Simple harmonic motion and static equilibrium

In summary, the conversation discusses a homework problem involving a spring in static equilibrium that is initially stretched by a hanging mass and then released with zero velocity. The goal is to determine the position of the mass after 4.2 seconds. The equations used are for equilibrium and a second-order, constant coefficient, homogeneous differential equation. The attempt at a solution involved setting the equation equal to zero for all time, but a change is suggested to set the initial position of the spring to 0.25m.
  • #1
oceanwalk
6
0

Homework Statement


In static equilibrium a spring is stretched 0.25 m by a hanging mass. If the system is initially moved up to the unstretched position and released with zero velocity, determine position of the mass (in m) after 4.2s. Take the equilibrium position to be zero and up the positive direction.

Homework Equations



at equilibrium
-kx+mg = 0
m(x double dot (t)) + kx(t) = 0

The Attempt at a Solution

 
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  • #2
oceanwalk said:

Homework Statement


In static equilibrium a spring is stretched 0.25 m by a hanging mass. If the system is initially moved up to the unstretched position and released with zero velocity, determine position of the mass (in m) after 4.2s. Take the equilibrium position to be zero and up the positive direction.

Homework Equations



at equilibrium
-kx+mg = 0
m(x double dot (t)) + kx(t) = 0
Do you know how to solve a 2nd-order, constant coefficient, homogeneous differential equation?
This is the initial value problem:
x''(t) + (k/m)x(t) = 0, with initial conditios x(0) = 0, x'(0) = 0
 
  • #3
Mark44 said:
Do you know how to solve a 2nd-order, constant coefficient, homogeneous differential equation?
This is the initial value problem:
x''(t) + (k/m)x(t) = 0, with initial conditios x(0) = 0, x'(0) = 0


Ok, I must not know how to do it correctly because I'm getting it to equal zero for all time.

Solving the equation you gave me:
x(t) = C2 sin(sqrt(a) t) + C1 cos(sqrt(a) t) = 0
using the initial conditions
C2 = 0 , C1 = 0

as such, x(t) = 0 for all time
 
  • #4
Try it with one change: x(0) = 0.25. This means that we are taking the origin to be the point where the spring is in static equilibrium. Raising it .25 m. and then letting it go means x(0) = .25 and x'(0) = 0.
 
  • #5


To determine the position of the mass after 4.2s, we can use the equation for simple harmonic motion: x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle. In this case, since the system is initially at rest, the phase angle φ is 0. The amplitude A can be found using the equation for static equilibrium: kx+mg = 0, where k is the spring constant and m is the mass. Solving for x, we get x = -mg/k = -0.25 m. Thus, the amplitude A is 0.25 m.

The angular frequency ω can be found using the equation ω = √(k/m), where k is the spring constant and m is the mass. Plugging in the values, we get ω = √(mg/k) = √(9.8/0.25) = 6.26 rad/s.

Therefore, the position of the mass after 4.2s can be calculated as x(4.2) = 0.25 cos(6.26 * 4.2) = 0.25 cos(26.3) = 0.25 * 0.898 = 0.2245 m.

In conclusion, after 4.2s, the mass will be at a position of 0.2245 m above the equilibrium position.
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object moves back and forth along a straight line, with its acceleration being directly proportional to its displacement from a fixed point. It can be seen in various systems such as a mass on a spring, a pendulum, and a vibrating guitar string.

2. How is simple harmonic motion related to static equilibrium?

Static equilibrium refers to the state of an object where it is not moving and the net force acting on it is equal to zero. Simple harmonic motion can occur in a system that is in static equilibrium, as long as the object remains at rest at the equilibrium position. This means that the forces acting on the object are balanced, and the object will not accelerate or change its position.

3. What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x = A*cos(ωt+φ), where x is the displacement of the object, A is the amplitude (maximum displacement), ω is the angular frequency, and φ is the phase constant. This equation can be used to describe the position, velocity, and acceleration of an object undergoing simple harmonic motion.

4. How is the period of simple harmonic motion related to its frequency?

The period of simple harmonic motion is the time it takes for the object to complete one full cycle of motion. It is inversely proportional to the frequency, which is the number of cycles completed per second. This means that as the frequency increases, the period decreases, and vice versa.

5. What factors affect the period of simple harmonic motion?

The period of simple harmonic motion is affected by the mass of the object, the force constant of the system, and the amplitude of the motion. A larger mass or a smaller force constant results in a longer period, while a larger amplitude results in a shorter period. The period is also independent of the initial velocity and the phase constant of the motion.

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