Classical Mechanics: Simple harmonic oscillator problem

[JordanGo]

Homework Statement

A simple harmonic oscillator with mass m = 1/2 and k = 2 is initially at the point
x = √3 when it is projected towards the origin with speed 2.
Find the equation of motion describing x(t).

Homework Equations

x=Asin(ωt+θ)

The Attempt at a Solution

At t=0, x=√3

√3=Asin(θ)

There is two unknowns and only one equation...I'm stuck

 

[gomboc]

You just need to use a little ingenuity to solve for A.

You know that the mass is projected towards the eq point from x=sqrt(3) with velocity 2. Since you know the spring constant and the mass, you can find the mass's total energy, and thus it's position (A) at maximum extension using conservation of energy.

 

[technician]

Do you recognise that ω = √k/m ?

 

[JordanGo]

K, well I did a lot of work on white board and my conclusion is:
x(t)=2sin(4t-pi/3)
Does this make sense? Do you need to see all my work?

 

[asteriatic]

I would approach this problem by first defining the variables and understanding the principles of a simple harmonic oscillator. The mass m and spring constant k are known values in this problem, and the initial position and velocity of the oscillator are also given. The equation of motion for a simple harmonic oscillator is x = Asin(ωt + θ), where A is the amplitude, ω is the angular frequency, and θ is the phase angle.

To solve for the equation of motion, we need to find the values of A, ω, and θ. We can use the initial conditions given in the problem to find these values. At t = 0, x = √3 and the oscillator is projected towards the origin with a speed of 2. This means that the initial velocity is v = 2 and the initial displacement is x = √3. We can use these values to find the amplitude A and the phase angle θ.

From the initial displacement, we can determine that A = √3, as the amplitude is the maximum displacement from the equilibrium position. From the initial velocity, we can use the equation v = ωAcos(ωt + θ) to solve for ω. Plugging in the values of v = 2 and A = √3, we get ω = 2/√3.

Now that we have the values of A and ω, we can write the equation of motion as x = √3sin((2/√3)t + θ). To solve for θ, we can use the initial displacement again. At t = 0, x = √3, so we can write the equation as √3 = √3sin(θ). This means that θ = 0, as sin(0) = 0.

Therefore, the equation of motion for this simple harmonic oscillator is x = √3sin((2/√3)t). This equation describes the position of the oscillator at any given time t, starting from the initial position of √3 and moving towards the origin with a speed of 2. This solution is consistent with the principles of classical mechanics and the properties of simple harmonic motion.