A mass attached to a spring hanging straight down will experience a force mg due to gravity. In this case, the equation for the vertical motion is my′′ + γy′ + ky = mg (Note that y is pointing down.) To find the general solution, find the null + particular solution. A particular solution is yp(t) = mg/k. Why? (a)  Determine the equilibrium position. How does it vary with mass and spring stiffness? (b)  Solve the equation with m = 3,γ = 0,k = 2,g = 10 and starting values y(0) = 0,y′(0) = 0. Sketch the solution and relate it to the hanging mass. (c)  What is the amplitude of the motion if the initial velocity is changed to y′(0) = 2?

icon
Related questions
Question
  1. A mass attached to a spring hanging straight down will experience a force mg due to gravity. In this case, the equation for the vertical motion is

    my′′ + γy′ + ky = mg
    (Note that y is pointing down.) To find the general solution, find the null + particular

    solution. A particular solution is yp(t) = mg/k. Why?

    1. (a)  Determine the equilibrium position. How does it vary with mass and spring stiffness?

    2. (b)  Solve the equation with m = 3,γ = 0,k = 2,g = 10 and starting values y(0) = 0,y′(0) = 0. Sketch the solution and relate it to the hanging mass.

    3. (c)  What is the amplitude of the motion if the initial velocity is changed to y′(0) = 2?

Expert Solution
steps

Step by step

Solved in 7 steps with 10 images

Blurred answer