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Harmonic motion- Two springs, One mass

  1. Jan 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Two springs each have spring constant k and equilibrium length ℓ. They are both stretched a distance ℓ and then attached to a mass m and two walls (which are 4 ℓ apart).
    At a given instant, the right spring constant is somehow magically changed to 3k (the relaxed length remains ℓ).

    The question is: what is the resulting motion of the mass?
    Take the initial position to be x = 0.
    a. Show that the radial frequency of the new system is $$ \omega = 2 \sqrt{\frac{k}{m}} $$
    b. Determine the new equilibrium position (relative to the original one).
    c. Introduce a coordinate y that measures the position away from the new equilibrium position. Formulate the initial conditions and determine the motion y(t).
    d. Determine the motion x(t) with respect to the original equilibrium position.

    2. Relevant equations

    $$ F(x)= -kx $$
    $$ x(t)= Acos(\omega t+\phi) $$
    $$ v(t)= −A \omega sin(\omega t +\phi) $$
    $$ a= −A \omega 2 cos(\omega t + \phi) $$

    3. The attempt at a solution
    Drawing the forces:
    $$ 3kℓ - kℓ = m \ddot{x} $$
    $$ 2kℓ = m \ddot{x} $$
    But from this equation I don't get the result given in the question.

    Any hint will be great :)

    b. The equilibrium position is where the forces on the mass is 0, meaning that both of the spring forces deleting each other, but I'm not sure how to determine it.

    I tried to describe the forces according to the displacement for every spring:
    Left spring: -k(ℓ+x)
    Right spring: 3k(ℓ-x)
    But I'm not sure how to use/solve the equation:

    3k(ℓ-x) = k(ℓ+x)
    2kℓ = 4kx
    x= 0.5ℓ

    I'm not sure if my way is correct.

    c. Does the mass moves in the y direction?

    d. Any hint will be great :)

    Thanks alot !
  2. jcsd
  3. Jan 12, 2016 #2


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    In (a), you are asked about subsequent motion. The extension l is only the initial state. When the displacement is x, what will the tensions in the springs be? (My inclination would be to solve b first, then use that as the x=0 for part d, then deduce a.)
    Your solution to b is correct.
    I think you are only supposed to consider motion in the x direction.
  4. Jan 13, 2016 #3
    I tried to solve using the new equilibrium position combining with the equations:

    The new equilibrium position is x= 0.5ℓ
    Therefor the Amplitude is $$ A=0.5ℓ $$
    $$ \dot{x}(0)= 0 $$
    $$ \dot{x}(0)= 0.5ℓ \omega cos(\phi) = 0$$
    $$ \phi = 0 $$

    The forces in the x(0)
    $$ 2kℓ = m \ddot{x}(0) $$
    $$ \ddot{x}(0) = 2kℓ/m $$
    $$ \ddot{x}(0) = a= − 0.5ℓ ω^2cos(ωt+ϕ) $$
    $$ 2kℓ/m = − 0.5ℓ ω^2cos(0) $$
    $$ ω^2 = 4k/m $$
    $$ ω = 2 \sqrt{k/m} $$

    $$ x(t)= 0.5ℓ cos(2 \sqrt{k/m}*t) $$

    Does is make sense ?
  5. Jan 13, 2016 #4


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    A few slips in there. It's not clear whether you are taking x(t) as the sine or the cosine of ##\omega t+\phi##. You have both ##\dot x## and ##\ddot x## using the cosine. Also you took cos(0) to be 0. Somehow this led to a negative value for ##\omega^2##, but you wisely ignored that.
    In terms of the problem as set, do you think your final equation is for x(t) or y(t)? Does it satisfy the initial conditions?
  6. Jan 13, 2016 #5
    Hey, Thanks so much !

    I noticed the mistake now, ##\dot x## with sine, it supposed to be:
    ##\dot x(0) = -0.5ℓωsin(ϕ)=0##
    ==> ## ϕ=0 ##

    about the negative value i wasn't sure if the equation is correct, should I put negative signs for both sides?
    The final equation is for x(t).
  7. Jan 13, 2016 #6


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    Looking at your final equation, where is the equilibrium? Where should it be for x?
  8. Jan 13, 2016 #7
    Corresponding with the initial x, the new equilibrium position is x=0.5ℓ.
  9. Jan 13, 2016 #8


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    Yes, but is that what your final equation in post #3 says? What, according to that equation, is the range of x?
  10. Jan 13, 2016 #9
    ohh i miss the displacement ? The spring should range from 0 to ℓ therefore:
    $$ x(t)= 0.5ℓ - 0.5ℓ cos(2 \sqrt{k/m}*t) $$
  11. Jan 13, 2016 #10


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