# Finding velocity and position of ##a(t)=−\omega^2(C_1\cos\theta+C_2\sin\theta)##

• fer Mnaj
In summary, the expression ##a(t) = −ω^2 (C_1 \cos θ + C_2 \sin θ)## refers to a one-dimensional simple harmonic oscillator problem, and can be solved for velocity and position by integrating it twice. The solution for velocity is ##v=\omega(C_2 \cos{\omega t}-C_1\sin{\omega t})+C##, where C is a constant of integration determined by the initial conditions.
fer Mnaj
Homework Statement
Given this expression : ##a(t)=−\omega^2(C_1\cos \theta+C_2\sin\theta)##

And with two initial conditions ##x(0)=x_0## and ##v(0)=v_0##

How to get velocity and position?

Furthermore, Can we get the values of ##C_1## and ##C_2##, what are these?
Relevant Equations
centripetal aceleration is the most likely ##a(t)=−\omega^2(\cos\theta+\sin\theta)##
this expression : ##a(t) = −ω^2 (C_1 \cos θ + C_2 \sin θ)##
I´ve never seen it before, where is it from?

It kinda looks like centripetal acceleration, but what exactly are ##C_1## and ##C_2##?

Can we calculate its velocity and position?

If I´ve got two initial conditions ##x(0)=x_0## and ##v(0)=v_0##I thought we could get velocity and position
with:

##\omega=\theta/t## IS constant

##dv/dt= −ω^2 (C_1 \cos θ + C_2 \sin θ)##

##\int dv=\int−ω^2 (C_1 \cos \omega t + C_2 \sin \omega t)dt##

##-\int1/w^2 dv=C_1\int \cos (\omega t) dt+ C_2\int\sin (\omega t) dt## is it correct?

Is it solvable?

Last edited:
It seems all right. Do integration.

What should it go in each side of the equation? velocity and position??

Velocity. Solve it first. Then in order to get position do time integration again.

Does the original expression have any vectors, that you might have omitted when writing it up? In it's current form its not clear what ##a(t)## refers to.

It looks like a one-dimensional simple harmonic oscillator problem.

etotheipi
vela said:
It looks like a one-dimensional simple harmonic oscillator problem.

Yes, you’re right. When I looked at it this morning I initially thought ##\theta## was a polar angle or something, for a particle moving in a weird force field, but yes it makes much more sense that it is actually the phase angle . Explains the ##-\omega^2## out the front

I also noticed it was the simple harmonic oscillator, but usually its solved from position to acceleration. Its taking me too much time to figure out in the opposite way

@fer Mnaj if you think about the general form for the equation of motion for a harmonic oscillator, you should be able to write down the function ##x(t)##.

But you can also just integrate the expression for ##a(t)## twice (do you know how to integrate trig functions?)

So I left it in: ##-\int1/w^2 dv=C_1\int \cos (\omega t) dt+ C_2\int\sin (\omega t) dt##
now let ##\omega t=u## so ##du/dt=\omega## and ##dt=du/\omega##
##-1/w^2\int dv=C_1\int \cos (u) du/\omega## + ##C_2\int\sin (u) du/\omega##
##-1/w^2\int dv=C_1/\omega\int \cos (u) du/## + ##C_2/\omega\int\sin (u) du##
##-1/w^2\int dv=C_1\omega^-1 sin(u)## - ##C_2\omega^-1 cos(u)##
##-1/w^2\int dv=C_1\omega^-1 sin(\omega t)## - ##C_2\omega^-1 cos(\omega t)##
Am I going right? what's next?

I missed the integration constants... how to aply them with the initial conditions i gave aa the begining?

$$-\frac{1}{\omega^2}[v]^v_{v_0}=\frac{C_1}{\omega}[sin\omega t]^t_0 - \frac{C_2}{\omega}[cos\omega t]^t_0$$
$$-\frac{1}{\omega^2}v_0=- \frac{C_2}{\omega}$$
So you can find the value of ##C_2##. Then further
$$-\frac{1}{\omega^2}\frac{dx}{dt}=\frac{C_1}{\omega}sin\omega t- \frac{C_2}{\omega}cos\omega t$$

anuttarasammyak said:
$$-\frac{1}{\omega^2}[v]^v_{v_0}=\frac{C_1}{\omega}[sin\omega t]^t_0 - \frac{C_2}{\omega}[cos\omega t]^t_0$$
$$-\frac{1}{\omega^2}v_0=- \frac{C_2}{\omega}$$
So you can find the value of ##C_2##. Then further
$$-\frac{1}{\omega^2}\frac{dx}{dt}=\frac{C_1}{\omega}sin\omega t- \frac{C_2}{\omega}cos\omega t$$
This is incorrect.

I'm actually shocked that no one here has yet been able to correctly integrate this simple first order ODE subject to an initial condition. The correct solution for the velocity is $$v=\omega(C_2 \cos{\omega t}-C_1\sin{\omega t})+C$$ where C is the constant of integration. Substituting the initial condition into this gives: $$v_0=\omega C_2+C$$So the constant of integration is:
$$C=v_0-\omega C_2$$. And thus the velocity is:
$$v=(v_0-\omega C_2)+\omega(C_2 \cos{\omega t}-C_1\sin{\omega t})$$

Last edited:
anuttarasammyak

## 1. How do you find the velocity and position of an object with the given acceleration function?

To find the velocity and position of an object with the given acceleration function, you can use the equations of motion. These equations relate the position, velocity, and acceleration of an object at a given time. Plug in the values for time and the given acceleration function to solve for the velocity and position.

## 2. What do the constants ##\omega##, ##C_1##, and ##C_2## represent in the given acceleration function?

The constant ##\omega## represents the angular frequency of the object's motion. ##C_1## and ##C_2## represent the initial position and velocity of the object, respectively. These constants are determined by the initial conditions of the object's motion.

## 3. Can this acceleration function be used for any type of motion?

Yes, this acceleration function can be used for any type of motion that follows a sinusoidal pattern. This includes simple harmonic motion, circular motion, and oscillatory motion.

## 4. How does the value of ##\omega## affect the motion of the object?

The value of ##\omega## affects the frequency of the object's motion. A larger value of ##\omega## results in a higher frequency and shorter period of motion, while a smaller value of ##\omega## results in a lower frequency and longer period of motion.

## 5. Can this acceleration function be used for both one-dimensional and two-dimensional motion?

Yes, this acceleration function can be used for both one-dimensional and two-dimensional motion. In one-dimensional motion, the object moves along a straight line, while in two-dimensional motion, the object moves along a curved path.

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