# I Period of a sine wave

1. Dec 4, 2017

### NickTheFill

Dear all
Something is bugging me. I hope you can help.

I read in texts that if f(t) = sin(t) then the period of the function is the time taken (secs) to complete one cycle.
I also read in texts that if f(t) = sin(2t) then the period of the halved.
No problems here. I see that (2*t) = (omega*t). Happy days.

I also read in texts that if f(x) = sin(x) then the period of the function is 2pi radians.
and that if f(x) = sin(2x) then the period of the function is pi radians.

How can this happen? How can we measure a period in radians? How can a full cycle be anything other than 2pi radians anyway?

I am going round in circles....

Nick

2. Dec 4, 2017

### Staff: Mentor

You've already said it: by running through the circle at doubled speed. In this case, the parameter $t$ is simply a measurement of time and not an angle. It is twice the angle so you reach the full circle after $\pi$ units of time. Either you consider the period as purely the distance (measured by whatever unit) between two maximums, or by an angle in which case the factor two has to be part of the drawing. It is basically the standard question with coordinates: does the scale change or the object? One can look at it both ways, but not both at the same time.

3. Dec 4, 2017

### NickTheFill

. My mind looks at this as twice the angular frequency not twice the angle, so the particle completes a full cycle of 2pi radians in half the time. If the period happens to be 2pi seconds then with twice the angular frequency the period is now pi seconds.

It is the reference to measuring periods in radians that i do not understand as a full cycle can only ever be 2pi radians.

If we are talking about measuring a distance, would that not be be wavelengths?

4. Dec 4, 2017

### Staff: Mentor

You combine two different concepts here: period and angle. Radians are simply a dimensionless quantity, a number. You can use it to measure angles, if a full circle is partitioned in $2\pi$ steps, or as the period of $\sin \omega t$, in which case you partitioned named period into only $\pi$ steps, if $\omega = 2$.
That's a possibility. But on a circle, this distance will depend on radius (usually $1$) and angular velocity $\omega$. Again, it is the concept of measurement you're dealing with: units and coordinates. It's as in the movie: Honey, I shrunk the kids. You can view it as a shrinkage from the point of an unchanged environment or as an enlargement, from the point of the shrunken object. Either way is possible, but not at the same time.

If we say $\varphi = 2 \pi [\text{radians}]$ is the angle of a full circle, then $\varphi = \pi [\text{radians}_{new}]$ is also a full circle, but this time, we measured in $\text{radians}_{new} = 2\cdot \text{radians}$, and the new radians is not the same angle as the old one. It is a different unit.

5. Dec 4, 2017

### PeroK

There's a difference between the period of a function and the period of the variable. When you talk about the period of a function that is the change in the variable that causes the function values to repeat.

6. Dec 4, 2017

### NickTheFill

If we say $\varphi = 2 \pi [\text{radians}]$ is the angle of a full circle, then $\varphi = \pi [\text{radians}_{new}]$ is also a full circle, but this time, we measured in $\text{radians}_{new} = 2\cdot \text{radians}$, and the new radians is not the same angle as the old one. It is a different unit.[/QUOTE]

Yes, I think this point is where my confusion lies. A radian is an SI derived unit and I treat it like a kg or a second, it is 57.3 degrees. So to say π radians can be the angle of a full circle is just not cricket!

If we say the x axis is the angle, then the period of one oscillation of a sine wave is only ever 2π radians and the only distance travelled is an angular distance.

If we say the x axis is time, then the period of one oscillation can be anything depending on the angular frequency of the oscillation.

7. Dec 4, 2017

### NickTheFill

Never thought about the period of a variable. I'm not sure my mind is up to that! I consider an independent variable to be just that and the dependent variable a function of it. It is the nature of the function that causes the periodic nature not the nature of the independent variable.