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(Special relativity) Two masses connected by spring

  1. May 26, 2016 #1
    1. The problem statement, all variables and given/known data
    Hi all--humanist here. Now that the semester is over, I am taking the opportunity to (attempt to) self-study introductory SR. This is problem 12 in chapter 1 of Special Relativity by AP French.

    (1-12) A body of mass m1+Δm is connected to a body of mass m2-Δm by a spring of spring constant k and negligible mass. The system is at rest on a frictionless table. A burst of radiation is emitted by the first body and absorbed by the second, changing the masses to m1 and m2 and setting the system into oscillations. If the time of transit of the radiation is negligibly small compared to the period of oscillation, show that the maximum extension of the spring is given by
    $$x=c \Delta m ({\frac{m_1 + m_2}{m_1 m_2 k} })^{\frac{1}{2}}$$

    2. Relevant equations
    E=cp
    E=mc2
    And I guess he's assuming we'll use Hooke's law. (?) (Or else potential energy ##U= \frac{1}{2} kx^2##.)

    3. The attempt at a solution
    The problem seems to call for an analysis of the kind French used to treat "Einstein's box" earlier in the chapter. The radiation at the first mass causes a recoil pushing it to the left, and when the radiation is absorbed on the right, it imparts a recoil there as well.

    I took the assumption about the time of transit to mean that, to a good approximation, we can assume that the recoil on the left and right happen "at the same time". (I may be wrong, though; it just seems like it would get heinously complicated without using this approximation.)

    Given this, I tried to write equations for the positions x1 and x2 as functions of time, after the recoil, using the above equation to express the momentum imparted, as well as Hooke's law for the acceleration.
    $$x_1(t)= \frac {-\frac{E}{c}} {m_1} t + \frac {k(x_2 - x_1 - X)} {2 m_1} t^2$$
    $$x_2(t)= \frac {\frac{E}{c}} {m_2} t - \frac {k(x_2 - x_1 - X)} {2 m_2} t^2$$
    where ##X## is the equilibrium position of the spring.

    Then I took ##x_2 - x_1 \equiv x_{21}##, which led to an equation in the reduced mass system $$x_{21} (t)= \frac{Et}{c \mu} - \frac {k t^2}{2 \mu} (x_{21} (t) - X)$$

    I did some other things, including changing variables to get rid of ##X##. Oh, and I took the derivative $$\dot x_{21} = \frac{E}{\mu c} - \frac{k}{2 \mu} (2 t x_{21} + t^2 \dot x_{21})$$
    then used the fact that at maximum extension ##\dot x_{21}## is zero. This yielded $$x_{21} = \frac {E}{kct}$$

    (My other thought was that at maximum extension, all the energy would be potential energy in the spring, which I could then equate with the energy "created" by ##\Delta m##, but this is not getting me the right result.)

    And now I am realizing that if I take the period of oscillation ##T=2 \pi \sqrt{\frac{\mu}{k}}##, divide it by four (because it is presumably one-fourth of a period to get to maximum in SHM?), and substitute this value for ##t## in the above equation for ##x_{21}##, I get the right answer except a factor of ##\frac{2}{\pi}##! $$x_{21}=\frac{\Delta m c}{\sqrt{\mu k}} \frac {2}{\pi}$$

    If you've managed to read this far and can put up with the confused thinking of a non-physicist I (congratulate you and) would be grateful for any guidance.
     
  2. jcsd
  3. May 27, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    Right.
    Where does that formula comes from? The t2 part would need a constant acceleration, which does not happen.

    You can set up differential equations and solve them, but conservation of energy is so much easier.
     
  4. May 27, 2016 #3
    D: Yes, that was silly of me.

    I think my difficulty is a more basic conceptual confusion, getting in the way of writing the conservation conditions. At the moment of maximum extension, there will be energy stored in the spring as well as the rest energy of the two masses, m1 and m2: ##\frac{1}{2} k x^2 + m_1 c^2 + m_2 c^2##. When they are at rest initially, they have rest energy ##m_1 + \Delta m## and ##m_2 - \Delta m##. But when you sum these, you will just get the rest energy of m1 and m2, because the ##\Delta m##s will cancel, and then when you equate these two expressions you get ##\frac{1}{2} k x^2 = 0##, which is wrong. Surely the (rest) masses must be less than ##m_1## and ##m_2## at maximum extension? Or perhaps my difficulty is whether ##m_1## and ##m_2## are supposed to be the masses including the initial kinetic energy imparted by the recoil? I just have little idea of what, physically, is supposed to be going on...

    I might be beyond the pale, here. Thank you for your reply, though.
     
  5. May 27, 2016 #4

    mfb

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    2016 Award

    Staff: Mentor

    Well, to have exact energy conservation during the exchange, you have to consider the recoil of the masses - if one emits ##\Delta m c^2## of energy it will lose more as mass, and the other will gain less as mass as it recoils as well. Those are second-order effects, however, I don't think you have to take them into account. Consider the momentum exchange for objects at rest, then let the masses move according to their momentum and treat everything else nonrelativistically.
     
  6. May 27, 2016 #5
    Dear lord, I spent so much time trying to even conceive of how I would take into account those second order effects mathematically, and then the answer turned out to be (barely) two lines long... :eek:
    Thank you for your help!
     
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