# Does changing phase constant also changes the mean position along with other things?

I am working on two different SHMs: $$P=a\sin\omega t \\ Q=a\sin(\omega t +\phi)$$ where $$\omega$$ = angular velocity, $$\phi$$ = phase constant, $$P,Q$$ = displacement at a instant, $$a$$ = amplitude

Now due to phase constant the mean position is changed.

Question: Does the positions ($$Q$$) where the maximum acceleration, zero acceleration, max velocity and zero velocity occur for the first time also change with respect to $$P$$?

The only difference between the $$P=a\sin\omega t$$ and the $$Q=a\sin(\omega t +\phi)$$ description is that you are observing the same motion but starting the clock, setting $$t=0$$, at difference parts of the motion.

So when $$t=0$$ you have $$P=0$$ and $$Q=a\sin \phi$$ and you might say that motion $$Q$$ is phase $$\phi$$ ahead of motion $$P$$ but the mean position over a period is zero for both descriptions of the motion.

The same argument can be followed through for velocity and acceleration.

If you have an object undergoing simple harmonic motion with a position $$x(t)$$ given by $$x(t) = A \sin (\omega t + \phi),$$

then the velocity of the object is: $$v(t) = \frac{\text{d}x}{\text{d}t} = A \omega \,\cos(\omega t + \phi),$$

and the acceleration is: $$a(t) = \frac{\text{d}v}{\text{d}t} = - A \omega^2 \, \sin(\omega t + \phi),$$

as it should be.

In your case, you have two different "position" functions, one for the oscillator $$P$$ and one for the oscillator $$Q$$. You can use the above formulae to calculate the quantities you are asked to calculate, and find if they are the same, or different.