# AC Waveform question

• newbie991

## Homework Statement

v = 60sin(458t + 60º)

Find the voltage of the waveform when t = 5ms

It may be a silly question but I've just started studying ac, just wondering how that + 60º affects the equation?

any help would be appreciated as I've looked everywhere for the method...

V = VmSin(2πft)

## The Attempt at a Solution

v = 60sin(458t + 60º)

=> v=60sin(458(5x10-3) + 60º)

=> v=60(0.039957 + sin60º) ? -----> not sure if this is correct

I would appreciate any reply

~Newbie

Welcome to PF, newbie991!

The 60º only sifts the angle, well 60 degrees. You've probably done trigonometry, where sin(∏+∏)=sin(2∏). You just got to be careful, though. Here the voltage is in the form v=60sin(ωt+60º).

Thanks, very interesting forum indeed! :)

Thanks for the reply, I've roughly plotted out this function and I am getting a different answer to the theoretical value.

what would you estimate it to be?

from the forumula I've extracted:

vmax=60 p-p = 120
f=72.892Hz
T=13.71ms

assuming all that is correct, i found v after 5ms to be between -40 and -20, although I am getting -11.66.

60 Sin (458 (5x10-3) + 1/3 ∏) = -11.66

I know there's something right in front of me I am not seeing!

would appreciate any further help!

~Newbie

vmax=60 p-p = 120

I'm not quite following you here. What does the last part, p-p = 120, mean?

When I graphically checked the answer, I got the same as plugging in the values to the given signal.

ah sorry that's kind of irrelevant that's the peak to peak voltage,

could you tell me what your answer was so i can see if I am on the right track?

or could you tell me where I am going wrong here:
60 Sin (458 (5x10-3) + 1/3 ∏)

ah sorry that's kind of irrelevant that's the peak to peak voltage,

could you tell me what your answer was so i can see if I am on the right track?

or could you tell me where I am going wrong here:
60 Sin (458 (5x10-3) + 1/3 ∏)

The way the problem was originally listed, it looks to all be in degrees. You just made an error with your calculator in multiplying 5ms * 458:

=> v=60sin(458(5x10-3) + 60º)

=> v=60(0.039957 + sin60º) ? -----> not sure if this is correct

Re-check 458 * 0.005 = ______

The way the problem was originally listed, it looks to all be in degrees. You just made an error with your calculator in multiplying 5ms * 458:

=> v=60sin(458(5x10-3) + 60º)

=> v=60(0.039957 + sin60º) ?

Re-check 458 * 0.005 = ______
__________________

thanks for the reply

458 * 0.005 = 2.29 :tongue:

ok this is exactly what I am entering:

rad mode:
60 sin (458(0.005) + 1/3 ∏)

and I am getting: -11.66159...

thanks for the reply

458 * 0.005 = 2.29 :tongue:

ok this is exactly what I am entering:

rad mode:
60 sin (458(0.005) + 1/3 ∏)

and I am getting: -11.66159...

sin(458 * 0.005) = 0.752

sin(PI/3) = 0.866

sin(458 * 0.005 + PI/3) = sin(2.29 + 1.047) = sin(3.337) = sin(3.337 - 3.142) = sin(0.195)

So the answer cannot be negative. Try doing the calculation in a different order, and also do the individual pieces to check your work.

Maybe your 1/3 PI isn't turning out to be PI/3...

sin(458 * 0.005) = 0.752

sin(PI/3) = 0.866

sin(458 * 0.005 + PI/3) = sin(2.29 + 1.047) = sin(3.337) = sin(3.337 - 3.142) = sin(0.195)

So the answer cannot be negative. Try doing the calculation in a different order, and also do the individual pieces to check your work.

Oops, my bad. I subtracted only one PI inside the sine... that's not right. Give me a sec...

sin(3.337) = -0.194

60 * sin(3.337) = -11.65

Are you sure that's not the right answer? If it's not, then maybe the whole original problem was in degrees after all.

Is the answer 60sin(2.29 + 60) = 53.1 ?

I get the same V=-11.66V as well. Is this an online assignment, or how do you know it's not the right answer?

If the problem was in degrees, I would assume the angular frequency would have the degree symbol added. After all, rad/s is the more commonly used unit.

Last edited:
I get the same V=-11.66V as well. Is this an online assignment, or how do you know it's not the right answer?

If the problem was in degrees, I would assume the angular frequency would have the degree symbol added. After all, rad/s is the more commonly used unit.

that is the correct answer

went over everything again and -11.66 is correct
was just a silly mistake on my part!

thanks a lot for your help Kruum and berkeman, much appreciated