Youngs Modulus and Standing Waves combined question

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SUMMARY

The discussion focuses on calculating the lowest frequency that generates a standing wave pattern in a compound wire made of aluminum and steel. The aluminum wire has a Young's Modulus of 7x1010 Pa and a density of 2600 kg/m3, while the steel wire has a Young's Modulus of 20x1010 Pa and a density of 7.80 g/cm3. The user attempts to equate the forces in both wires using the formula Y = (F/A)/(ΔL/L) but struggles with the change in length for each wire. The discussion emphasizes the importance of calculating linear mass density (μ) from volume mass density (ρ) and cross-sectional area (A) to find the frequency.

PREREQUISITES
  • Understanding of Young's Modulus and its application in material science.
  • Knowledge of wave mechanics, specifically standing waves and frequency calculations.
  • Familiarity with linear mass density (μ) and its relationship with mass, density, and length.
  • Basic algebra and physics principles related to forces in materials.
NEXT STEPS
  • Calculate the linear mass density (μ) for both aluminum and steel wires using the formula μ = ρA.
  • Explore the relationship between frequency, wave velocity, and wavelength in standing waves.
  • Investigate the effect of varying cross-sectional area on the frequency of standing waves.
  • Learn about the principles of wave interference and nodes in standing wave patterns.
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Students and professionals in physics, particularly those studying wave mechanics, material properties, and engineering applications involving composite materials.

phoebz
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The figure shows an aluminum wire of length L 1 = 60.0 cm, cross-sectional area 1.00 x 10-2 cm2, and density 2600 kg/m3, joined to a steel wire of density 7.80 g/cm3 and the same cross-sectional area. The compound wire, loaded with a block of mass 10.0 kg, is arranged so that the distance L 2from the joint to the supporting pulley is 86.6 cm. Transverse waves are set up on the wire by an external source of variable frequency; a node is located at the pulley. [Hint: You may want to calculate the ratio of the number of loops in the steel compared to the aluminum. Then surmise which numbers will give you the desired lowest frequency.]

-What is the lowest frequency that generates a standing wave pattern that has the joint as one of the nodes?

*Youngs Modulus --> Aluminium: 7x10^10
Steel: 20x10^10 kg/m^3

My Attempt:

Y= (F/A)/(ΔL/L)
-The force put on both strings will be equal so I tried setting the two wire equations equal as
YAΔL/L (for aluminium) = YAΔL/L (for steel)

I was trying to find the length of the other string but I got stumped because I did have the change in length for either string.

I needed the length of each string for the equation μ=mass/length because v=√(F/μ) and frequency equals velocity/λ.

I'm not sure I even knew where I was going with this. Please help!
 

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You can get the linear mass density μ from the volume mass density ρ and the cross-sectional area A. Consider a section of a wire of length L. Can you find an expression (in symbols, not numbers) for the mass of the section in terms of ρ, A, and L? Use this expression for the mass to determine an expression for μ in terms of ρ and A.
 

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