A short extract from my textbook (which I'm finding a little hard to believe) is as follows:

If the general solution to the differential equation of motion of a simple pendulum is given by $$\theta=A\cos(\omega t)+B\sin(\omega t)\tag{1}$$ and initial conditions are $$\theta=\theta_0\quad\text{and}\quad\omega=0\qquad\qquad\text{when}\quad t=0$$

then equation $(1)$ reduces to $$\theta=\theta_0\cos(\omega t)\tag{2}$$

To check $(2)$ is correct I substitute the initial conditions into $(1)$: $$\begin{align}\theta_0 &=A\cos(0\cdot 0)+B\sin(0\cdot 0)\\&=A\cdot 1+B\cdot 0\\&=A\end{align}$$

Therefore equation $(1)$ becomes $$\theta=\color{blue}{\theta_0}\cos(\omega t)+\color{red}{B\sin(\omega t)}$$

So the simple question I have is; Why is the $\color{red}{B\sin(\omega t)}$ not present in $(2)$?

Or, put in another way: How can I prove that $B=0$?


The initial conditions say that at $t=0$, $$\theta=\theta_0=A\cos(0)+B\sin(0)=A$$ Also $\dot\theta=-A\omega\sin(\omega t)+B\omega\cos(\omega t)$ so that at $t=0$, $$\dot\theta=0=-A\omega\sin(0)+B\omega\cos(0)=B\omega$$ So there you have it: $A=\theta_0$ and $B=0$. One potential for confusion I see is that you were using the same symbol for instantaneous angular velocity of the pendulum as for constant natural angular frequency. I used $\dot\theta$ for the former and $\omega$ only for the latter.

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  • $\begingroup$ Perfect answer, thanks very much. $\endgroup$ – BLAZE Jun 15 '16 at 5:20

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