# How to find the initial phase and amplitude of a particle undergoing SHM when I know the initial position and velocity?

According to my book, we can find the amplitude $$A$$ and the initial phase $$\delta$$ if we know the initial displacement $$x$$ and the velocity of the particle at $$x$$, $$v$$. However, my book doesn't give any example of this concept. Still, I understand what my book is trying to say. Suppose, the initial displacement is $$5m$$ and the velocity is $$0ms^{-1}$$. Voila! we can simply deduct from the information given that the amplitude is $$5m$$ since the velocity is $$0$$ at the amplitude. Furthermore, putting the values in $$x=A\sin(\omega t+\delta)\implies 5=5\sin\delta\implies \delta=\frac{\pi}{2}$$ we can find the initial phase $$\delta$$ as well. However, I'm unable to solve more complex problems.

If the initial displacement is $$5m$$ and the velocity of the object is $$5ms^{-1}$$ at that point, how will I be able to find the amplitude and the initial phase?

The general equation for position of a particle performing SHM is of type

$$x=A\sin(\omega t+\delta)\tag{1}$$

Let initial position be $$\alpha$$, therefore

$$\alpha=A\sin(\delta)\tag{2}$$

Let velocity of particle at $$x=\alpha$$ be $$\beta$$

$$\beta=A\omega\cos(\delta)\tag{3}$$

Now you have two equations and two unknowns. Solve them and you may find $$A$$ and $$\delta$$

• I don't have two unknowns; I have three: $A$, $\delta$ & $\omega$. How do I find the value of $\omega$? Commented Jul 17, 2021 at 15:31
• @AbuSafwan I think you will be provided that Commented Jul 17, 2021 at 16:00
• @AbuSafwan Once you have $\delta$ you can divide equations (2) and (3) to obtain $\alpha / \beta = \tan(\delta) / \omega$, which you can solve for $\omega$.
– ummg
Commented Jul 17, 2021 at 17:15
• @ummg but how to find $\delta$ Commented Jul 17, 2021 at 17:32
• @lalittolani Oh, oops. Read this to quickly. We need to know one of the parameters $A$, $\delta$, $\omega$ before hand (or have some other constraint). Then we can solve for the other two.
– ummg
Commented Jul 17, 2021 at 19:14

You have already the equation for the position. $$x(t)=A\sin(\omega t+\delta) \tag{1}$$

From this you get the velocity by differentiating with respect to time $$t$$. $$v(t)=\dot{x}(t)=A\omega\cos(\omega t+\delta) \tag{2}$$

From equations (1) and (2) you get by putting $$t=0$$: $$x(0)=A\sin(\delta) \tag{3}$$ $$v(0)=A\omega\cos(\delta) \tag{4}$$

Now you have two equations for two unknowns ($$A$$ and $$\delta$$). So now you need to resolve equations (3) and (4) for $$A$$ and $$\delta$$. I will leave this to you as an exercise.

• But in (3), I don't know the value of $\omega$ Commented Jul 17, 2021 at 15:16