Simple harmonic motion of a particle

[~christina~]

Homework Statement

A particle moves along x-axis It is initially at 0.270m moving with velocity of 0.140m/s
and acceleration of -0.320m/s^2
Suppose it moves with constant acceleration for 4.50s.

Find it's:
a) position and velocity at end of time interval

next assume it moves with simple harmonic motin for 4.50s and x= 0 is it's equillibrium position.

Find it's:

b) position and velocity at end of this time interval

Homework Equations

$$x(t)= A cos (\omega t + \phi )$$
$$v(t)= - \omega A sin (\omega t + \phi )$$
$$a(t)= - \omega ^2 A cos (\omega t + \phi)$$

The Attempt at a Solution

well I know for the first situation that:

$$x_o$$= 0.270
$$v_o$$= 0.140m/s
$$a_o$$= -0.320m/s^2

and the acceration remain constant for 4.50s

to find a) the postiion and velocity at end of time interval...

I was thinking that I would take the period (T) to equal T= 4.50s?
and would I use 4.51 seconds and use the velocity as a function of time equation

$$v(t)= - \omega A sin (\omega t + \phi )$$

though I need the $$\phi$$...hm..

and I think I'm lost here.

can someone Please help me out with this?

 

[Littlepig]

so, a) it's no problem as is an 1 dimension problem, with movement equations...

about b), it's a whole new problem:
you will need to know the position and velocity at $$t_0=4.50$$
so, you can write: $$t_1=0$$ at $$x_1=0$$(equilibrium point)
by a), you calculated velocity at $$t_0$$, so, it's the particles velocity at $$t_1$$

I guess this should help...

 

[Moetasim]

I would approach this problem by first understanding the concept of simple harmonic motion. Simple harmonic motion is a type of motion in which the restoring force is directly proportional to the displacement from equilibrium and is directed towards the equilibrium position. This type of motion is commonly observed in systems such as springs, pendulums, and mass-spring systems.

In this given problem, the particle is initially at 0.270m and is moving with a velocity of 0.140m/s and an acceleration of -0.320m/s^2. This means that the particle is initially moving away from the equilibrium position with a decreasing speed, indicating that it is experiencing a restoring force directed towards the equilibrium position.

For part a), we can use the equations x(t) = x_o + v_o*t + (1/2)*a*t^2 and v(t) = v_o + a*t to find the position and velocity at the end of the time interval of 4.50s. Plugging in the given values, we get x(4.50) = 0.270 + (0.140*4.50) + (1/2)*(-0.320)*(4.50)^2 = 0.090m and v(4.50) = 0.140 + (-0.320)*(4.50) = -1.500m/s.

For part b), we are now assuming that the particle is moving with simple harmonic motion for 4.50s and x=0 is its equilibrium position. This means that the equations x(t) = A*cos(ωt+φ) and v(t) = -ωA*sin(ωt+φ) apply, where A is the amplitude and ω is the angular frequency. To find these values, we can use the given information that the particle moves with a constant acceleration for 4.50s. We know that acceleration is the second derivative of position with respect to time, so we can set the equation for acceleration a(t) = -ω^2A*cos(ωt+φ) equal to the given acceleration of -0.320m/s^2 and solve for ω. This gives us ω = 2π/T = 2π/4.50 = 1.396 rad/s.

Now, we can use the equations x(t) = A*cos(ωt+φ) and v(t) =