# Can the equivalence principle be safely used in non-relativistic mechanics?

Yes, it can be. In classical mechanics the principle is nothing but the statement that inertial and gravitational masses are identical. As a consequence all inertial forces can be mathematically interpreted as gravitational forces using the standard mathematical machinery of Newtonian mechanics. Your solution is correct.

If we write the acceleration of the pendulum as a combination of the uniform acceleration ($$a_u$$) and the acceleartion in the rest frame ($$a_r$$) of the train we get $$\vec{a} = m \vec{a_u} + m \vec{a_r} = \vec{F_r} + \vec{F_g}$$ with $$F_g$$ the gravitational force acting on the pendulum and $$F_r$$ the reaction force of the string the pendulum is attached to. What we now want to find is an expression for $$a_r$$, we write $$m \vec{a_r} = \vec{F_r} - m(\vec{g} -\vec{a_u})$$ This is of course quite straightforward. If I understand your problem correct, you are wondering whether the reaction force $$\vec{F_r}$$ is now different then when the pendulum would swing in a uniform gravitational field with gravitational acceleration $$\vec{g}-\vec{a_u}$$.

One should remember how this reaction force is determined when solving the E.O.M. for a pendulum, it is uniquely determined by requiring that $$\vec{a_r} \perp \vec{v}$$ and thus that the pendulum swings on a circle. There is no difference in our case, we must still have $$\vec{a_r}\perp\vec{v}$$ and will find therefore exactly the same reaction force as we would for a pendulum in a $$\vec{g}-\vec{a_u}$$ gravitational field.