Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In one of my high school maths questions the example given to find the maximum displacement of a Simple Harmonic Motion where $ x=2+4\cos \left (2t + \frac{\pi}{3} \right ) $ and the motion lies in the interval: $-2 \leq x\leq 6$ is:

  1. let $x=2+4\cos \left (2t + \frac{\pi}{3} \right )= 6$
  2. then $\cos \left (2t + \frac{\pi}{3} \right )= 1$
  3. $2t + \frac{\pi}{3} = 2\pi$
  4. $t = \frac{5\pi}{6}$

I am getting confused on the transition from line 2 to line 3. I'm confused as to how getting rid of the cos turns 1 into $2\pi$. Could someone please explain this at a high school level?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

Remember: $$\cos x=1\Longleftrightarrow x=2k\pi\,,\,k\,\,\text{an integer}$$so$$\cos\left(2t+\frac{\pi}{3}\right)=1\Longleftrightarrow 2t+\frac{\pi}{3}=2k\pi$$

Assuming, as surely is the case, that it must be $\,t\geq 0\,$ , we get that$$2k\pi>\frac{\pi}{3}\Longrightarrow k=1,2,3,...$$and we can choose $k=1\Longrightarrow 2t+\frac{\pi}{3}=2\pi$

share|improve this answer

You normally would apply the $\arccos$ function to both sides - see here. This would yield $2t+\tfrac {\pi} 3=0$, however, and the question implies (or perhaps tacitly assumes) a positive value of $t$. Note that, observing the cosine function, we see that it equals $1$ whenever it's argument is $0,2\pi,4\pi,...$ in general whenever it's argument is $n\cdot 2\pi$, where $n$ is an integer.

There are actually infinitely many values of $t$ for which $\cos(2t+\tfrac{\pi} 3)=1$, but the one provided is the smallest positive value.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.