Problem solving Periodic Function

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Homework Help Overview

The discussion revolves around a problem involving a periodic function represented graphically, with specific questions regarding voltage values at certain points, the period of the function, and the average value over one period. The context includes a voltage value of 70 Volts and a graph that is drawn to scale.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of slopes and the determination of voltage values at specific points on the graph. There is an attempt to clarify the equation of the line based on given points. Questions arise regarding the definition of the period for the function and how to calculate the average value over one period.

Discussion Status

Some participants have made progress in answering the initial questions but continue to seek clarification on the period and average value calculations. Guidance has been offered regarding the correct interpretation of the graph and the need to consider areas above and below the x-axis when calculating averages.

Contextual Notes

There are indications of confusion regarding the definitions and calculations related to periodic functions, particularly in distinguishing between harmonic functions and the specific function in question. Participants also express concern about the constraints of their homework submissions.

dGasim
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Homework Statement


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)?


Homework Equations


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

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?

Thanks in advance,
Gasim Gasimzada
 

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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.
 
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?
 
dGasim said:
A=70 is given. 2.5 was the answer from previous question (number between 2.0 and 3.0). What I tried to do was to actually take the domain for the slope [2.0, 2.5] which will lead me to the range [70,0]. Can I do something like this?
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).
 
Oh. Thank you. I just saw the problem. i put y0=0 but it should be y0=70. Thanks for the hint!
 
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?

Thanks in advance
 
dGasim said:
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?

Thanks in advance
You can do if you like, but there's no need to.

Well then, what do we mean by the period?
 
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:
dGasim said:
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.
Looks good to me :biggrin:

So, how are you shaping up for Q4?
 
  • #10
Sorry to bother you guys with this post but I still cannot find the last question about average function.


Homework Equations


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

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 homework assignment because I have only one try left.

EDIT: Oops. Posted at the same time :)

Thanks,
Gasim Gasimzada
 
  • #11
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
Worked! i was adding it in my previous tries. Thank you very much for your help! I love this forum :)
 
  • #13
dGasim said:
Worked! i was adding it in my previous tries. Thank you very much for your help! I love this forum :)

A pleasure!
 

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