Titus 2/10

Application of Ball Spring Model

• Concept: Using the ball spring model to determine the stiffness of atomic bonds in a solid, specifically copper.

• Current Knowledge:

  • Diameter of a copper atom

  • Number of bonds and springs along a chain

  • Number of atoms in a cross-sectional area.

Model of Chains as Springs

• Chains as Springs: Treat chains as single springs arranged in parallel.

• Effective Stiffness in Parallel:

  • When multiple springs are placed in parallel, their combined stiffness increases.

  • Example Scenario: Stretching one spring vs. two springs in parallel.

    • Stretching two springs required more force, indicating increased stiffness.

  • Mathematical Representation:

    • Effective Stiffness ( k_{eff} = n \cdot k ) (where ( n ) is the number of springs and ( k ) is individual spring stiffness).

Experiment to Demonstrate Stiffness

• Experiment Setup: Two identical springs are stretched together.

• Observation: Requires greater force to stretch both springs the same distance compared to one spring.

• Conclusion: Adding springs in parallel results in greater effective stiffness.

Handling Non-Identical Springs

• Different Stiffnesses: If springs are not identical, their stiffnesses would still be summed.

• Reasoning: For homework and tests, identical springs are primarily considered, simplifying calculations.

Example Problem of Stiffness

• Problem Statement: Hanging a mass on a spring stretches it a certain distance.

• Single Spring Scenario:

  • The original spring stretches ( s = mg/k )

• Parallel Springs:

  • Adding a second spring in parallel reduces the stretch for the same force applied.

  • If stiffness doubles, stretch distance is halved.

Series vs. Parallel Arrangement of Springs

• Effect of Arranging in Series:

  • Series arrangement makes the overall system less stiff.

  • Stretch leads to less force required because each spring stretches only half as much.

  • Equivalent Stiffness: ( k_{eff} = k/n )

Further Example Calculations

• Two Springs Example:

  • Hanging one kg from two springs (series) stretches less than hanging from one. Calculation confirmed that stretch is halved:

  • If ( s = mg/k ), with two parallel springs, the effective stiffness doubles, reducing stretch.

Application to Atomic Bonds

• Task at Hand: Determine the stiffness of a single interatomic bond, using concepts from previous experiments.

• Parameters Provided: Applied force of 98 Newtons stretches wire by 1.51 millimeters.

  • Calculation Framework: ( k = mg/s )

  • Resulting in stiffness calculation as ( k \approx 64900 , ext{N/m} ).

Transition to Parallel Chains in Wire

• Wire as Parallel Chains: Model the wire as multiple parallel chains of springs.

• Effective Stiffness Notes:

  • Effective stiffness of wire given by ( k_{wire} = n \cdot k_{chain} ).

• Number of Parallel Chains Determination:

  • Amount of parallel chains calculated from atomic structure (1.91 x 10^{13} chains).

  • Stiffness of a single chain will be ( k_{chain} = rac{k_{wire}}{n} ).

Conclusion and Next Steps

• Final Step: Divide the wire's stiffness by the number of chains to find a single bond's stiffness.

• Next Class: Continue evaluation of calculations and validate the outcomes.