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The phase and phase constant in a displacement time equation show from where the particle has started.

In my school textbook, first the displacement equation was given as :- $$x= A\sin(\omega t+\phi)$$

where $\phi$ is the phase constant.

But then it said if the particle is at extreme position then we add $\pi/2$ because obviously displacement is maximum at $\pi/2$ So now the equation at extreme should be :- $$x=A\sin\left(\omega t+\frac{\pi}{2}\right)$$ $$x=A\cos(\omega t)$$

But in my textbook the equation is :- $$x=A\cos(\omega t + \phi')$$

It says that $\phi '$ is another arbitrary constant. But technically $\phi$ is $\sin ^{-1} (x/A)$, here $x$ will be $A$ and we get $\pi/2$ so no constant remains. But what is this $\phi '$ constant and on which thing it depends?

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3 Answers 3

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OK, you already understood the most important part. This is that $$ \sin{(wt + \frac{\pi}{2})} = \cos{(wt)} $$ This implies, that the two following equations are equally valid

  • $x(t) = A \sin{(wt + \phi)}$ or
  • $x(t) = A \cos{(wt + \phi^\prime)}$

to define the position of an oscillator. If we like to start at the max. amplitude at $t=0$ we can either write

  • $x(t) = A \sin{(wt + \pi/2)}$, where we used $\phi = \pi/2$ or
  • $x(t) = A \cos{(wt)}$, where we used $\phi^\prime = 0$.

Thus, no matter what, you will always obtain $\phi = \phi^\prime + \pi/2$.

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    $\begingroup$ Please trade your $w\to\omega$ to make it nice... $\endgroup$ Commented Feb 2, 2020 at 14:43
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    $\begingroup$ @ZeroTheHero: Thank you very much for your compliment, but I don't feel that my $w$ are ugly :) Furthermore, I don't want to bring this answer to the top of the stack just by eliminating this minor "defective appearance". $\endgroup$
    – Semoi
    Commented Feb 2, 2020 at 15:11
  • $\begingroup$ the OP did use $\omega$, not $w$. Anyways agreed it's not worth sending back to the top now. $\endgroup$ Commented Feb 2, 2020 at 15:23
  • $\begingroup$ @ZeroTheHero FWIW, the OP used non-MathJax $\Omega$ which I edited to $\omega$. $\endgroup$
    – user253029
    Commented Feb 2, 2020 at 15:30
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    $\begingroup$ @FakeMod much nicer with $\omega$ or $\Omega$... :D $\endgroup$ Commented Feb 2, 2020 at 15:31
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It depends on the initial conditions. If its at maximum displacement at $t=0$ then the equation is $x=A\cos{\omega t}$. If it is at equilibrium position and maximum velocity at $t=0$ then $x=A\sin{\omega t}$. In general, the solution is $x=a\cos{\omega t} + b\sin{\omega t}$ which simplifies to either $A\cos{(\omega t + \phi ')}$ or $A\sin{(\omega t +\phi)}$ using the harmonic addition formula.

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At $t=0$ the displacement $x$ is not necessarily $0$: $$ x(0)=A\sin(\phi) \tag{1} $$ In addition, the velocity at $t=0$ $$ \dot{x}(0)=\omega A \cos(\phi) \tag{2} $$ (1) and (2) are two equations for your 2 unknowns $A$ and $\phi$. Thus, as you alluded to $$ \frac{x(0)}{\dot x(0)}= \omega\tan(\phi) $$ from which you can determine $\phi$, and plug it back into either (1) or (2) to obtain $A$ if you need the amplitude.

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