Phase Shift in Sinusoidal Functions: Solving for Φ in Different Cases

[isukatphysics69]

Homework Statement

Homework Equations

Asin(wt + Φ)

The Attempt at a Solution

let t = 0

Case A:
-.10 = .20sin(Φ)
sin^-1(-.10/.20) = Φ = -0.52

Case B:
.10 = .10sin(Φ)
sin^-1(.10/.10) = Φ = 1.57Case C:
-.05 = .05sin(Φ)
sin^-1(-.05/.05) = Φ = -1.57Case D:
0 = .20sin(Φ)
sin^-1(0/.20) = Φ = 0

Now i think i have to determine which ones are in which quadrant. i have to determine whether to subtract pi or not. But how to determine?​

 

[Chandra Prayaga]

What was the question? No question was stated. Also, the graph is not clear enough to make out anything.

 

[isukatphysics69]

What was the question? No question was stated. Also, the graph is not clear enough to make out anything.

 

[isukatphysics69]

here is a better pic

 

[Merlin3189]

You can see from your graphs which quadrant they are in.
Since you use the sine function, A sin( ωt ), the first quadrant is the section that starts at 0 and rises to A, the second starts at A and falls to 0, the third starts at 0 and falls to -A, and the fourth starts at -A and rises to 0. Equally you can look at the time values: if T is the period, then first quadrant is from t=o to t=T/4, second from t=T/4 to T/2, etc.
Just look at the intercept on the displacement axis and see which part of the cycle it corresponds to.

 

[isukatphysics69]

You can see from your graphs which quadrant they are in.
Since you use the sine function, A sin( ωt ), the first quadrant is the section that starts at 0 and rises to A, the second starts at A and falls to 0, the third starts at 0 and falls to -A, and the fourth starts at -A and rises to 0. Equally you can look at the time values: if T is the period, then first quadrant is from t=o to t=T/4, second from t=T/4 to T/2, etc.
Just look at the intercept on the displacement axis and see which part of the cycle it corresponds to.
View attachment 225241

So i ranked it as c < d < b < a
some of them i thought i would not have to add pi ( i made a mistake in op i think it is add pi) and this was incorrect. some of them are not in any quadrant so there is no no need to add. i added pi to a, and that seemed like the only one that needed it

 

[Merlin3189]

 

[isukatphysics69]

View attachment 225243

yes i agree with this

 

[Merlin3189]

Your original answers were: A= -π/6 , B=π/2, C= -π/2 and D= 0.
Looking at my diagram, two of these are not correct. What have you now said they are?

Edit: in case you are not familiar with these values;
π = 3.14, π/2 = 1.57, π/6 = 0.53 approx.

 

[isukatphysics69]

Your original answers were: A= -π/6 , B=π/2, C= -π/2 and D= 0.
Looking at my diagram, two of these are not correct. What have you now said they are?

b and a were incorrect

 

[isukatphysics69]

not a actually a is correct

 

[isukatphysics69]

i need to just step away from this for a few and go get something to eat

 

[Merlin3189]

Yeah. I needed to sleep as well.

I disagree with a. (I'll leave the other error for now)
Your calculation sin^-1(-.10/.20) = Φ = -0.52 gives one numerically possible result. There is another. You picked the wrong one.

You can see this from the graph -

I've put the origin in the centre and written the phase scale in decimal radians as you used.
You can see that your value of Φ = -0.52 does not correspond with the position of A.
There are two phase angles which give sin(Φ) = -0.5
One is Φ = -0.52 and its aliases 2πN - 0.52 (where N is an integer)
The other is where A is.

Another method you may have used to solve trig equations, implied by your term "quadrant" could be this diagram:

Where you can see that the other solution to Φ = sin^-1(-0.5) is Φ = π + 0.52