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## Homework Statement

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 m

_{1}+Δm is connected to a body of mass m

_{2}-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 m

_{1}and m

_{2}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

$$x=c \Delta m ({\frac{m_1 + m_2}{m_1 m_2 k} })^{\frac{1}{2}}$$

## Homework Equations

E=cp

E=mc

^{2}

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

## 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 x

_{1}and x

_{2}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.