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Homework Help: Young's Modulus problem

  1. Jan 28, 2017 #1


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    1. The problem statement, all variables and given/known data
    A particle mass [itex] m [/itex] moves in a straight line on a smooth horizontal table, and is connected to two points [itex] A [/itex] and [itex] B [/itex] by light elastic springs of natural lengths [itex] 2l_{o} [/itex] and [itex] 3l_{o} [/itex], respectively, and modulus of elasticity [itex] λ [/itex]. The points [itex] A [/itex] and [itex] B [/itex] are a distance [itex] 6l_{o} [/itex] apart. Show that the equation of motion can be written as [tex] m \ddot{x} = \frac{\lambda}{6l_{o}}(12l_{o}-5x) [/tex]
    where [itex] x [/itex] is the displacement of the particle from [itex] A [/itex] measures positive towards [itex] B [/itex]
    2. Relevant equations
    [tex] F = kx [/tex]
    [tex] \lambda = \frac{x}{l_{o}} [/tex]
    [tex] \frac{F}{A} = \lambda \frac{x}{l_{o}} [/tex]

    3. The attempt at a solution
    I'm not sure what to do here. I understand what the question is asking but I'm not sure how to go about it. It's asking for the equation of motion so does that mean I have to relate Hooke's Law with Young's modulus? The problem I'm having is that the equation that I was trying to solve the problem with has area in it. But we are talking about springs. So that's what makes me think I need to find a relationship between spring constant and modulus equation. The [itex] (12l_{o}-5x) [/itex] part, is that from [itex] F= k(x - x_{o}) [/itex]

    I've tried to relate hooke's law using this formula I read online [tex] k = \frac{\lambda A}{l} [/tex]

    if you could point me in the right direction I'd be very grateful.
  2. jcsd
  3. Jan 28, 2017 #2
    A picture would be helpful. If I understand the problem, points A and B are 6L apart. Spring "A" = 2L and spring "B" = 3L, so before the problem even begins, the springs must be in an initially stretched state 2L + 3L + Xo = 6L correct? or X (initial combined stretch of the springs = L)?
  4. Jan 28, 2017 #3


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    Looking at the thing to be proved, λ has the dimension of force. So it is not what is properly called modulus of elasticity. Rather, it is the modulus multiplied by the cross sectional area.
  5. Jan 30, 2017 #4
    The "relevant equations" seem to imply Lambda is 2 different things? The second equation implies lambda is some form of strain, the 3rd equation implies lambda is the modulus of elasticity (Young's Modulus). The solution equation implies the motion is undamped free vibration --> no velocity or damping terms are included.
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