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NVAMediumJEE 2023Superposition Principle & Standing Waves

JEE Physics 2023 Question with Solution

A wire of density 8×103 kg/m38 \times 10^3 \, \text{kg/m}^3 is stretched between two clamps 0.5 m0.5 \, \text{m} apart. The extension developed in the wire is 3.2×10−4 m3.2 \times 10^{-4} \, \text{m}. If Y=8×1010 N/m2Y = 8 \times 10^{10} \, \text{N/m}^2, the fundamental frequency of vibration in the wire will be _____ Hz.

Answer

Correct answer:80

Step-by-step solution

Standard Method

Given: density ρ=8×103 kg/m3\rho = 8 \times 10^3 \, \text{kg/m}^3, length ℓ=0.5 m\ell = 0.5 \, \text{m}, extension Δℓ=3.2×10−4 m\Delta \ell = 3.2 \times 10^{-4} \, \text{m}, and Young's modulus Y=8×1010 N/m2Y = 8 \times 10^{10} \, \text{N/m}^2.

Find: the fundamental frequency of vibration.

For a stretched wire, the fundamental frequency is

f=12ℓTμf = \frac{1}{2\ell}\sqrt{\frac{T}{\mu}}

where TT is the tension and μ\mu is the linear mass density.

Using Young's modulus,

Y=stressstrain=T/AΔℓ/ℓY = \frac{\text{stress}}{\text{strain}} = \frac{T/A}{\Delta \ell/\ell}

so

T=YAΔℓℓT = Y A \frac{\Delta \ell}{\ell}

Also, linear mass density is

μ=ρA\mu = \rho A

Substituting these into the frequency formula,

f=12ℓYA(Δℓ/ℓ)ρAf = \frac{1}{2\ell}\sqrt{\frac{Y A (\Delta \ell/\ell)}{\rho A}}

Hence, AA cancels and we get

f=12ℓYΔℓρℓf = \frac{1}{2\ell}\sqrt{\frac{Y \Delta \ell}{\rho \ell}}

Now substitute the values:

f=12×0.58×1010×3.2×10−48×103×0.5f = \frac{1}{2 \times 0.5}\sqrt{\frac{8 \times 10^{10} \times 3.2 \times 10^{-4}}{8 \times 10^3 \times 0.5}}

Since 12×0.5=1\frac{1}{2 \times 0.5} = 1,

f=8×1010×3.2×10−44×103f = \sqrt{\frac{8 \times 10^{10} \times 3.2 \times 10^{-4}}{4 \times 10^3}}

Simplifying,

f=6.4×103f = \sqrt{6.4 \times 10^3} f=80 Hzf = 80 \, \text{Hz}

Therefore, the fundamental frequency of vibration in the wire is 80 Hz80 \, \text{Hz}.

Why area cancels

Given: the material density and Young's modulus are known, but the cross-sectional area is not given.

Find: how to evaluate frequency without knowing the wire radius or area.

The key observation is that both tension and linear mass density depend on the same cross-sectional area AA:

T=YAΔℓℓ,μ=ρAT = Y A \frac{\Delta \ell}{\ell}, \qquad \mu = \rho A

Therefore,

Tμ=YA(Δℓ/ℓ)ρA=YΔℓρℓ\frac{T}{\mu} = \frac{Y A (\Delta \ell/\ell)}{\rho A} = \frac{Y \Delta \ell}{\rho \ell}

So the unknown area cancels automatically.

Then the frequency becomes

f=12ℓYΔℓρℓf = \frac{1}{2\ell}\sqrt{\frac{Y \Delta \ell}{\rho \ell}}

which depends only on the given quantities. This is why the problem is solvable even though the wire thickness is not provided.

Common mistakes

  • Using μ=ρ\mu = \rho is incorrect because ρ\rho is volume mass density in kg/m3\text{kg/m}^3, whereas μ\mu is linear mass density in kg/m\text{kg/m}. Use μ=ρA\mu = \rho A instead.

  • Substituting the extension as the length or vice versa is wrong. The wire length is ℓ=0.5 m\ell = 0.5 \, \text{m} and the extension is only Δℓ=3.2×10−4 m\Delta \ell = 3.2 \times 10^{-4} \, \text{m}. Keep these quantities distinct in the strain term Δℓ/ℓ\Delta \ell / \ell.

  • Trying to find the cross-sectional area numerically is unnecessary. In T/μT/\mu, the factor AA cancels, so introducing extra assumptions about radius or diameter only complicates the solution.

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