# Simple Oscillator: find x(t)

1. Oct 18, 2004

### cj

At time t = T/2 the speed of a simple oscillator of angular frequency

$$\omega = \omega_0$$

has maximum amplitude U and positive value.

Find x(t).

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The textbook answer for this problem is:

$$x(t) = -\frac {U}{\omega_0}sin(\omega_0t)$$

where

$$U = \omega_0A}$$
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This is confounding me.

For a simple sin solution (I'm assuming zero phase shift), wouldn't x at t = T/4 have to be positive, i.e.:

$$x = \frac {U}{\omega_0}$$

and NOT negative (as given by the alleged solution):

$$x = -\frac {U}{\omega_0}$$

2. Oct 18, 2004

### Integral

Staff Emeritus
What is the definition of the coordinate system?

3. Oct 18, 2004

### cj

Vertical axis = displacement (x), velocity (v) and acceleration (a).

Horizontal axis = time, t

4. Oct 18, 2004

### Integral

Staff Emeritus
I am trying to get you to think about the definition of what is meant by negative and positive. Have you given us the complete problem?

5. Oct 18, 2004

### cj

Yes, the posted question and textbook answer (which I guess could be wrong) is verbatim.

Maybe I don't understand the meaning of the negative sign.

I'm imagining a displacement-speed-acceleration relationship as given in the graph at the top of the following linked page:

Sample

The bottom line is that if x(t) is as given by the textbook, it would clearly have a negative value (relative to the equilibrium position at t = T/4. This seems incorrect since even junior high school students understand that at t = T/4 the value for a sine function is clearly a maximum and clearly positive.

Right?