Question

A wire of density \(8 × 10^3\) \(kg/m^3\) is stretched between two clamps 0.5 m apart. The extension developed in the wire is \(3.2 × 10^{–4}\) m. If Y = \(8 × 10^{10}\) \(N/m^2\) , the fundamental frequency of vibration in the wire will be ________Hz.

Answer 1

Correct Answer: 80

Step 1: Formula for Frequency 

The frequency of a vibrating string is given by:

\[ f = \frac{1}{2\ell} \sqrt{\frac{T}{\mu}}, \]

where:

• \( \ell \): Length of the string
• \( T \): Tension in the string
• \( \mu \): Linear mass density of the string.

Step 2: Substitute Tension in Terms of Young’s Modulus

The tension \( T \) can be expressed in terms of Young’s modulus (\( Y \)) as:

\[ T = Y \cdot A \cdot \frac{\Delta \ell}{\ell}, \]

where:

• \( Y \): Young’s modulus
• \( A \): Cross-sectional area of the string
• \( \Delta \ell \): Extension of the string
• \( \ell \): Original length of the string.

Substitute \( T \) into the frequency formula:

\[ f = \frac{1}{2\ell} \sqrt{\frac{Y \cdot A \cdot \Delta \ell}{\ell \cdot \mu}}. \]

Step 3: Substitute the Given Values

Substitute the numerical values:

• \( \ell = 0.5 \, \text{m} \)
• \( Y = 8 \times 10^{10} \, \text{Pa} \)
• \( A = 3.2 \times 10^{-4} \, \text{m}^2 \)
• \( \mu = 8 \times 10^{-3} \, \text{kg/m} \)
• \( \Delta \ell = 0.5 \, \text{m} \).

Substitute into the formula:

\[ f = \frac{1}{2 \cdot 0.5} \sqrt{\frac{8 \times 10^{10} \cdot 3.2 \times 10^{-4} \cdot 0.5}{0.5 \cdot 8 \times 10^{-3}}}. \]

Step 4: Simplify the Expression

Simplify the terms inside the square root:

\[ f = 1 \cdot \sqrt{\frac{8 \times 10^{10} \cdot 3.2 \times 10^{-4} \cdot 0.5}{4 \times 10^{-3}}}. \]

\[ f = \sqrt{\frac{1.28 \times 10^{8}}{4 \times 10^{-3}}}. \]

\[ f = \sqrt{3.2 \times 10^{10}}. \]

Simplify further:

\[ f = 80 \, \text{Hz}. \]

Final Answer:

The frequency of the vibrating string is \( f = 80 \, \text{Hz} \).