Undamped Harmonic Motion (ODE problem)

In summary, the equation of motion for a 24-lb weight attached to a spring and released from a point 3 inches above the equilibrium position is \frac{d^{2}x}{dt^2}+96x=0. To solve for the constants C_{1} and C_{2}, set x(0)=(-1/4) and x'(0)=0.
  • #1
leroyjenkens
616
49

Homework Statement


A 24-lb weight, attached to the end of a spring, stretches it 4 inches. Find the equation of motion if the weight is released from rest from a point 3 inches above the equilibrium position.


Homework Equations


[tex]\frac{d^{2}{x}}{dt^2}+\frac{k}{m}x=0[/tex]
[tex]F=ma[/tex]


The Attempt at a Solution



First I found the spring constant k using [itex]F=ma=ks[/itex] at the point of equilibrium where the weight isn't moving. s is the displacement the weight causes on the spring, which is 4 inches, or 1/3 feet. So that equation turns out to be the weight equals the mass times the acceleration of gravity, which equals the spring constant times the displacement the weight caused, which is [itex]24=k(\frac{1}{3})[/itex].

I found mass, m, by using [itex]F=ma[/itex] which is [itex]24lb=m(32\frac{ft}{s^2})[/itex] which gives me [itex]\frac{3}{4} slug[/itex]

Putting that into my relevant equation at the top, it gives me [itex]\frac{d^{2}x}{dt^2}+96x=0[/itex] And using that one method (I forgot the name), I get [itex]m^2+96=0[/itex] which gives me [itex]m=\pm\sqrt{96}i[/itex]

I use that for [itex]C_{1}cos\sqrt{96}t+c_{2}sin\sqrt{96}t[/itex]

Now I have to find initial conditions to solve for [itex]C_{1}[/itex] and [itex]C_{2}[/itex]

My initial position is [itex]\frac{1}{4}ft[/itex] above the equilibrium point, so does that mean I set that equation equal to [itex]-\frac{1}{4}[/itex] and then solve for when t = 0?

And to get the other one, do I need to find the equation for the velocity by deriving the position equation? I'm not sure what I would set that equation equal to.

Thanks.
 
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  • #2
leroyjenkens said:

Homework Statement


A 24-lb weight, attached to the end of a spring, stretches it 4 inches. Find the equation of motion if the weight is released from rest from a point 3 inches above the equilibrium position.


Homework Equations


[tex]\frac{d^{2}{x}}{dt^2}+\frac{k}{m}x=0[/tex]
[tex]F=ma[/tex]


The Attempt at a Solution



First I found the spring constant k using [itex]F=ma=ks[/itex] at the point of equilibrium where the weight isn't moving. s is the displacement the weight causes on the spring, which is 4 inches, or 1/3 feet. So that equation turns out to be the weight equals the mass times the acceleration of gravity, which equals the spring constant times the displacement the weight caused, which is [itex]24=k(\frac{1}{3})[/itex].

I found mass, m, by using [itex]F=ma[/itex] which is [itex]24lb=m(32\frac{ft}{s^2})[/itex] which gives me [itex]\frac{3}{4} slug[/itex]

Putting that into my relevant equation at the top, it gives me [itex]\frac{d^{2}x}{dt^2}+96x=0[/itex] And using that one method (I forgot the name), I get [itex]m^2+96=0[/itex] which gives me [itex]m=\pm\sqrt{96}i[/itex]

I use that for [itex]C_{1}cos\sqrt{96}t+c_{2}sin\sqrt{96}t[/itex]

Now I have to find initial conditions to solve for [itex]C_{1}[/itex] and [itex]C_{2}[/itex]

My initial position is [itex]\frac{1}{4}ft[/itex] above the equilibrium point, so does that mean I set that equation equal to [itex]-\frac{1}{4}[/itex] and then solve for when t = 0?

And to get the other one, do I need to find the equation for the velocity by deriving the position equation? I'm not sure what I would set that equation equal to.

Thanks.

Yes, if x(t) is the position then you know x(0)=(-1/4) (if you want displacements above equilibrium to be negative). Since it's release "from rest" you also know velocity at t=0 is 0. So set x'(0)=0.
 

1. What is undamped harmonic motion?

Undamped harmonic motion is a type of motion in which an object oscillates back and forth at a constant frequency and amplitude without any external forces acting on it. This type of motion is described by a second-order differential equation known as the undamped harmonic oscillator equation.

2. What is the difference between undamped and damped harmonic motion?

The main difference between undamped and damped harmonic motion is the presence of external forces. In undamped harmonic motion, there are no external forces acting on the object, while in damped harmonic motion, there is an external force (usually friction) that causes the oscillations to gradually decrease over time.

3. How is undamped harmonic motion described mathematically?

Undamped harmonic motion is described by a second-order differential equation, where the acceleration of the object is proportional to its displacement from its equilibrium position, and in the opposite direction. This equation can be written as: x'' + ω^2x = 0, where x is the displacement, x'' is the second derivative of x with respect to time, and ω is the angular frequency of the oscillations.

4. What is the natural frequency in undamped harmonic motion?

The natural frequency in undamped harmonic motion is the frequency at which the object will oscillate without any external forces acting on it. It is determined by the equation ω = √(k/m), where k is the spring constant and m is the mass of the object.

5. How can undamped harmonic motion be applied in real-world situations?

Undamped harmonic motion can be observed in various real-world situations, such as a pendulum, a mass-spring system, or a vibrating guitar string. It is also used in engineering and physics to model the behavior of systems such as bridges, buildings, and electronic circuits.

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