Question

A given metal wire has length 1m, linear density \( 0.6 \, \text{kg/m} \) and uniform cross-sectional area \( 10^{-7} \, \text{m}^2 \) is fixed at both ends. The temperature of wire is decreased by \( 40^\circ C \). The fundamental frequency of the transverse wave is \( \gamma = 2 \times 10^{11} \, \text{N/m}^2 \), coefficient of linear expansion of metal is \( 1.2 \times 10^{-5} / ^\circ C \). The frequency is

• A. 0.5 Hz
• B. 2 Hz
• C. 1 Hz
• D. 2.5 Hz

Hint:

When dealing with frequency on a stretched wire, consider both the tension and the linear density, and remember that temperature change affects the tension.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
The fundamental frequency \( f \) of a transverse wave on a stretched wire is given by the formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] where \( L \) is the length of the wire, \( T \) is the tension, and \( \mu \) is the linear density of the wire. The tension \( T \) is affected by the change in temperature, as the wire expands or contracts. The change in tension is proportional to the change in length due to temperature change. Step 2: Calculating the change in length.
The change in length \( \Delta L \) is given by: \[ \Delta L = L \alpha \Delta T \] where \( \alpha \) is the coefficient of linear expansion and \( \Delta T \) is the temperature change. Substituting values: \[ \Delta L = 1 \times 1.2 \times 10^{-5} \times (-40) = -4.8 \times 10^{-4} \, \text{m} \] Step 3: Finding the fundamental frequency.
Using the values provided and solving for the frequency, we find: \[ f = 2 \, \text{Hz} \] Step 4: Conclusion.
Thus, the fundamental frequency is 2 Hz, corresponding to option (B).