Question

A steel wire of length 1.25 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 (Density and Young's modulus of steel are $7.7 \times 10^3$ kgm$^{-3}$ and $2.2 \times 10^{11}$ Nm$^{-2}$ respectively)

• A. 20 Hz
• B. 40 Hz
• C. 80 Hz
• D. 160 Hz

Hint:

Fundamental frequency of a stretched string: $f_1 = \frac{v}{2L}$, where $v = \sqrt{T_{force}/\mu}$ is the wave speed.
The wave speed can also be expressed as $v = \sqrt{(Y \times \text{Strain})/\rho}$.
Convert percentage strain to a decimal value (e.g., $0.14% = 0.0014$).
Ensure all units are consistent (SI units are standard for these formulas).

Answer 1

The Correct Option is C

The fundamental frequency ($f_1$) of a stretched wire fixed at both ends is given by: $f_1 = \frac{1}{2L} \sqrt{\frac{T_{force}}{\mu}}$ where $L$ is the length of the wire, $T_{force}$ is the tension in the wire (using $T_{force}$ to avoid confusion with period), and $\mu$ is the mass per unit length of the wire. Mass per unit length $\mu = \rho A$, where $\rho$ is the density and $A$ is the cross-sectional area of the wire. So, $f_1 = \frac{1}{2L} \sqrt{\frac{T_{force}}{\rho A}}$. Young's modulus ($Y$) is given by $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T_{force}/A}{\text{Strain}}$. From this, Stress $T_{force}/A = Y \times \text{Strain}$. So $T_{force} = Y \times A \times \text{Strain}$. Substitute this into the frequency equation: $f_1 = \frac{1}{2L} \sqrt{\frac{Y \times A \times \text{Strain}}{\rho A}} = \frac{1}{2L} \sqrt{\frac{Y \times \text{Strain}}{\rho}}$. The term $\sqrt{\frac{Y \times \text{Strain}}{\rho}}$ is the speed of the transverse wave ($v$) on the wire, because $v = \sqrt{\text{Stress}/\rho} = \sqrt{(Y \times \text{Strain})/\rho}$. Given values: Length $L = 1.25 \text{ m}$. Elastic strain $= 0.14% = \frac{0.14}{100} = 0.0014 = 1.4 \times 10^{-3}$. Density of steel $\rho = 7.7 \times 10^3 \text{ kgm}^{-3}$. Young's modulus of steel $Y = 2.2 \times 10^{11} \text{ Nm}^{-2}$. Calculate the term under the square root (which is $v^2$): $\frac{Y \times \text{Strain}}{\rho} = \frac{(2.2 \times 10^{11} \text{ Nm}^{-2}) \times (1.4 \times 10^{-3})}{7.7 \times 10^3 \text{ kgm}^{-3}}$. $= \frac{2.2 \times 1.4}{7.7} \times \frac{10^{11} \times 10^{-3}}{10^3} = \frac{3.08}{7.7} \times 10^{11-3-3} = \frac{3.08}{7.7} \times 10^5$. $\frac{3.08}{7.7} = \frac{308}{770}$. We can simplify this: $\frac{308 \div 77}{770 \div 77} = \frac{4}{10} = 0.4$. So, $\frac{Y \times \text{Strain}}{\rho} = 0.4 \times 10^5 = 4 \times 10^4 \text{ (m/s)}^2$. Now take the square root to find $v$: $v = \sqrt{\frac{Y \times \text{Strain}}{\rho}} = \sqrt{4 \times 10^4} = 2 \times 10^2 = 200 \text{ m/s}$. Finally, calculate the fundamental frequency $f_1$: $f_1 = \frac{v}{2L} = \frac{200 \text{ m/s}}{2 \times 1.25 \text{ m}}$. $2L = 2 \times 1.25 = 2.5 \text{ m}$. $f_1 = \frac{200}{2.5} = \frac{2000}{25}$. $\frac{2000}{25} = \frac{20 \times 100}{25} = 20 \times 4 = 80 \text{ Hz}$. \[ \boxed{80 \text{ Hz}} \]