Here is what i am given:
The oscillations of a pendulum are described by the equation: $$\ddot{\phi}+\frac{g}{l}\sin{\phi}=0$$ where $\phi$ is the angle between the pendulum and the vertical axis, $l$ is the length of the pendulum, and $g$ is the acceleration of gravity. The pendulum of a longcase clock swings to a maximum angle of $5^{\angle}$ from the vertical. How many seconds does the clock gain or lose each day (1day=86400sec) if the clock is adjusted to keep perfect time when the angular swing is infintesimaly small? Hints: (1) Since the swing angle in the problem are small, simplify the equation for the pendulum while still keeping it nonlinear. (2)The relavite loss or gain of the clock is proportional to the relative change in the frequencies of the oscillations. (3) Don't forget to convert angles to radians.
So i'm having trouble with the last few steps of the problem. I'll catch you up on where i am.
Since the angle is small i used the first 2 terms of the MacLauren series to estimate $\sin\phi=\phi-\frac{\phi^3}{6}$ so our problem becomes $$\ddot{\phi}-\omega^2(\frac{\phi^3}{6}-\phi)=0$$ where $\omega^2=\frac{g}{l}$ and $\omega$ is the angular frequency in $s^{-1}$. I dropped $\omega^2$ for now and will introduce it at the end since it is a constant and inserted a small parameter $\epsilon$. Rearranging gives: $$\ddot{\phi}+\phi=\epsilon\frac{\phi^3}{6}$$ From here i followed the steps for a nonlinear oscillator (like the Van der Pol oscillator) where we look for a solution in the form $\phi(t)=a(t)\cos(t+\psi(t))$ with $\dot{\phi}(t)=-a(t)\sin(t+\psi{t})$ where a(t) is the amplitude which is a function of time $t$ and $\psi(t)$ is a (phase?) constant dependent on initial conditions. I'm skipping a lot of steps here but a vague description of what i did was: differentiate, substitute into original problem, solve for $a'(t)$ and $\psi'(t)$ and then take their averages over one period ($2\pi$). If you're familiar with the method of averaging and certain solution methods for the Van der Pol or Wayleigh oscillators you probably understand what i did. I end up with $a'_{avg}(t)=0$ so that $a(t)=A$ (some constant dependent on initial conditions) and $$\psi_{avg}'(t)=-\epsilon\frac{(a(t))^2}{16}$$. This is a separable differential equation to which i find solution $$\psi(t)=\epsilon\frac{(a(t))^2}{16}(t+k)$$ where $k$ is an integration constant $k=\psi(0)$ (i think). And since we determined that $a(t)=A$ i use this substition in $\psi$ also.
Substituting this solution of $\psi$ back into our desired solution form gives $$\phi(t)=A\cos\left(t-\epsilon\frac{A^2}{16}(t+k)\right)$$ And since Simple Harmonic Oscillators generally have solutions in the form of $x(t)=A\cos(\omega t)$ i reinsert $\omega$ inside the solution to get $$\phi(t)=A\cos\left(\omega(t-\epsilon\frac{A^2}{16}(t+k))\right)$$
So that is where i am... I'm very UNconfident about those last couple steps and assumptions. I don't know how i'm supposed to find the loss (or gain) in time over the course of day. I'm also a bit unsure about that integration constant $k$. One of my classmates said that $k=0$ but i don't know how he got that and he can't justify it. I'm not sure how to use the given information to find a solution.
I'm leaning toward this: Since we are given that "the clock swings to a maximum angle of $5^{\angle}$ from the vertical" it must be that $\phi(0)=5^{\angle}$ since the clock must have its maximum amplitude at $t=0$. But then i run into the problem with the "$k$" so i think i made some incorrect assumptions in the last few steps. Also, i'm not sure how that would help me find the time gain/loss per day.
Is anyone familiar with this equation and this method for solving it?
I think your elimination of $\omega$ and introduction of $\epsilon$ causes confusion. Leave those in there all the way and then you should get $$\dot\psi=-\frac{\omega a^2}{16}$$ Consequently $$\phi=a\cos\left(\omega t-\frac{\omega a^2}{16}(t+k)\right)$$ That $k$ is just a phase offset and doesn't affect the outcome. What you find then is that $$\omega^{\prime}=\omega\left(1-\frac{a^2}{16}\right)$$ and this is consistent with Wikipedia. Then the number of seconds gained in a day is $$\Delta t=\frac{\Delta\omega t}{\omega t}N_0=\frac{\omega^{\prime}t-\omega t}{\omega t}86400=-\frac{a^2}{16}86400=-\frac1{16}\left(\frac{5\pi}{180}\right)^286400=-41.1$$