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

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Waves & Sound

A wire of density

8 \times 1 0^{3} kg/m^{3}

is stretched between two clamps 0.5 m apart. The extension developed in the wire is

3.2 \times 1 0^{- 4} m

. If

Y = 8 \times 1 0^{10} N/m^{2}

, the fundamental frequency of vibration in the wire will be

____

Hz.

Answer: 80

The fundamental frequency of a stretched wire is

f = \frac{1}{2 L} \frac{T}{μ}

, where

T

is tension and

μ

is linear mass density. The tension is

T = Y A \frac{Δ L}{L}

and

μ = ρ A

. Thus

f = \frac{1}{2 L} \frac{Y Δ L}{ρ L} = \frac{1}{2 \times 0.5} \frac{8 \times 1 0 ^{10} \times 3.2 \times 1 0 ^{- 4}}{8 \times 1 0 ^{3} \times 0.5} = \frac{8 \times 3.2 \times 1 0 ^{6}}{8 \times 0.5} = 6400 = 80

Hz.

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Integer answer