A wire of density 8 × 10³ kg / m³ is stretched between two clamps 0.5 m apart. The extension developed in the wire is 3.2 × 10⁻⁴ m. If Y=8 × 10¹⁰ N / m², the fundamental frequency of vibration in the wire will be ___________ Hz.

Solution

Physics sir · 4 steps

Get the tension from Young's modulus. The area then cancels when you divide by mass per unit length.

1. Tension from strain. The stretched wire pulls back with T = YA(Δ L)/L. Its mass per unit length is μ = ρ A. The cross-section A is not given, and you do not need it.

   v = √(T/μ) = √((Y A (Δ L/L))/(ρ A)) = √(Y/ρ·(Δ L)/L)

2. Put in numbers. Strain = (3.2 × 10⁻⁴)/0.5 = 6.4 × 10⁻⁴. Then (8 × 10¹⁰ × 6.4 × 10⁻⁴)/(8 × 10³) = 6400 m² s⁻², so v = 80 m s⁻¹.
3. Fundamental mode. With both ends clamped, the fundamental has λ = 2L = 1 m. So f = v/2L = 80/1 Hz.
4. So. The fundamental frequency is 80 Hz. Here f and v have the same number only because 2L = 1 m.

The idea to keep. When a wire's tension comes from its own stretch, the wave speed is v = √(Y × strain/ρ). The cross-section always cancels, so do not go looking for it.

Answer from the NTA final answer key. Solution by PrepWiser.