Simple Harmonic Motion in x direction

[Yubsicle]

Homework Statement

A simple harmonic oscillator, with oscillations in the x direction, has velocity given by: $$v_{x} = (2.2 \frac {\mathrm{m}} {\mathrm{s}}) \sin [(6.9 \frac {\mathrm{rad}} {\mathrm{s}}) t]$$.
Find the values of $\omega , A, f , T ,$ and $\phi$

Homework Equations

$$v_{x} = \frac {dx}{dt}$$
$$x = A \cos (\omega t - \frac{\pi}{2}) = A \sin \omega t$$
$$\omega = \frac{2\pi}{T} = 2 \pi f$$

The Attempt at a Solution

I integrated v with respect to t and got $$x = (-0.3188\mathrm{m})\cos [(6.9\frac{rad}{s} t)] $$
(I'm assuming that the integration constant C = 0, please correct me if I'm wrong). The first problem I have is that A is negative, and my understanding is that amplitude can only be positive. Do I take the absolute value of -0.3188 and get $A = 0.3188$, did I make a mistake, or do I need to do more steps?
The second problem is that the formula is $x = A \cos (\omega t - \frac{\pi}{2})$ but i don't have a $-\frac{\pi}{2}$ anywhere. Again, is there an easy fix, did I mess up, or have I not completed all the steps needed? Thanks in advance.

 

[Orodruin]

What happened to your general phase shift $\phi$?

 

[Yubsicle]

Oh, it looks like I misread my textbook. Apparently $x = A \cos(\omega t - \frac{\pi}{2})$ only works if $x = 0$ when $t = 0$, and otherwise it's $x = A \cos(\omega t + \phi)$. So I was wrong in assuming that C = 0.
That brings me to $$(-0.3188 \mathrm{m})\cos [(6.9\frac{\mathrm{rad}}{\mathrm{s}})t]+C$$ , which is missing $\phi$ and I really don't know what to do about C. Where do I go from here?

 

[Orodruin]

So I was wrong in assuming that C = 0.

Not really. The integration constant is related to the average displacement. This can be chosen to be zero. The general statement for $x$ with this choice is what you gave, $x = A \cos(\omega t + \phi)$. Try applying some trigonometric identities to this.

 

[Yubsicle]

I think I get it now. Plugging $t = 0$ into the equation given by the problem gives $v = 0$, so $\sin(\omega t+ \phi) = 0$ if $\phi = 0$ and so $x = A \cos\omega t$ . However, I still get $A = -0.3188$. Do I just take the absolute value, since the cosine function has the same "height" on both sides of the y-axis (not sure how to word it)?

 

[rude man]

The Attempt at a Solution

I integrated v with respect to t and got $$x = (-0.3188\mathrm{m})\cos [(6.9\frac{rad}{s} t)] $$
(I'm assuming that the integration constant C = 0, please correct me if I'm wrong).

Not relevant to the problem, but OK.

The first problem I have is that A is negative, and my understanding is that amplitude can only be positive.

Right.

Do I take the absolute value of -0.3188 and get $A = 0.3188$, did I make a mistake, or do I need to do more steps?

You could but you can also wind up with an expression with A>0. Consider -cos(x) = sin(x + φ) & you pick the correct φ.

The second problem is that the formula is $x = A \cos (\omega t - \frac{\pi}{2})$

? I don't think that is a correct solution for x(t) regardless of the constant of integration or the sign of A.

 

[Orodruin]

I think I get it now. Plugging $t = 0$ into the equation given by the problem gives $v = 0$, so $\sin(\omega t+ \phi) = 0$ if $\phi = 0$ and so $x = A \cos\omega t$ . However, I still get $A = -0.3188$. Do I just take the absolute value, since the cosine function has the same "height" on both sides of the y-axis (not sure how to word it)?

I suggest you do what I proposed and apply trigonometric identities to the general expression.

 

[haruspex]

I still get A=−0.3188A. Do I just take the absolute value

No, just taking the absolute value, and keeping all else the same, yields a different equation, so will not match the data. The equation with a negative amplitude is a valid description of x, but doesn't fit the canonical form. I.e., by convention, we generally arrange that the amplitude is positive by adjusting the phase.

 

[Yubsicle]

Alright, I used the trig identity $\cos(\theta + \pi) = -\cos(\theta)$ and got $\phi = \pi \:\mathrm{rad}$ and $A = 0.32\:\mathrm{m}$ (rounded to 2 sig figs), as well as $\omega = 6.9\:\mathrm{rad/s} , T = 0.91\:\mathrm{s}, f = 1.1\:\mathrm{Hz}$. Is my use of trig identities, and the value for $\phi$ correct?

Consider -cos(x) = sin(x + φ) & you pick the correct φ.

Does it matter that I used $\cos(\theta + \pi) = -\cos(\theta)$ instead of the one you suggested?

 

[haruspex]

Is my use of trig identities, and the value for ϕ correct?

Yes.

Does it matter that I used $\cos(\theta + \pi) = -\cos(\theta)$ instead of the one you suggested?

No, that's fine.

 

[Yubsicle]

Thanks to everyone for taking the time to help.