Calculating time from two voltages

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In summary: Hi Spreco, do this: u1=u2, 50 sin(314t + ∏/3)=100 sin(314t - ∏/6) => sin(Pi t + Pi/3)=2 sin(Pi t - Pi/6),sin(a+b) = sin a cos b + cos a sin b, then:sin(Pi t) cos(Pi/3) + cos(Pi t) sin(Pi/3) = 2 sin(Pi t) cos(Pi/6) - 2 cos(Pi t) sin(Pi/6)sin(Pi t)= a, cos(Pi t)= b,sqrt(3) /
  • #1
Spreco
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I've gotten this for homework from my Electrotechnics class

Homework Statement



You are given two voltages: u1=50 sin(314t + ∏/3) and u2=100 sin(314t - ∏/6). You need to calculate the phase difference (pretty easy, already did that it's 90 degrees) and time when the two voltages have the same value. The solution in the book says 3,16ms.

Homework Equations



General formula for calculating the voltage is u = Um sin(ωt +- θ).

The Attempt at a Solution



So, because it says that I need to find the time both voltages are the same value, u1 must be equal to u2. So, u1=u2 50 sin(314t + ∏/3)=100 sin(314t - ∏/6). Now the problem is how do I get the 't' out of the sine and aren't they going to cancel each other out because they're same?

Please ask me anything else you need to solve this problem. THank you.
 
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  • #2
Hi Spreco! Welcome to PF!

Spreco said:
I've gotten this for homework from my Electrotechnics class

Homework Statement



You are given two voltages: u1=50 sin(314t + ∏/3) and u2=100 sin(314t - ∏/6). You need to calculate the phase difference (pretty easy, already did that it's 90 degrees) and time when the two voltages have the same value. The solution in the book says 3,16ms.

Homework Equations



General formula for calculating the voltage is u = Um sin(ωt +- θ).

The Attempt at a Solution



So, because it says that I need to find the time both voltages are the same value, u1 must be equal to u2. So, u1=u2 50 sin(314t + ∏/3)=100 sin(314t - ∏/6). Now the problem is how do I get the 't' out of the sine and aren't they going to cancel each other out because they're same?

Please ask me anything else you need to solve this problem. THank you.

What cancels what? You have 50 sin(314t + ∏/3)=100 sin(314t - ∏/6) or sin(314t + ∏/3)=2sin(314t - ∏/6). You cannot solve this or is it something else that troubles you?
 
  • #3
Hi Spreco

Welcome to Physicsforums!

Spreco said:
I've gotten this for homework from my Electrotechnics class

Homework Statement



You are given two voltages: u1=50 sin(314t + ∏/3) and u2=100 sin(314t - ∏/6). You need to calculate the phase difference (pretty easy, already did that it's 90 degrees) and time when the two voltages have the same value. The solution in the book says 3,16ms.

Homework Equations



General formula for calculating the voltage is u = Um sin(ωt +- θ).

The Attempt at a Solution



So, because it says that I need to find the time both voltages are the same value, u1 must be equal to u2. So, u1=u2 50 sin(314t + ∏/3)=100 sin(314t - ∏/6). Now the problem is how do I get the 't' out of the sine and aren't they going to cancel each other out because they're same?

Please ask me anything else you need to solve this problem. THank you.

Rewrite u1=50 sin(314t + ∏/3) such that instead of sine ,you get a cosine .I mean convert sine into cosine.

What do you get ?
 
Last edited:
  • #4
Hi Spreco, do this: u1=u2, 50 sin(314t + ∏/3)=100 sin(314t - ∏/6) => sin(Pi t + Pi/3)=2 sin(Pi t - Pi/6),
sin(a+b) = sin a cos b + cos a sin b, then:
sin(Pi t) cos(Pi/3) + cos(Pi t) sin(Pi/3) = 2 sin(Pi t) cos(Pi/6) - 2 cos(Pi t) sin(Pi/6)
sin(Pi t)= a, cos(Pi t)= b,
sqrt(3) / 2 a + 1/2 b = sqrt(3) a - b
-1/2 a = 3/2 b
-a = 3 b
sin x = -3 cos x, now specify any value of cos x s.t. 3 cos x<1, which means cos x< 1/3, -70.5< x < 70.5°, or 109< x <251 and then the equation has a solution. In particular you find the values of t from x= 180 * t. The
 
  • #5
The task was to find the time, and after some thinking and ideas I solved it and got the solution. Thanks for all the help and ideas you gave me.
http://i.imgur.com/EwR4GOp.jpg (The picture is high resolution, you can zoom in really close to see the details)
 

FAQ: Calculating time from two voltages

How do you calculate time from two voltages?

Calculating time from two voltages involves using the equation t = (v2 - v1) / a, where t is the time in seconds, v1 and v2 are the voltages in volts, and a is the acceleration in meters per second squared.

What is the purpose of calculating time from two voltages?

The purpose of calculating time from two voltages is to determine the time it takes for an object to travel between two points based on its acceleration and the difference in voltage readings at those points.

Can this method be used for any type of motion?

Yes, this method can be used for any type of motion as long as the acceleration remains constant and the voltage readings accurately reflect the position of the object.

Is it necessary to have two voltage readings to calculate time?

Yes, having two voltage readings is necessary to accurately calculate time using this method. One voltage reading represents the starting point and the other represents the ending point of the object's motion.

Are there any limitations to calculating time from two voltages?

One limitation to calculating time from two voltages is that it assumes a constant acceleration, which may not always be the case in real-life scenarios. Additionally, any errors in the voltage readings or the equation used can result in inaccurate calculations.

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