# Deriving x(t) into A sin (\omega*t + \varphi)

• krnhseya
In summary, x(t) is equal to e^{-\zeta\omega t}\left[(a_1+a_2)\cos\left(\omega\sqrt{1-\zeta^2}t\right)+a_1-a_2\sin\left(\omega\sqrt{1-\zeta^2}t\right)\right].
krnhseya

## Homework Statement

Derive...

x(t)=(exp^(-$$\zeta$$$$\omega$$t))*(a1(exp^(i$$\omega$$sqrt(1-$$\zeta^2$$)*t)))+a2(exp^(-i$$\omega$$sqrt(1-$$\zeta^2$$)*t))))

into

x(t)=(exp^(-$$\zeta$$$$\omega$$t))*(A sin ($$\omega$$*t + $$\varphi$$))

n/a

## The Attempt at a Solution

I've managed to get x(t) = (exp^(-$$\zeta$$$$\omega$$t))*((a1(cos$$\omega$$*t) + i sin ($$\omega$$*t))+(a2(cos$$\omega$$*t) - i sin ($$\omega$$*t))) then when i simplify things...sin terms cancel out and i end up geting...

exponential term * (a1+a2) * (2cos $$\omega$$*t)

Hi,

Is the $\zeta$ a constant here?

krnhseya said:
I've managed to get x(t) = (exp^(-$$\zeta$$$$\omega$$t))*((a1(cos$$\omega$$*t) + i sin ($$\omega$$*t))+(a2(cos$$\omega$$*t) - i sin ($$\omega$$*t))) then when i simplify things...sin terms cancel out and i end up geting...

exponential term * (a1+a2) * (2cos $$\omega$$*t)

No, the sine terms do not cancel out. You have the following.

$$x(t)=e^{-\zeta\omega t}\left[(a_1+a_2)\cos\left(\omega\sqrt{1-\zeta^2}t\right)+(a_1-a_2)\sin\left(\omega\sqrt{1-\zeta^2}t\right)\right]$$.

The only way the sine term will vanish is if $a_1=a_2$, and you did not say that that is the case in your problem statement.

Tom Mattson said:
Hi,

Is the $\zeta$ a constant here?

No, the sine terms do not cancel out. You have the following.

$$x(t)=e^{-\zeta\omega t}\left[(a_1+a_2)\cos\left(\omega\sqrt{1-\zeta^2}t\right)+(a_1-a_2)\sin\left(\omega\sqrt{1-\zeta^2}t\right)\right]$$.

The only way the sine term will vanish is if $a_1=a_2$, and you did not say that that is the case in your problem statement.

Hello. It is a constant.
Yeah, I figured out that I made a mistake.

that's as far as i can go but i need to combine that into sin like I've posted above.

And, there should be a bracket infront of "sqrt" so exponential part it outside the whole thing.

 I think I got it...Thank you very much! :)

Last edited:

## 1. What is the equation for deriving x(t) into A sin (\omega*t + \varphi)?

The equation is x(t) = A sin (\omega*t + \varphi), where A represents the amplitude, \omega represents the angular frequency, t represents time, and \varphi represents the phase shift.

## 2. How do you determine the amplitude, angular frequency, and phase shift in the equation?

The amplitude, A, is the maximum displacement of the wave from its equilibrium position. The angular frequency, \omega, is the rate at which the wave oscillates, and is equal to 2π divided by the period of the wave. The phase shift, \varphi, is the horizontal translation of the wave from its original position. These values can be determined by analyzing the given function or by using data from a graph of the function.

## 3. What is the physical significance of the amplitude in this equation?

The amplitude represents the maximum displacement of the wave, which can be related to the intensity or strength of the wave. In the context of physics, it can also represent the maximum amount of energy carried by the wave.

## 4. How does the angular frequency affect the behavior of the wave?

The angular frequency determines the rate at which the wave oscillates. A higher value of \omega means that the wave will oscillate more quickly, while a lower value will result in a slower oscillation. This parameter is closely related to the period and frequency of the wave.

## 5. Can the phase shift affect the shape of the wave?

Yes, the phase shift determines the horizontal translation of the wave. This means that it can shift the entire wave to the left or right, which can change its shape. For example, a positive phase shift will shift the wave to the left, while a negative phase shift will shift it to the right. However, the amplitude and angular frequency will remain the same.

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