Two wires of steel of the same length are stretched on a sonometer. The tension in the first and second wire are 8 Kg-wt and 2 kg-wt respectively. Find the ratio of the fundamental notes emitted by two wires when the diameter of cross-section of the first wire is half that of the second wire.?

Question

Answer

Understanding the Problem
To find the ratio of the fundamental frequencies emitted by two steel wires under different tensions and diameters, we need to consider the physical properties that affect the frequency of vibration.
Key Parameters
- Tension (T): The force applied to the wire.
- Length (L): The same for both wires.
- Mass per unit length (μ): This depends on the diameter and density of the wire.
- Fundamental frequency (f): Given by the formula f = (1/2L) * √(T/μ).
Calculating Mass per Unit Length
- For the first wire (d1 = 1/2 d2):
- μ1 = (π/4) * (d1^2) * ρ
- μ1 = (π/4) * (1/4 d2^2) * ρ = (π/16) * d2^2 * ρ
- For the second wire:
- μ2 = (π/4) * (d2^2) * ρ
Calculating Frequencies
- Frequency of the first wire:
- f1 = (1/2L) * √(T1/μ1)
- f1 = (1/2L) * √(8g/((π/16) * d2^2 * ρ))
- Frequency of the second wire:
- f2 = (1/2L) * √(T2/μ2)
- f2 = (1/2L) * √(2g/((π/4) * d2^2 * ρ))
Finding the Ratio of Frequencies
- The ratio of frequencies is given by:
- f1/f2 = √[(T1 * μ2)/(T2 * μ1)]
Substituting the values:
- f1/f2 = √[(8g * (π/4) * d2^2 * ρ)/(2g * (π/16) * d2^2 * ρ)]
This simplifies to:
- f1/f2 = √[(8 * (4))/(2 * 1)] = √(16) = 4
Conclusion
Thus, the ratio of the fundamental notes emitted by the two wires is 4:1.

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