Oscillations
Force law and energy in SHM
Find periods of spring systems and the kinetic and potential energies in SHM.
What makes a body move in SHM? A restoring force that is proportional to the displacement and always points back towards the mean: F = –kx.
- The minus sign means the force is always opposite to the displacement: pull the body right and it is pushed left; push it left and it is pushed right.
- A force like –kx² would not do: x² is positive on both sides, so the force would point the same way on both sides, and the body would never come back.
Since F = ma, a = –(k/m)x. Compare with a = –ω²x: ω = √(k/m) and T = 2π√(m/k). The test for SHM is therefore simple: a must equal –(a positive constant) × x. Then the period is T = 2π√|x/a|.
A mass on a spring is the standard example. For a 0.5 kg block on a 200 N m⁻¹ spring, ω = √(200/0.5) = 20 rad s⁻¹ and T = 2π/20 ≈ 0.31 s. The period does not depend on how far you pull it.
Hang the spring vertically and gravity cancels out. At rest the spring is stretched by d, with kd = mg; from there, an extra stretch x gives a net restoring force of exactly –kx, so T is still 2π√(m/k). If a hanging mass stretches its spring by 2.5 cm, then k/m = g/d and T = 2π√(0.025/9.8) ≈ 0.32 s.
Springs can be combined. The sure test is what is shared:
- Parallel (they share the same stretch): k = k₁ + k₂. A block between two springs, one on each side, is parallel too: as one stretches, the other squeezes by the same amount.
- Series (end to end, so the same force passes through each): 1/k = 1/k₁ + 1/k₂.
- With 100 and 300 N m⁻¹ springs and a 2 kg block: parallel gives 400 N m⁻¹ and T ≈ 0.44 s; series gives 75 N m⁻¹ and T ≈ 1.03 s.
Energy in SHM: the potential energy is U = ½kx² and the kinetic energy is K = ½k(A² – x²). Their sum is E = ½kA², which stays constant when there is no friction.
- At the mean position the energy is all kinetic; at the extremes it is all potential.
- K and U each go through their whole cycle twice per oscillation, so each has period T/2.
- Example: the 0.5 kg block on the 200 N m⁻¹ spring with A = 6 cm has E = ½ × 200 × 0.06² = 0.36 J. At x = 3 cm, U = 0.09 J and K = 0.27 J.
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