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. Numerical answer type. Work it out, then open the answer.

Answer: 80. Stress T/A = Y × strain = 8×10¹⁰ × 3.2×10⁻⁴0.5 = 5.12×10⁷ N/m² v = √(T/μ) = √((T/A)/(ρ)) = √((5.12×10⁷)/(8×10³)) = 80 m/s n = v/2L = (80)/(2×0.5) = 80 Hz

A wire of density 8 × 10³ kg/m³ is stretched between two clamps 0.5 m… | JEE Main 2023 (11 Apr, Shift 2)

Q69JEE Main 2023 · 11 APR · SHIFT 2Medium

Stationary Waves & Vibrating Strings

A wire of density $8 \times 10^{3}\text{ kg/m}^3$ is stretched between two clamps $0.5\text{ m}$ apart. The extension developed in the wire is $3.2 \times 10^{-4}\text{ m}$ . If $Y = 8 \times 10^{10}\text{ N/m}^2$ , the fundamental frequency of vibration in the wire will be ___________ $\text{Hz}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 80

Stress $\dfrac{T}{A} = Y \times \text{strain} = 8\times10^{10} \times \dfrac{3.2\times10^{-4}}{0.5} = 5.12\times10^{7}\text{ N/m}^2$
$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{T/A}{\rho}} = \sqrt{\dfrac{5.12\times10^7}{8\times10^3}} = 80\text{ m/s}$
$n = \dfrac{v}{2L} = \dfrac{80}{2\times0.5} = 80\text{ Hz}$

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Stationary Waves & Vibrating Strings question from JEE Main 2023 (11 Apr, Shift 2)

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