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.

Answer:
Athena: 3 days ago

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 $$\frac{t_1}{t_2}=\sqrt[4]{1+\left(\frac{a}{g}\right)^2}.$$