports 40.05 cm apart. In its fundamental mode, it vibrates with a frequency 1000/64 Hz. Find the Young\'s modulus of the wire in the form X × 1 0 9 N/m² and fill the value of X :

TestHub · Accepted Answer

μ = 3.2 × 1 0 − 3 /0.40 = 8 × 1 0 − 3 kg/m. v = 2 L f = 2 × 0.4005 × 15.625 = 12.52 m/s. T = μ v 2 = 8 × 1 0 − 3 × 156.6 = 1.253 N. Strain = 0.05/40 = 1.25 × 1 0 − 3 . Y = T / ( A ⋅ strain ) = 1.253/ ( 1 0 − 6 × 1.25 × 1 0 − 3 ) ≈ 1.0 × 1 0 9 N/m², so X = 1 .

Physics - Wave on String Question with Solution | TestHub

Options:

Doubts & Discussion