# Qustion on sinewave

1. Dec 18, 2011

### greg997

If the expression forsinewave is y=Asin(ωt±α)
then α is the phase angle.
But what is time displacement looking at that expression?
Thank you

2. Dec 18, 2011

### Mentallic

Time displacement? I assume you mean how long it takes for the sine wave to repeat a cycle, or its [STRIKE]frequency[/STRIKE].

All sine waves repeat every $2\pi$ units. So what we're looking for is

$$wt_1+\alpha+2\pi$$

to be equal to

$$wt_2+\alpha$$

and we want to find t2 in terms of t1, t2>t1

So equating each expression and simplifying:

$$wt_1+\alpha+2\pi=wt_2+\alpha$$

$$wt_2=wt_1+2\pi$$

$$t_2=t_1+\frac{2\pi}{w}$$

So clearly from this, we can see that the time it takes from the first point in a cycle (t1) to the next (t2) takes $2\pi/w$ time.

Also to get a more intuitive understanding of this, just think about the length of a complete cycle for sin(x), then sin(2x), sin(x/3) - which is the same as sin(1/3*x) etc.

edit: meant to say period, not frequency.

Last edited: Dec 18, 2011
3. Dec 18, 2011

### greg997

Thank you for your really good explanation.
But I am not sure this is the answer to the question I was given. I am not sure.
Can frequency and time period T be called time displacement? Is it not about that phase angle?
Thanks any way

4. Dec 18, 2011

### technician

Is it ω that is worrying you?
ω is an 'angular velocity' it is the number of cycles (given in radians) completed per second
1 cycle is 2∏ radians and therefore ω = 2∏/T where T is the time for 1 cycle.... the time period

So T = 2∏/ω

and frequency f = 1/T = ω/2∏

Hope this helps

5. Dec 18, 2011

### LCKurtz

You haven't given the definition of "time displacement". If you write$$sin(\omega t \pm \alpha) = \sin(\omega(t \pm \frac \alpha \omega))$$ is it the $\pm\frac \alpha \omega$ you want? (Notice that is a question; I'm just guessing here).