Average of a sinusodial Function

[YoshiMoshi]

Homework Statement

Find the average of a sinusoidal function over 1.5 cycles

2880*sin(wt - 30 degrees)

Homework Equations

The Attempt at a Solution

Alright so

1/( (3*pi)/2 - 0) * integral[0, (3*pi)/2] 2880*sin(wt - 30 degrees) dt
I shift it over to the origin because it shouldn't effect the average no matter the phase?
(5760/(3pi)) * integral[0, (3*pi)/2] 2880*sin(wt) dt
- (5760/(3pi))*cos(t)|[0, (3*pi)/2]
- (5760/(3pi))*(0-1)
1920/pi

Does this look ok?

 

[berkeman]

I shift it over to the origin because it shouldn't effect the average no matter the phase?

That doesn't sound right to me. 1.5 cycles means 1 cycle which averages to zero, but then the other 1/2 cycle can be either positive or negative, which give very different averages.

Are you sure you are writing out the whole problem? Is a t=0 start specified?

 

[epenguin]

Quote: Does this look ok?

Even if it was OK you can't say it looks OK . I recently found it is not all that hard to use Tex for integrals and is only way for them to look presentable.

But are you sure one and a half cycles is 3π/2 ?

I don't think your answer should have π in the answer, I mean does sin(π) or sin or cos of any easy fraction of π have it?

Optionally simplify by the thought that
one and a half is, er,
One...
and a half.

I don't think it's right that the thing is not changed by phase though it was tempting to think so. Because some of a half-cycle is positive and some negative, and as you change phase some of the positive is becoming negative with no compensation.

I suggest you draw yourself a picture of this sine and its limits and you will see this.

 

[Ray Vickson]

Homework Statement

Find the average of a sinusoidal function over 1.5 cycles

2880*sin(wt - 30 degrees)

Homework Equations

The Attempt at a Solution

Alright so

1/( (3*pi)/2 - 0) * integral[0, (3*pi)/2] 2880*sin(wt - 30 degrees) dt
I shift it over to the origin because it shouldn't effect the average no matter the phase?
(5760/(3pi)) * integral[0, (3*pi)/2] 2880*sin(wt) dt
- (5760/(3pi))*cos(t)|[0, (3*pi)/2]
- (5760/(3pi))*(0-1)
1920/pi

Does this look ok?

Questions:
(1) Why do you choose an upper limit of 3*pi/2? What happened to w?
(2) Are you really sure that you can shift the origin? Certainly, for a whole number of cycles you could do it, but have you actually proven that you can do it for a fractional number of cycles?

Anyway, my answer does not agree with yours.

Finally: never mix up units the way you are doing; either use degrees all the way, or radians throughout--with the latter strongly preferred, as it makes integration and differentiation easier. So, re-write 30 degrees in terms of radians.