Question:

# Mathematical pendulum in accelerating frame of reference

Lily: 3 days ago

An aquintance of mine, who is a first year physics student was given a simple task as a homework-like task, which is about determining the ratio of periods between two equal-parameter mathematical pendulums, where one is in an inertial frame, and the other is on a train wagon accelerating with constant $a$.

As equations of motion, assuming small-amplitude vibrations, I got $$\frac{d^{2}}{dt^{2}}\varphi(t)+\frac{g}{l}\varphi(t)=-\frac{a}{l},$$ where $\varphi$ is the pendulum's angular displacement from the vertical, and $l$ is the pendulum's length.

This is an ordinary, second order, linear, inhomogenous, constant coefficient differential equation, which I solved and got $$\varphi(t)=A\cos(\omega t+\alpha)-\frac{a}{g}$$ as its general solution, and making the substitution, this is a correct solution.

According to this solution, however, the pendulum has the same period as an unaccelerated one, and according to the solution manual (which we do have, just not a method of solution), the solution is $$\frac{t_1}{t_2}=4\sqrt{1+(\frac{a}{g})^2}$$where $t_1$ is the unaccelerated pendulum's period, and $t_2$ is the accelerated one's period. This, however, seems to be contradictory with my solution of the differential equation.

Any help is greatly appreciated in how to solve this problem.

For the first pendulum, we obtain harmonic motion with angular frequency $\omega_1=\sqrt{g/l}$.
For the second pendulum, the combined acceleration of the cart and gravity produce an effective acceleration with magnitude $g_e=\sqrt{g^2+a^2}$ at an angle $\tan^1(a/g)$. It is as if gravity has been tilted and increased in strength. This is the only difference, though, so the motion of this pendulum is also harmonic and has angular frequency $$\omega_2=\sqrt{\frac{g_e}{l}}=\sqrt{\frac{\sqrt{g^2+a^2}}{l}}=\sqrt{g/l}\sqrt{\sqrt{1+\left(\frac{a}{g}\right)^2}}=\omega_1\sqrt[4]{1+\left(\frac{a}{g}\right)^2}.$$ Since the frequency and the period are inversely proportional, we conclude that $$\frac{t_1}{t_2}=\sqrt[4]{1+\left(\frac{a}{g}\right)^2}.$$