Max speed of wave on a string from elastic limit given density

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SUMMARY

The maximum speed of transverse wave pulses in a steel wire is determined by the elastic limit and density of the material. Given that the elastic limit of steel is 2.7 X 109 Pa and its density is 7.86 X 103 kg/m3, the wave speed can be calculated using the formula v = √(T/μ). By relating tension (T) to the elastic limit and density, one can derive the maximum wave speed before exceeding the elastic limit.

PREREQUISITES
  • Understanding of wave mechanics and transverse waves
  • Familiarity with the concepts of tension and mass density
  • Knowledge of the relationship between pressure, force, and area
  • Basic algebra for manipulating equations
NEXT STEPS
  • Calculate wave speed using the formula v = √(T/μ) with given parameters
  • Explore the relationship between tension and elastic limit in materials
  • Investigate the effects of cross-sectional area on wave propagation
  • Learn about different materials and their elastic limits for comparative analysis
USEFUL FOR

Physics students, engineers, and material scientists interested in wave mechanics and the properties of materials under stress.

lizzyb
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How do I set this up? "The elastic limit of a piece of steel wire is 2.7 X 10^9 Pa. What is the maximum speed at which transverse wave pulses can propagate along this wire before this stress is exceeded? (The density of steel is 7.86 X 10^3 kg/m^3)

I know [tex]v = \sqrt{\frac{T}{\mu}}[/tex] so I guess I'd solve for T? And how do I pull the force from the elastic limit without the area?

thanx!
 
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lizzyb said:
How do I set this up? "The elastic limit of a piece of steel wire is 2.7 X 10^9 Pa. What is the maximum speed at which transverse wave pulses can propagate along this wire before this stress is exceeded? (The density of steel is 7.86 X 10^3 kg/m^3)

I know [tex]v = \sqrt{\frac{T}{\mu}}[/tex] so I guess I'd solve for T? And how do I pull the force from the elastic limit without the area?

thanx!
Hi.

Notice that [itex]\rho = { mass \over length \times area}[/itex] and [itex]\mu = {mass \over length}[/itex] so that [itex]\rho = {\mu \over area}[/itex] where, by "area" I mean the cross sectional area.

Also, Pressure = Force over area, so [itex]P_{max} = {T_{max} \over area}[/itex]. With this you should be able to rewrite the speed in terms of the pressure and the volume mass density [itex]\rho[/itex].

I hope this helps

Patrick
 
yes it helped! i hope i remember it! thanks!
 

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