NCEES PROBLEM 115

[eng787]

WHY SOLUTION IS INTEGRATING SIN@ instead of SIN(wt) given in problem. Any explanation ??????

 

[Chickman]

WHY SOLUTION IS INTEGRATING SIN@ instead of SIN(wt) given in problem. Any explanation ??????

I think it should be 1/t * -cos(wt)/(2pi/t) which is half of the answer they provide.

 

[Chickman]

WHY SOLUTION IS INTEGRATING SIN@ instead of SIN(wt) given in problem. Any explanation ??????

I'm rusty but it may contain some trick in regards to modications to both period and angular frequency. The period is certainly pi, but maybe angular frequency is now simply 1

Obviously the w in the sinusoid has no bearing to the evaluation of the cosine. ..

Any smary people who can put this to rest? It is averaging zero for a portion of it.

 

[Chickman]

Sorry to beat this into the ground, but I think the actual integral is ...

(1/T) * -cos(wt)/w evaluated from 180 to 45 or pi to pi/4.

From my understanding the relationships are as follows ...

T =1/f, w=2*pi*f, therefore w=2*pi/T

this should cancel out the T and leave the result as ... -cos(wt)/(2*pi) which is roughly half of the solution they provide.

Seems to me that the average value looks more appropriate at approximately 0.6, but would like to know for future reference whether modifying the sine wave does away with the basic frequency, angular frequency, and period rules. I do agree that the pi is the period but they should state somewhere that w (omega) is 1 rad/s.

 

[170B]

I was wondering this last night. Any more? I doubt the ncees answeer is wrong.

 

[gte636i]

If you'll look at the x axis in the problem the units are already in (wt) not in (t), therefore you only need to evaluate the integral of the sin function (crosses at 0 so it's a sin) between 45 and 180 degrees and multiply by the reciprocal of the period.

In other words the 45 degrees given in the problem is already w*t, the w has already been factored in.