Phase of a sine wave from a plot How can I find the phase of a sine wave $y=A\cdot \sin(\omega_{n}t+\phi)$ from its plot?

Its phase is $\phi=89^{\circ}$ but how do I find it?
 A: The phase is the distance that the rising zero-crossing is moved to the left of the $y$-axis.
In your example we can't see anything to the left of $x=0$, but instead we can find the next rising zero-crossing and subtract from $2\pi$.
In your graph it looks like there are rising zero-crossings at about $x=450$ and $x=1100$ (though it is hard to read them precisely on that graph). So a full wave of length $1100-450=650$ corresponds to $2\pi$ of phase and the phase offset of the curve is then given by
$$ 450\frac{2\pi}{1100-450} + \phi = 2\pi $$
or in other words
$$ \phi = 2\pi( 1-\frac{450}{1100-450} ) \approx 1.93 \approx 110^\circ $$
The fact that we don't get $89^\circ$ is due to errors in estimating the zero crossings at 450 and 1100. Using an actual ruler instead of just eyeballing as I did would improve precision.
If you want the phase in degrees, you can just use $360^\circ$ instead of $2\pi$ during the entire calculation.
A: You can find the phase by checking by how much degrees the 'original sine curve' is shifted to the right.
The original sine curve: $sin(\phi)$ is $0$ when $\phi$ is $0$
This one is $0$ at $89$ degrees, hence, that is the phase
A: Concentrate just on your formula $f(t)=A\cdot \sin(\omega_{n}t+\phi)$. $A$ is the peak of the curve, so it is evident that $A=10 \cdot 10^{-5}$. Also, you can see from the graph that 
$$f(0) = 10 \cdot {10^{ - 5}}\,\,\, \to \,\,\,\sin \left( \phi  \right) = 1\,\,\, \to \,\,\,\phi  = {\pi  \over 2}\,\,\,\,or\,\,\,\,\phi  = {90^ \circ }$$
${1^ \circ }$ error is due to some reading-error from the graph; however, it seems acceptable. 
