Question

A steel wire of length \(1.25\ \text{m}\) is stretched between two rigid supports. The tension in the wire produces an elastic strain of \(0.14\%\). The fundamental frequency of the wire is
\[ (\text{Density and Young's modulus of steel are }7.7\times10^3\ \text{kg m}^{-3}\text{ and }2.2\times10^{11}\ \text{N m}^{-2}\text{ respectively}) \]

• A. \(20\ \text{Hz}\)
• B. \(40\ \text{Hz}\)
• C. \(80\ \text{Hz}\)
• D. \(160\ \text{Hz}\)

Hint:

For a stretched wire, \[ f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}. \] If strain is given, use \[ \frac{T}{\mu}=\frac{Y\times \text{strain}}{\rho}. \] This eliminates the need to know the cross-sectional area of the wire.

Answer 1

The Correct Option is C

Step 1: Find wave speed in wire using elastic strain and density.
Elastic strain \( = \Delta L/L = 0.14\% = 1.4\times10^{-3} \). Stress \( = Y\times\text{strain} = 2.2\times10^{11}\times1.4\times10^{-3} = 3.08\times10^8\text{ N/m}^2 \).
\( v = \sqrt{\frac{Y\times\text{strain}}{\rho}} = \sqrt{\frac{3.08\times10^8}{7.7\times10^3}} = \sqrt{4\times10^4} = 200\text{ m/s} \)

Step 2: Fundamental frequency.
\( f = \frac{v}{2L} = \frac{200}{2\times1.25} = 80\text{ Hz} \)

\[ \boxed{f = 80\text{ Hz}} \]