Voltage of Electrical Outlet

[harpazo]

The voltage in a household electrical outlet is given by V = 170 cos (120pi•t), where V is measured in volts and t in seconds.

A. Specify the amplitude and the frequency for this oscillation.

B. Graph the function given above for two complete cycles beginning at t = 0.

Part A

Amplitude = |170|
A = 170

Let F = frequency

F = 1/T

I first found the period to be 1/60 = T.

F = 1/(1/60)

F = 60 cycles/sec

Part B

I am having trouble graphing the given function by hand. I created a table for (t, V).

What values should I select for t to calculate for V?

At first I decided to let t be 1, 2, 3, 4, and 5. I got crazy numbers for V values. I then decided to let t be 0, 1/120, 1/60, 1/40, 1/30. When I did, the value for V was 170 for every fraction selected.

What values should I use for t to calculate for V?

 

[Dr.Peterson]

You found that the period is 1/60. To graph a sinusoid, I first plot what can be called the key points, or quarter-cycle points, namely the start of a cycle and each 1/4 cycle from there (5 points make one cycle). A quarter cycle is 1/4 of 1/60 = 1/240; so find the values at t = 0, 1/240, 2/240 = 1/120, 3/240 = 1/80, and 4/240 = 1/60. Then continue for another cycle, just repeating the values you find, so you don't really need a table for these last four points.

Then fill in the curve between the points.

 

[harpazo]

You found that the period is 1/60. To graph a sinusoid, I first plot what can be called the key points, or quarter-cycle points, namely the start of a cycle and each 1/4 cycle from there (5 points make one cycle). A quarter cycle is 1/4 of 1/60 = 1/240; so find the values at t = 0, 1/240, 2/240 = 1/120, 3/240 = 1/80, and 4/240 = 1/60. Then continue for another cycle, just repeating the values you find, so you don't really need a table for these last four points.

Then fill in the curve between the points.

You said:

"Then continue for another cycle, just repeating the values you find, so you don't really need a table for these last four points."

What fraction starts the second cycle?

 

[Dr.Peterson]

You said:

"Then continue for another cycle, just repeating the values you find, so you don't really need a table for these last four points."

What fraction starts the second cycle?

Take a guess!

(That's the only way to learn, believe it or not: Try using what you know, so you exercise your mind, just as your muscles grow only as you try using them, even not knowing whether you can do it.)

Hint: the second cycle begins where the first cycle ends.

Hint 2: the length of each cycle is called the period, so a new cycle starts every time t increases by one period.

 

[harpazo]

Take a guess!

(That's the only way to learn, believe it or not: Try using what you know, so you exercise your mind, just as your muscles grow only as you try using them, even not knowing whether you can do it.)

Hint: the second cycle begins where the first cycle ends.

Hint 2: the length of each cycle is called the period, so a new cycle starts every time t increases by one period.

I noticed that you stopped at 4/240 = 1/60.
Increasing the numerator by 1 leads to 5/240, 6/240, 7/240, 8/240, 9/240. Of course, each fraction must be reduced.

 

[Dr.Peterson]

I noticed that you stopped at 4/240 = 1/60.
Increasing the numerator by 1 leads to 5/240, 6/240, 7/240, 8/240, 9/240. Of course, each fraction must be reduced.

So the first cycle is from 0 to 1/60, the second from 1/60 to 2/60 = 1/30, and so on. What you wrote doesn't quite show that you stopped at the right place; keep in mind that the first point in one cycle is the last point in the previous cycle, so 1/60 is the first of the five points for the second cycle, and 8/240 = 1/30 is the last.

I personally find it most useful to keep the common denominator throughout my work, and only reduce when I am required to report a value (or when reducing it, as I did for 4/240, shows that I am where I need to be, constituting a check).

 

[harpazo]

So the first cycle is from 0 to 1/60, the second from 1/60 to 2/60 = 1/30, and so on. What you wrote doesn't quite show that you stopped at the right place; keep in mind that the first point in one cycle is the last point in the previous cycle, so 1/60 is the first of the five points for the second cycle, and 8/240 = 1/30 is the last.

I personally find it most useful to keep the common denominator throughout my work, and only reduce when I am required to report a value (or when reducing it, as I did for 4/240, shows that I am where I need to be, constituting a check).

Ok. I will try graphing this function using your notes as a guide.

Note:

1. From now on, I will post one question at a time. I will keep this problem. There's nothing I can do about all the new threads posted in the last week or two.

2. I will show my work (right or wrong) to all new threads.

3. I will break down each question posted into small parts requesting help as needed along the way.

4. It make no sense to bounce from one topic to another ALL IN ONE DAY. For example, if I need help in terms of solving a particular trig equation, it makes no sense to post a thread about finding asymptotes of trig functions ON THE SAME DAY. This only leads to further confusion. Agree?

5. There is no need to rush through the David Cohen textbook, especially that I am currently unemployed due to COVID-19.