Q.Explain what is meant by the phase and the phase constant of a particle executing S.H.M. Two particles execute S.H.M. of the same amplitude and frequency with a phase difference of $\pi/2$ between them. Describe their relative motion when one of them is at its extreme position.

The phase of an S.H.M. at time $t$ is the argument $(\omega t+\phi)$ of the sine or cosine function describing its motion; it is the single quantity that fixes BOTH the particle's instantaneous displacement and the direction it is currently moving. The phase constant $\phi$ is simply the value of this phase at the instant $t=0$, fixed by the particle's own displacement and velocity at that starting instant.

If two particles execute S.H.M. of the same amplitude $A$ and the same angular frequency $\omega$, but with a phase difference of $\pi/2$ between them -- say $x_1(t)=A\sin(\omega t)$ and $x_2(t)=A\sin(\omega t+\pi/2)=A\cos(\omega t)$ -- then whenever particle 1 reaches an extreme position ($x_1=\pm A$, momentarily at rest), particle 2 is at that very same instant passing through ITS OWN mean position ($x_2=0$) with its maximum speed $A\omega$. Equally, when particle 1 is passing through its own mean position at maximum speed, particle 2 is at that instant sitting at one of its own extreme positions, momentarily at rest. The two particles are thus always a quarter-cycle apart in their motion -- exactly the same relationship that exists between the displacement and the velocity of a SINGLE S.H.M. particle.

Recall the definitions of phase and phase constant, then substitute a $\pi/2$ phase difference into the general S.H.M. expressions and compare the two particles' states at the same instant.

• Confusing a phase difference of $\pi/2$ with one of $\pi$ (which instead makes the two particles always exactly opposite in displacement, not in quadrature).
• Describing the two particles as moving with different amplitudes or frequencies, when a pure phase difference changes neither.