Understanding AC Waveforms: How Does Phase Shift Affect Voltage?

[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...

Homework Equations

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

 

[Kruum]

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).

 

[newbie991]

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

 

[Kruum]

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.

 

[newbie991]

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 ∏)

 

[berkeman]

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 = ______

 

[newbie991]

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

ok this is exactly what I am entering:

rad mode:
60 sin (458(0.005) + 1/3 ∏)

and I am getting: -11.66159...

 

[berkeman]

thanks for the reply

458 * 0.005 = 2.29

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.

 

[berkeman]

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

 

[berkeman]

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...

 

[berkeman]

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 ?

 

[Kruum]

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.

 

[newbie991]

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