# Simple Harmonic Motion acceleration problem

1. Nov 28, 2005

### NutriGrainKiller

This is a problem right out of the book. My professor worked it out in class but now that I look over it I'm not sure why he chose the method he did. This is the problem exactly as it is stated in the book:
From this I was able to derive several other components that will be very helpful for solving this problem. Here are the givens:

f=2.5Hz
T=.4sec

Problem #1
Now, I get confused in several places. First, my professor states that as a given Xi=+1.1cm and Vi=-15cm/s. Can someone explain to me [in words] how he took the position at t=0 and assumed them values of Xi and Vi?
Problem #2
We were given these three formulas:
{keep in mind 'o' is phase offset or "phi"}
x(t) = A*Cos(wt+o)
v(t)=A*w*Sin(wt+o)
a(t)=A*w^2*Cos(wt+o)

Now from this my professor said this: a = -w^2*X => ai = -w^2*Xi -this ends up being -2.71 m/s^2
Problem #3
He continued to find "phi" or the phase offset. Here is what he did: arctan(Vi/WXi). This equation is right out of the book, and so far I have only seen it in this exact form. I understand how it was derived, but will it ever be needed in a different form?
Thanks guys!

2. Nov 29, 2005

### daniel_i_l

Xi and Vi are just the initial positions and velocities, initial would be at time t=0.
In 2 I think that he used the position equation -x(t)- to find the A and then used the acceleration as function of time equation and set the time to 0 and the A to what he found from the x(t) equation.

3. Nov 29, 2005

### andrevdh

The rotating object was just caught at these specific conditions with the stopwatch. That is when the stopwatch was started the object were at this position with this velocity. Remember that the object is oscillating to and fro along the x-axis.

Last edited: Nov 29, 2006
4. Nov 29, 2005

### NutriGrainKiller

for the first one i misread the problem, turns out i was more tired than i thought last night heh.

my calculator currently is not working, it got very wet on the way to class..so I cannot solve these problems. but here is the work for the second one:

I took x(t) equation and plugged in the values and solved for A, which ended up being 1.1/Cos(15.71*.4). Let's assume this value is stored into 'A'

took the a(x) equation and solved: a(x)=A*16.71^2*Cos(15.71*.4)

is this right?

5. Nov 29, 2005

### andrevdh

The last two equations should both have a negative sign after the = symbol (the derivative of cos is -sin). If one sets the time t=0 in the first two one gets
$$x_i=A\cos(\phi)$$
and
$$v_i=-A\omega\sin(\phi)$$
by dividing the second equation by the first we arrive at
$$\phi=\arctan(\frac{-v_i}{x_i\omega})$$
from which the phase offset can be calculated, which I do not get the same as you (15.71*.4). How did you arrive at these values? Is it in degrees or radians?