Topic: Oscillatory Motion
Concepts: Rotational Newton’s 2nd law, Simple harmonic motion, Small angle approximation

---

Solution:

We first find the original oscillation frequency $f$.

If we displace the rod by a small angle $\theta$, then the spring gets stretched by $x=l\theta$. The spring force $F$ is

and the corresponding torque $\tau$ is

where we used the small angle approximation $\cos\theta \approx 1$ and the fact that $L \gg l$ to assume the spring force acts downward. Applying Newton’s 2nd law $\tau=I\alpha$,

This is of simple harmonic form ($\ddot{z}=-\omega^2z$) so we can identify the angular frequency as

Then the frequency $f$ is

If the spring is moved to the midpoint of the rod, then the moment arm $l$ is halved to $l’=l/2$ for the force and torque calculation. The moment of inertia $I$ is unchanged since the rod is still rotating around the same point. Because we found $f \propto l$, we have

so the answer is A.

Topic: Oscillatory Motion
Concepts: Rotational Newton’s 2nd law, Simple harmonic motion, Small angle approximation

---

Solution:

We found in the previous problem that

When the spring is moved to the right of the rod, we have a constant spring force $F=-kl$ pointing to the right (using our given assumption $L \gg l$). If we displace the rod by a small angle $\theta$, the restoring torque $\tau$ is

Applying Newton’s 2nd law $\tau=I\alpha$,

This is the same equation of motion as that followed by the original system so the frequency $f’=f$ is the same as before. Thus, the answer is C.