# Initial phase of Simple Harmonic Oscillator

Starting with the following solution for SHO: $$x(t) = A\sin(\omega t+\varphi)\Rightarrow x_{0}=A\sin(\varphi)$$ $$\dot{x}(t) = v(t) = \omega A\cos(\omega t+\varphi) \Rightarrow v_{0}=\omega A\cos(\varphi)$$ I get the following initial phase: $$\Rightarrow \tan(\varphi) = \frac{x_{0} \omega}{v_{0}}$$

But starting with: $$x(t) = A\cos(\omega t+\varphi)\Rightarrow x_{0}=A\cos(\varphi)$$ which is a also a valid solution for the differential equation of the SMO, $$\dot{x}(t) = v(t) = -\omega A\sin(\omega t+\varphi) \Rightarrow v_{0}=-\omega A\sin(\varphi)$$ I get this initial phase: $$\Rightarrow \tan(\varphi) = -\frac{v_{0}}{x_{0} \omega}$$

Which one should be used? Are both correct?

• If you are describing the motion with two different functions, then the phases at t = 0 will not be the same. Mar 26 at 15:09

Congratulations, you have discovered an identity. Since $$\sin\left(x-{\pi\over2}\right)=-\cos x,\\ \cos\left(x-{\pi\over2}\right)=\sin x,$$ I would write this identity as, $$\tan\left(x-{\pi\over2}\right)={\sin\left(x-{\pi\over2}\right) \over\cos\left(x-{\pi\over2}\right)} =-\cot x,$$but of course you have found $$-1/\tan x$$ which is a valid expression for the same number.