TITLE: Find $\varphi_{1}$ from $q_{1}=A_{1}\sin(\omega t+\varphi_{1})$
QUESTION [0 upvotes]: I have $q_{1}=A_{1}\sin(\omega t+\varphi_{1})$ where $q_{1}=0$ and $\dot q_{1}=v_{0}$ and I must find $\varphi_{1}$.
I know that $\varphi_{1}$ must be zero but I must demonstrate it first. Any help, please?

I said that $$q_{1}=A_{1}\sin(\omega t+\varphi_{1}) \implies  A_{1}\sin(\omega t+\varphi_{1})=0 \tag 1 $$
$$\dot q_{1}=v_{0} \implies \dot q_{1}=A_{1}\cos(\omega t+\varphi_{1})\omega \implies A_{1}\cos(\omega t+\varphi_{1})\omega =v_{0}\tag 2$$
I multiply the first equation with $\omega$ and than I made sum of the  $(1)$ and $(2)$ equations (the quadric equation) knowing that $q_{1}=0$ and $\dot q_{1}=v_{0}$. But my $\varphi$ it's not going to be zero. Any help will be very well received.

REPLY [0 votes]: You must make some assumption on $v_{0}$, otherwise your second equation (2) is useless, it doesn't tell us anything. I have assumed $v_{0} \neq 0$ in my answer below.
From (1), $A_{1} \sin(\varphi_{1}) = 0 \Leftrightarrow \left\{ \begin{array}{c} A_{1} = 0 \\ \text{or } \varphi_{1} = 0 \text{ (mod $\pi$)}. \end{array} \right.$
From (2), $\omega A_{1} \cos(\varphi_{1}) = v_{0} \neq 0 \Leftrightarrow \left\{ \begin{array}{c} \omega \neq 0 \\ \text{and } A_{1} \neq 0 \\ \text{and } \varphi_{1} \neq \frac{\pi }{2} \text{ (mod $\pi$)}. \end{array} \right.$
Since $A_{1} \neq 0 $ must be satisfied in (2), we must have $\varphi_{1} = 0 \text{ (mod $\pi$)}$ in order to satisfy (1).
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