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Hooke's Law

Extension and Compression
  • T = kx
  • T is the magnitude of tension
  • x is the extension or compression
  • k is the spring constant
  • Combining spring constants k:
    • Springs in parallel:

      • kT = k1 + k2 + ⋅⋅⋅
    • Springs in series:

      • 1/kT = 1/k1 + 1/k2 + ⋅⋅⋅
Modulus of Elasticity
  • T = λx/l
  • λ is the modulus of elasticity
  • l is the natural length
  • Relating the spring constant and the modulus of elasticity: λ = kl
Scenarios
  • If a mass is hanging at one end of a spring with the other attached to a fixed point, the tension in the spring must be equal to the weight of the object
  • If two springs are attached between two fixed points and both are extended, the tension in both must be the same in order for there to be no overall net force
  • If two springs are attached to fixed points and an object in the middle, tensions and frictional force must act in such a way that the overall force on the mass is equal to 0
  • If a spring is attached to an object on a rough surface, the frictional force acts in the direction opposing tension in the spring (preventing it to return to its original shape)
  • If a mass is on an incline and held at rest by a spring, the tension in the spring must be equal to the component of the weight parallel to the slope
Elastic Potential Energy
  • Work done in stretching (or compressing) a string or spring is given by W = ½kx²
  • For a scenario, form an equation by conservation of energy and you can include elastic potential energy, kinetic energy, gravitational potential energy, and work done against friction
Examples
  1. A particle P of mass 0.35 kg is attached to the mid-point of a light elastic string of natural length 4 m. Ends of the string are attached to fixed points A and B. P hangs in equilibrium 0.7 m vertically below the mid-point M of AB

    1. Find the tension in the string and hence show that the modulus of elasticity of the string is 25 N
      • Find the extension of the string by finding length AP and PB using right angled triangles:
        • AP = PB = √(2.4² + 0.7²) = 2.5 m
        • 2.5 + 2.5 – 4 = 1 m
      • The vertical component of tension in the string is equal to the weight as the system is in equilibrium:
        • 2T cos θ = mg
        • 2T × (0.7/2.5) = 0.35 × 10
        • ∴ T = 6.25
      • Find the modulus of elasticity by using the calculated information:
        • T = λx/l
        • λ = 6.25 × 4 = 25
    2. P is now held at 1.8 m vertically below M, and released. Find the speed with which P passes through M
      • Find the spring constant k:
        • k = λ/l = 25/4 = 6.25
      • Find the extension of the string
        • AP = PB = √(2.4² + 1.8²) = 3 m
        • 3 + 3 – 4 = 2 m
      • Form an equation by the conservation of energy:
        • e.p.e. = gain in k.e. + gain in g.p.e + e.p.e at M
        • (½ × 6.25 × 2²) = (½ × 0.35 × v²) + (0.35 × 10 × 1.8) + (½ × 6.25 × 0.8²)
        • 12.5 = 0.175v² + 6.3 + 2
        • v² = 24
        • ∴ v = ±4.90 ms-1
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