A steel wire of diameter 8 × 10⁻⁴ m is fixed at one end and wrapped around a tuning peg of diameter 5 mm at the other. The vibrating length is 0.06 m. Initially the wire is taut but unstressed. The peg is turned by one-quarter turn. Find the fundamental frequency. Take ρ = 7800 kg m⁻³ and Y = 20 × 10¹⁰ N m⁻².
Physics · Wave Motion
JEE Mains Advanced

Question

A steel wire of diameter 8 × 10⁻⁴ m is fixed at one end and wrapped around a tuning peg of diameter 5 mm at the other. The vibrating length is 0.06 m. Initially the wire is taut but unstressed. The peg is turned by one-quarter turn. Find the fundamental frequency. Take ρ = 7800 kg m⁻³ and Y = 20 × 10¹⁰ N m⁻².

Physics · Wave Motion

G1 Physics · Accepted Answer

Step 1: Quarter turn of a 5 mm diameter peg increases length by ΔL = (¼)πd = π(0.005)/4. Step 2: Strain = ΔL/L and stress = Y·strain; tension is stress×area. Step 3: With μ=ρA and f₁=(1/2L)√(T/μ), the area cancels and f₁ = (1/2L)√[YΔL/(ρL)] ≈ 1.08×10⁴ Hz.