Simplify the equation - question about the angle

AI Thread Summary
The discussion revolves around understanding the derivation of the angle φ = π/3 in the equation x(t) = 5e(-t/5)(√3 sin(t) + cos(t)). The user initially struggles to see how this value is obtained from the coefficients of the sine and cosine functions. It is clarified that φ can be calculated using the relationships between the coefficients, specifically that cos(φ) = √3/2 and sin(φ) = 1/2, leading to φ = π/3. The importance of mastering these concepts for future studies in physics or electrical engineering is emphasized. The user expresses gratitude for the clarification and acknowledges the need for more practice.
Vital
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Homework Statement


Hello!
Please, take a look at the following equation and help me to understand where the authors got the value of φ = π/3. I don't see where it is derived from as no additional conditions are given.

Homework Equations


x(t) = 5e(-t/5) cos(t) + 5e(-t/5) √3 sin(t)

The Attempt at a Solution



Here is how I proceed:
x(t) = 5e(-t/5) (cos(t) + √3 sin(t))

Now, given the formula:
f(x) = a sin(wx) + b cos(wx) + B (w > 0) is the same as
f(x) = √a2+b2 sin(wx + φ) + B

In my case, B = 0, so I rewrite the expression as:
x(t) = 5e(-t/5) ( √3 sin(t) + cos(t) )

x(t) = 5e(-t/5) √√32+12 sin(t + φ)

x(t) = 10e(-t/5) sin(t + φ)

But in the book they have:
x(t) = 10e(-t/5) sin(t + π/3)

Where did they get φ = π/3 from?

Thank you!
 
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x(t) = 5e(-t/5) ( √3 sin(t) + cos(t) )=##10e^{-t/5}\left(\sin {t }\left(\frac{\sqrt{3}}{2}\right)+\cos{t}\left(\frac{1}{2}\right)\right)##

##\sin{(t+\phi)}=\sin t \cos{\phi}+\cos{t}\sin{\phi}##

What is the value of ##\phi##?
 
Chestermiller said:
x(t) = 5e(-t/5) ( √3 sin(t) + cos(t) )=##10e^{-t/5}\left(\sin {t }\left(\frac{\sqrt{3}}{2}\right)+\cos{t}\left(\frac{1}{2}\right)\right)##

##\sin{(t+\phi)}=\sin t \cos{\phi}+\cos{t}\sin{\phi}##

What is the value of ##\phi##?
Ah! I see. I should have computed the value of φ based on the coefficients' values:

coefficient1 x cos(φ) = √3
coefficient2 x sin(φ) = 1
(coefficient1 x cos(φ))2 + (coefficient2 x sin(φ))2 = coefficient2
hence, coefficient = 2, and thus cos(φ) = √3/2 and sin(φ) = ½, and of course φ = π/3
Now I see how the value appeared.
 
Vital said:
Ah! I see. I should have computed the value of φ based on the coefficients' values:

coefficient1 x cos(φ) = √3
coefficient2 x sin(φ) = 1
(coefficient1 x cos(φ))2 + (coefficient2 x sin(φ))2 = coefficient2
hence, coefficient = 2, and thus cos(φ) = √3/2 and sin(φ) = ½, and of course φ = π/3
Now I see how the value appeared.

You have been shown this more than once already in other threads. It is something you need to learn and commit to memory, especially if you are going into Phhsics or Electrical Engineering.
 
Ray Vickson said:
You have been shown this more than once already in other threads. It is something you need to learn and commit to memory, especially if you are going into Phhsics or Electrical Engineering.
It takes practice and time to remember these notions. I am doing my best, and as I am learning on my own I don't have enough practice. Happy to be able to ask here. :-)
 
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