# How to find the time a pendulum reaches max angle given that its position is $3\cos(2t)+\sin(2t)$?

Kept the equation simple in the title, it's a bit more complicated

The position of the pendulum over time, $$y(t)$$, is described by the equation $$y=0.3cos(2.214t) + 0.135sin(2.214t)$$

3 questions:

1. When does the pendulum first reach its maximum angle from vertical (hint: you might want to use an inverse trig in your answer)

2. what is the max angle?

3. how long after reaching its max angle until the pendulum reaches max deflection in the other direction (hint: where is the next critical point)?

So...I can guess that finding the time when velocity hits 0 would give me the maximum angle of the pendulum, but the equation has two trig functions so I don't know how to solve for $$t$$...

Since $$y=0.3\cos(2.214t) + 0.135(\sin(2.214t)$$, we have

$$y'(t) = 2.21 * 0.3 * -sin(2.214t) + 0.135 * 2.214 * cos(2.214t) = -0.663sin(2.214t) + 0.3cos(2.214)$$

That's as far as I can get. I cant see any trig identities that could help me isolate t here. Even if I did have such an identity, I don't know how I'd answer part 3 which asks you to find the second time that the velocity = 0 (that makes no sense since solving the velocity equation for 0 should just give you a single t value)

• The single t value will be in one period, or maybe a half period. But it will oscillate. Jun 1 '20 at 1:47
• @TheChaz2.0 Sorry, not following...which of the questions are you answering and what do you mean? Jun 1 '20 at 2:00
• Set the last equation $=0$, move some things around, and you should get $$\frac{\sin(2.214t)}{\cos(2.214t)} = \tan(2.214t) = \frac{1}{221}$$ Jun 1 '20 at 2:49
• The tangent function hits that value ($\frac{1}{221}$) an infinite number of times, each corresponding to a different value of t. (For a related thought, consider the equation $\sin(x) = 1$, which has a solution at 90 degrees, plus any integer number of full circles/rotations) Jun 1 '20 at 2:51
• @TheChaz2.0 That helps a lot and got me the correct answers, thanks so much! Jun 1 '20 at 13:58

There are some formulas that can help combine sin and cos functions, I'm not going to prove this formula because there are plenty of proofs you can find online. Also I do not think the second question is doable without some more context, I'm assuming that y describes elevation, relative to an arbitrary standard, so it is impossible to find the length of the pendulum, and therefore the angle, but we can still find its maximum elevation. $$asinx+bcosx = \sqrt{a^2+b^2}cos(x-\alpha);tan(\alpha)=\frac{a}{b}$$ Using this formula gives: $$y=0.135sin(2.214t)+0.3cos(2.214t) = \sqrt{0.3^2+0.135^2}cos(2.214t-arctan(\frac{0.135}{0.3}))$$ $$y\approx0.3289cos(2.214t−0.4228)$$ Taking the derivative to get the velocity formula of: $$y'=-0.3289\times2.214sin(2.214t−0.4228)$$ Now it should be apparent why there are multiple points where velocity (in the y direction) equates to zero (a sin wave has infinite zeros). Solving for t gives: $$0=-0.3289\times2.214sin(2.214t−0.4228)$$ $$0=sin(2.214t−0.4228)$$ $$k\pi=2.214t−0.4228$$ $$t=\frac{k\pi+0.4228}{2.214}$$ This gives us when the y component of velocity equals zero, which includes both when the pendulum peaks and when it is at the bottom of its cycle. So to identify which is which, evaluate $$y(t)$$ for some integer values of $$k$$. Evaluating at $$k=0$$ gives you the first maximum, and $$k=1$$ gives the first minimum. Because increasing integer values of $$k$$ should give alternating maximums and minimums, one can conclude that even numbers of $$k$$ give maximums and odd values of $$k$$ give minimums. So to answer question one, we just need to evaluate $$t$$ at $$k=0$$, which gives: $$t_0=\frac{0\times\pi+0.4228}{2.214}$$ $$t_0=\frac{0.4228}{2.214}=0.1910$$ The time it takes to get to the next peak should be $$t_2-t_0$$, so: $$t_2-t_0=\frac{2\times\pi+0.4228}{2.214}-\frac{0\times\pi+0.4228}{2.214}$$ $$t_2-t_0=\frac{2\times\pi}{2.214}=2.838$$ *sorry in advance if anything is rounded incorrectly
• Thank you!! This clears a lot of things up. Everything makes perfect sense. Just one question: why did you choose to subtract $t_2 - t_0$ instead of $t_1 - t_0$ ? Going by the answer to the question, they actually wanted $t_1 - t_0$ apparently. Not sure why since the question is pretty vague, but I'm wondering what you had in mind Jun 1 '20 at 14:44
• $t_1$ is where $k=1$ and it will give the point where the pendulum is at its lowest point, so not the opposite highest point. Jun 3 '20 at 17:42
• Why would $t_1$ give the lowest point? At the lowest point (i.e. at the very middle of the swing) the velocity would be nonzero, right? Jun 4 '20 at 23:12