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

Question

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?

Answer 1

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 $\delta$

Answer 2

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.