Problem solving Periodic Function

In summary: I love it too. All the best with your studies.In summary, the conversation discusses a problem involving a graph with a given voltage and various questions regarding the numerical values and period of the function. The conversation also includes a mistake in determining the equation of the line and a hint for finding the period. The final question involves finding the average function by subtracting the area below the x-axis from the area above the x-axis.
  • #1
dGasim
16
0

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
 

Attachments

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  • #2
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
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
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).
 
  • #5
Oh. Thank you. I just saw the problem. i put y0=0 but it should be y0=70. Thanks for the hint!
 
  • #6
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
 
  • #7
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?
 
  • #8
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:
  • #9
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


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

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
[itex]f_{ave} = 128.33333 / 4 = 32.0833333[/itex]

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!
 

Related to Problem solving Periodic Function

1. What is a periodic function?

A periodic function is a mathematical function that repeats its values at regular intervals. This means that for every x-value, there is a corresponding y-value, and the pattern of these values repeats itself over and over again.

2. How do you identify a periodic function?

A periodic function can be identified by looking for a repeating pattern in the graph or equation. The function should have at least one cycle that repeats itself infinitely in both the positive and negative direction.

3. What is the period of a periodic function?

The period of a periodic function is the length of one complete cycle. It is the distance between two consecutive peaks or two consecutive troughs on the graph. This value is often denoted as "T" in mathematical equations.

4. How do you solve problems involving periodic functions?

To solve problems involving periodic functions, you can use the properties of the function, such as the amplitude, period, and phase shift, to create an equation that represents the problem. Then, you can use this equation to find the desired solution.

5. What are some real-life applications of periodic functions?

Periodic functions are commonly used in fields such as physics, engineering, and economics to model and predict natural phenomena, such as the motion of a pendulum or the fluctuations in stock market prices. They are also used in music and sound engineering to describe the periodic vibrations of sound waves.

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