SHM · Lecture 7 of 26
The phasor diagram: SHM phase questions in one picture
The phasor diagram turns SHM phase questions in JEE into angles on a circle, and gives v max = A ω and a max = A ω².
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What this lecture covers
Uniform circular motion builds the phasor diagram, where the phase of the SHM equation is an angle on the circle and the shadow gives x at every t. Speed ω r and acceleration ω² r towards the centre project to v max = A ω and a max = A ω², peaking at opposite places. Phase π puts the particle at the mean position with v = −A ω, and phase 3 π/2 reads the same way.
In this lecture
- Play from SHM so far: the shadow of the circle
- Play from The phasor diagram and the phase angle
- Play from Setting up the reference circle
- Play from The shadow gives x at every t
- Play from Why we use the circle
- Play from Speed and acceleration on a circle
- Play from Max speed A ω, max a A ω²
- Play from Phase π and phase 3 π/2 on the circle
- Play from A quarter turn takes T/4
- Play from Recap: the phasor diagram