# Problem solving Periodic Function

1. Aug 22, 2011

### dGasim

1. The problem statement, all variables and given/known data
Hello,
I am having problem solving this problem (picture below). The figure IS drawn to scale. A = 70 Volts

Q1. What is the numerical value of V(2) on the graph (in volts)?
Q2. What is the numerical value of point B on the graph (in volts)?
Q3. What is the period for the function V3?
Q4. What is the average for the function V3 (over one period of time)?

2. Relevant equations
f(x) = y0 + k(x-x0)
period = 2pi/n

3. The attempt at a solution
For the Q1 and Q2 I have tried to find the slope:
deltaX = x1-x0 = 2.5 - 1.0 = 1.5
deltaY = y1-y0 = 0 - 70 = -70
f(x) = 0 + (-70/1.5)(x-1.0)
f(2.0) = (-70/1.5)(1.0) = -70/1.5 (which is not correct)

I went the same way for Q2. If I understand Q1 i'll do Q2 very easily.

for Q3 and Q4 I don't know how to do them at all. How should I apply period function into this graph?

#### Attached Files:

• ###### v3b_graph1.gif
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2. Aug 23, 2011

### Hootenanny

Staff Emeritus
Welcome to Physics Forums.

Where did you pull those numbers from? I can see only A and B on the graph.

When you compute deltaY, you have deltaY=0 -70. However, y1 (point B) isn't zero.

3. Aug 23, 2011

### dGasim

A=70 is given.

I found 2.5 from the previous question:
In the decreasing portion of the graph, what is the value of the time t (in s) for which V3(t)=0. Answer: 2.5

What I tried to do was to actually take the domain for the slope [1.0, 2.5] which will lead me to the range [70,0]. Can I do something like this?

4. Aug 23, 2011

### Hootenanny

Staff Emeritus
Okay. So what you've done now makes sense!

You have correctly worked out the slope of the line. However, you've made a mistake when it comes to determining the equation of the line.

Note that the line goes through the point (1,70).

5. Aug 23, 2011

### dGasim

Oh. Thank you. I just saw the problem. i put y0=0 but it should be y0=70. Thanks for the hint!

6. Aug 23, 2011

### dGasim

Alright I have found the answer to Question 1 and 2 (should i post answer here?) but I still don't know how to find the period. Could you please help me with that?

7. Aug 23, 2011

### Hootenanny

Staff Emeritus
You can do if you like, but there's no need to.

Well then, what do we mean by the period?

8. Aug 23, 2011

### dGasim

the 3rd question asks for the period of the function. I don't understand it because its not a harmonic function or anything to have a period. How should I attack the question in this case?

EDIT: Found the answer. I just thought that a period is time when it makes once cycle which is when it gets back to its amplitude. So i got the answer as 4.

Last edited: Aug 23, 2011
9. Aug 23, 2011

### Hootenanny

Staff Emeritus
Looks good to me

So, how are you shaping up for Q4?

10. Aug 23, 2011

### dGasim

Sorry to bother you guys with this post but I still cannot find the last question about average function.

2. Relevant equations
$f_{ave} = 1/(b-a) \int_a^b \! f(x) \, dx$

3. The attempt at a solution
1. I found the area under the curve in range from 0 to 4 (the period):
A = 1 * 70 + 1.5 * 70 * 0.5 + (23.333333 * 0.5 * 0.5) + 0 = 128.333333

2. Then I plugged the area into the average function formula
$f_{ave} = 128.33333 / 4 = 32.0833333$

3. What am I doing wrong? 23.33333 is the absolute value of B. Should it be -23.33333? I can't try it on my hw assignment because I have only one try left.

EDIT: Oops. Posted at the same time :)

Thanks,

11. Aug 23, 2011

### Hootenanny

Staff Emeritus
You need to subtract the area that is below the x-axis from the area that is above the x-axis.

HINT: You can ignore the segment between [3,4] seconds.

12. Aug 23, 2011

### dGasim

Worked!! i was adding it in my previous tries. Thank you very much for your help! I love this forum :)

13. Aug 23, 2011

### Hootenanny

Staff Emeritus
A pleasure!