Question

For the simple harmonic motion of x = 2sin(3t + 3Pi/4) I need to find the time and position where velocity = 0. I got to the stage where I have cos(3t + 3Pi/4) = 0... but apparently the next step is 3t + 3Pi/4 = 3Pi/2 and I don't understand why cos^-1(0) = 3Pi/2 in this instance, as opposed to just Pi/2. Could someone please help me out?

Answer 1

So you have as the position function \(x(t)=2\sin\left(3t+\dfrac{3\pi}{4}\right)\), which gives you the velocity function \(x'(t)=2\cos\left(3t+\dfrac{3\pi}{4}\right)\cdot3=6\cos\left(3t+\dfrac{3\pi}{4}\right)\).

Solving for \(t\) in \(x'(t)=0\):
\[6\cos\left(3t+\dfrac{3\pi}{4}\right)=0\\
\cos\left(3t+\dfrac{3\pi}{4}\right)=0\\
3t+\frac{3\pi}{4}=\cos^{-1}0\\
3t+\frac{3\pi}{4}=\frac{n\pi}{2},~~~~\text{where }n\text{ is an integer}\]
There's no reason why you should consider \(\dfrac{3\pi}{2}\) as opposed to \(\dfrac{\pi}{2}\), since you're right about both working. It could be that you're restricted to a certain domain of \(t\), such as \(\pi<t<2\pi\).

Answer 2

Hmm, according to my textbook, there isn't any other given information and it has 3t + 3Pi/4 = 3Pi/2 without any other restrictions or explanations, apart from the original graph that I gave you. Could you explain why I'd get nPi/2 as my answer? That could help.

Additionally, in another question, it has a particle moving in SHM with the equation x = 2 + 4cos(2t + Pi/3). Then it asks for the first time when the particle is at maximum displacement (so when x = 6) and it has t = 5Pi/6, opposed to the answer I got which was t = -Pi/6 (which I know is wrong as t can't be negative here... but I'm still now sure about how I should be correctly approaching it).

Answer 3

still not sure*

Answer 4

Sorry, I made a slight error: Cosine is zero for any ODD integer multiple of pi/2.

If \(\theta=\pm\dfrac{\pi}{2},\pm\dfrac{3\pi}{2},\pm\dfrac{5\pi}{2},\cdots=\pm\dfrac{(2n+1)\pi}{2}\), then \(\cos\theta=0\).

I'm not sure what reasons your textbook/instructor has for that first question...

As for your second question, you've already answered it: the "first" time \(x\) reaches its max, with respect to "positive" increase in time, is at 5pi/6.

Answer 5

I think the question meant to have the FIRST time velocity is 0 - in that case, and considering the initial phase is 3Pi/4, is that why cos^-1(0) = 3Pi/2?
Is it because Pi/2 is before the initial phase and the next odd-n multiple of nPi/2 is 3Pi/2?
Thanks for your help :)

Answer 6

Ah, that explains it! and you're welcome