Calculating strain from wave speed and tension in a wire

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Homework Help Overview

The discussion revolves around calculating strain in an aluminum wire based on wave speed and tension. The problem involves understanding the relationship between wave speed, tension, and material properties such as Young's modulus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate strain using given parameters and equations related to wave speed and Young's modulus. Some participants question the correctness of unit conversions and the application of formulas.

Discussion Status

Participants are actively checking calculations and providing feedback on unit consistency. There is a progression in the discussion as participants refine their understanding of the relationships between the variables involved.

Contextual Notes

Participants emphasize the importance of including units in calculations to avoid mistakes. The original poster's assumptions about the wire's conditions and properties are also under consideration.

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Homework Statement



Just wanted to check my work on this one.

An aluminum wire is clamped at each end under zero tension at room temperature. The tension in the wire is increased by reducing the temperature which results in a decrease in the wire's equilibrium length. What strain (ΔL/L) will result in a transverse wave speed of 100 m/s? Take the cross-sectional area of the wire to be 5.0 x 10-6 m2. The density of aluminum is ρ = 2.7 x 103 kg/m3 and Young's modulus is Y = 6.8 x 1010 N/m2.

Homework Equations


[/B]
v = √(τ/μ), F/A = E(ΔL/L)

The Attempt at a Solution


[/B]
100 m/s = √(τ/μ) μ = ρ x A = 2.7 x 103 kg/m3 x 5.0 x 10-6 m2 = 1.35 x 10-2 kg/m. τ = v2μ = 1 x 104 m2/s2 x 1.35 x 10-2 kg/m = 135 N = F/A. F/A = E(ΔL/L), 135 N = 6.8 x 1010(ΔL/L). ΔL/L = 2.0 x 10-9.

Does this look correct?
 
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135 N ≠ F/A

Check your units.
 
Oh, right...F/A would be 135 N / 5.0 x 10-6 m2 = 2.7 x 107 N/m2?
 
Correct.

I always strongly recommend to include all your units in all your calculations. That way you can catch easy mistakes like this.
 
So ΔL/L = 4.0 x 10-4?
 
Looks about right.
 

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