A wire of density 8 × 10 3 kg m - 3 is stretched between two clamps 0 . 5 m apart. The extension developed in the wire is 3 . 2 × 10 - 4 m . If Y = 8 × 10 10 N m - 2 , the fundamental frequency of vibration in the wire will be _____ Hz

Step-by-step solution

Using the relation of Young's modulus, T A = Y Δ L L     ⇒ T = Y Δ L L × A The linear mass density is   μ = m L . So, T μ = Y Δ L A L m L = Y ( Δ L ) × L A L ( m ) = Y Δ L L × 1 ρ Substituting the values, T μ = 8 × 10 10 × 3 . 2 × 10 - 4 0 . 5 × 1 8 × 10 3 = 6 . 4 × 10 3 ⇒ T μ = 64 × 10 2 The fundamental frequency is given by f = 1 2 L T μ . ⇒ T μ = 8 × 10 = 80 &#16