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Two springs and a mass

  1. Apr 11, 2016 #1
    1. The problem statement, all variables and given/known data
    the two springs are identical and have the same constant k. they also have the same contracted length l0. a force of ##k\cdot l_0## has to be applied in order to start stretching the springs from length l0 and on.
    L>l0, ##mg>k\cdot l_0##.
    The ring slides frictionless on the vertical column.
    Draw a graph of the force the spring applies as a function of it's elongation and express the force each spring applies as a function of it's length l.
    Use the graph and write the expression for the elastic energy as a function of the spring's length l.
    When the connection point between the springs A is at yA, what's the force the pipe exerts on the ring?
    find y0, the location of the weight at equilibrium.
    Express the potential energy of the 2 springs and weight system as a function of the mass's location y.
    The weight is released from rest from point y=l0, where will it stop the first time?
    What's it's maximum speed during it's movement?
    Snap1.jpg
    2. Relevant equations
    Elastic energy of spring: ##E=\frac{1}{2}kx^2##
    Constant of an equivalent spring ##\over{k}## of two springs in a row: ##\frac{1}{\over{k}}=\frac{1}{k_1}+\frac{1}{k_2}##

    3. The attempt at a solution
    Snap2.jpg The elastic energy:
    $$E=(l-l_0)kl_0+\frac{1}{2}k(l-l_0)^2=\frac{1}{2}k(l^2-l_0^2)$$
    The force the horizontal spring applies: ##F_1=k\left( \sqrt{L^2+y_A^2} \right)##
    The vertical one: ##F_2=k(y-y_A-l_0)##
    $$F_{pipe}=F_1\cos\alpha=k\left( \sqrt{L^2+y_A^2} \right)\frac{L}{\left( \sqrt{L^2+y_A^2} \right)}=kL$$
    The location of the mass at equilibrium y0 is found with 2 methods: the equivalent spring and a direct method.
    The vertical force at yA the horizontal spring applies is:
    $$F_{ver}=F_1\cos\alpha=k\left( \sqrt{L^2+y_A^2} \right)\frac{y_A}{\sqrt{L^2+y_A^2}}=ky_A$$
    This is an equation of a linear spring, so the equivalent constant of the the "two" vertical springs is ##\frac{1}{2}k##
    Using this spring y0 is:
    $$mg=\frac{k}{2}y_0\;\rightarrow\; y_0=\frac{2mg}{k}$$
    With the direct method is use the vertical force the horizontal spring exerts:
    $$mg=F_1\sin\alpha\;\rightarrow\;mg=ky_A\;\rightarrow\;y_{0A}=\frac{mg}{k}$$
    The other spring stretches also the same length so in total ##y_0=\frac{2mg}{k}##
    The potential energy of the horizontal spring:
    Since the vertical force of the horizontal spring spring at yA has the same constant k as the lower spring: ##y_A=\frac{y}{2}##
    $$E_1=\frac{1}{2}k\left[ \left( \sqrt{L^2+y_A^2} \right)^2-l_o^2 \right]=\frac{k}{2}\left[ L^2+\frac{y^2}{4}-l_o^2 \right]$$
    The potential energy of the vertical one:
    $$E_1=\frac{1}{2}k\left[ (y-y_A)^2-l_0^2 \right]=\frac{k}{2}\left( \frac{y^2}{4}-l_0^2 \right)$$
    I decide to make the gravitational energy of m relative to the equilibrium point ##\frac{2mg}{k}##:
    $$E_p=mg\left[ y-\frac{2mg}{k} \right]$$
    m is released from l0, where will it stop first? the potential energy at l0 is:
    $$(E_1+E_2)_{l_0}=\frac{k}{2}\left[ L^2+\frac{l_0^2}{4}-l_0^2+\frac{l_0^2}{4}-l_0^2 \right]$$
    $$(E_1+E_2)_{l_0}=E_p=\frac{k}{2}\left[ L^2-\frac{3}{2}l_0^2 \right]$$
    This energy equals the gravitational energy:
    $$E_p=mg\cdot y\;\rightarrow\;y=\frac{k}{2mg}\left[ L^2-\frac{3}{2}l_0^2 \right]$$
    The mass makes a harmonic motion: ##y=A\cos(\omega t+\theta)##
    $$A=\frac{2mg}{k}-l_0,\;\omega=\sqrt{\frac{2m}{k}}$$
    The velocity is max when m passes through the equilibrium point, where ##\sin=0##:
    $$\dot x=-\omega A\sin(\omega t+\theta)\;\rightarrow\; \dot x=\sqrt{\frac{2m}{k}}\left( \frac{2mg}{k}-l_0 \right)$$
     
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  3. Apr 11, 2016 #2

    haruspex

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    This is a strange specification. Is it in effect saying the relaxed length is zero, but some physical constraint prevents the length ever being less than l0?
     
  4. Apr 11, 2016 #3
    Yes, the springs are identical. the coils of the springs cling to one another and prevent the spring from shortening.
     
  5. Apr 11, 2016 #4

    haruspex

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    Ok, but do you agree that the relaxed length is effectively zero, so if the total length of the spring is x (>l0) then the tension is kx. I don't think that's what your equations say.
     
  6. Apr 12, 2016 #5
    The force the vertical spring applies: ##F_2=k(y-y_A)##, and since i found that ##y_A=\frac{y}{2}##:
    $$F_2=\frac{k}{2}y$$
    Since i guess the energy i made right, so that is the only change.
    I guess everything else is correct, is it? since all the rest is based on energy
     
  7. Apr 13, 2016 #6

    haruspex

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    Yes, the final equations look right. As you found, the spring at an angle effectively acts as two springs, one vertical and one horizontal, of the same constant k. The horizontal one has no effect on the motion, so the problem reduces to two vertical springs in series with a combined constant k/2.
     
  8. Apr 14, 2016 #7
    Thank you very much Haruspex. is the rest O.K.?
     
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