← all cardsTopic 10 · Oscillations · Equations & the two systems
Equations of motion — pick by the start
Starts at equilibrium: x = x₀ sin ωt. Starts at an extreme: x = x₀ cos ωt. Differentiate for v, again for a. Calculator in radians, always.
x = x₀ sin ωt → v = ωx₀ cos ωt → a = −ω²x₀ sin ωtWhat this actually means
The choice of sin or cos is set entirely by where the motion is at t = 0. Starting at equilibrium: sin (begins at zero, rising). Starting at an extreme: cos (begins at maximum).
Differentiate to descend the ladder from x to v to a. Each step multiplies the amplitude by ω and shifts the phase by a quarter cycle.
Radians, always. Degree mode silently wrecks every substituted value, and the error propagates through the whole question.
The trap
Choosing sin vs cos without checking the t = 0 position — every later part inherits the error.
Prove it — watch it be true
- Set the start-convention toggle to cos: the x–t graph opens at a peak — the extreme-release case
- Flip the toggle to sin: the identical motion now opens at zero and rising — the equilibrium-release case
- In either setting take gradients down the x → v → a graphs with the tangent construction: each step multiplies the amplitude by ω and shifts a quarter cycle