Question

A uniform wire of length \(L\), diameter \(D\) and density \(\rho\) is stretched under tension \(T\). The relation between its fundamental frequency \(f\), length \(L\) and diameter \(D\) is

• A. \(f\propto \frac{1}{LD}\)
• B. \(f\propto \frac{1}{\ell \sqrt{D}}\)
• C. \(f\propto \frac{1}{D^{2}}\)
• D. \(f\propto \frac{1}{\ell D^{2}}\)

Hint:

Write the linear density in terms of diameter and density, then substitute into the fundamental frequency formula for a given tension.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The fundamental frequency is \(f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}\), where \(\mu\) is the mass per unit length.

Step 2: Linear Density
\(\mu=\rho\cdot\frac{\pi D^{2}}{4}\).

Step 3: Substitute
\(f=\frac{1}{2L}\sqrt{\frac{4T}{\rho\pi D^{2}}}=\frac{1}{LD}\sqrt{\frac{T}{\pi\rho}}\).

Step 4: Compare the options
For fixed \(T\) and \(\rho\), \(f\propto\frac{1}{LD}\).

Final Answer:
The frequency is inversely proportional to the product of length and diameter. \[ \boxed{\text{(A) }f\propto\dfrac{1}{LD}} \]