Mass of Pivoted Rod in Spring Oscillation

The image displays a physical pendulum system consisting of a uniform rod of mass $M$ and length $L$, pivoted at its geometric center. A spring with constant $k$ is attached horizontally to the bottom end of the rod. The system is shown at a small angular displacement $\theta$.

Answer

The correct option is (B). By equating the restoring torque of the spring to the product of the rotational inertia and angular acceleration, we find that the mass $M$ is proportional to $kT^2/\pi^2$ with a coefficient of $3/4$.

Explanation

Known: Rotational inertia $I = \frac{1}{12}ML^2$, spring constant $k$, period $T$, pivot at $L/2$. Find: Mass $M$.

1. Analyze the Restoring Torque When the rod is displaced by a small angle $\theta$, the end of the rod moves a distance $x \approx \frac{L}{2}\theta$ (using the small angle approximation $\sin \theta \approx \theta$). The spring exerts a force $F = -kx = -k(\frac{L}{2}\theta)$. The torque $\tau$ is the product of this force and the lever arm (which is $L/2$). $\tau = -F \cdot r = -k\left(\frac{L}{2}\theta\right) \cdot \frac{L}{2} = -\frac{kL^2}{4}\theta$ The torque is proportional to the angular displacement and acts in the opposite direction.

2. Apply Newton's Second Law for Rotation We use the rotational form of Newton's second law, $\tau = I\alpha$, where $\alpha = \frac{d^2\theta}{dt^2}$. $I\alpha = -\frac{kL^2}{4}\theta \implies \frac{1}{12}ML^2 \frac{d^2\theta}{dt^2} = -\frac{kL^2}{4}\theta$ This formula relates the angular acceleration to the restoring torque of the spring.

3. Determine Angular Frequency $\omega$ Rewrite the equation in the standard form for Simple Harmonic Motion (SHM): $\frac{d^2\theta}{dt^2} + \omega^2\theta = 0$. $\alpha = -\frac{\frac{kL^2}{4}}{\frac{ML^2}{12}}\theta = -\frac{3k}{M}\theta$ From this, we identify the squared angular frequency: $\omega^2 = \frac{3k}{M}$ The angular frequency depends only on the spring constant and the mass, as the length $L^2$ cancels out.

4. Relate Period $T$ to Mass $M$ The relationship between the period and angular frequency is $T = \frac{2\pi}{\omega}$, or $T^2 = \frac{4\pi^2}{\omega^2}$. $T^2 = 4\pi^2 \left(\frac{M}{3k}\right)$ This equation links the observable period to the physical parameters of the system.

5. Solve for Mass and Unit Check Rearrange the equation to isolate $M$: $M = \frac{3kT^2}{4\pi^2} = \frac{3}{4}\frac{kT^2}{\pi^2}$ Unit Check: $[k] = N/m = kg/s^2$. $[T^2] = s^2$. Thus, $(kg/s^2) \cdot s^2 = kg$. The dimensions are consistent with mass.

Option	Formula	Reason for Error
(A)	$\frac{1}{4} \frac{kT^2}{\pi^2}$	Forgotten factor of 3 from the $1/12$ inertia term.
(B)	$\frac{3}{4} \frac{kT^2}{\pi^2}$	Correct derivation of torque and inertia.
(C)	$\frac{3}{2} \frac{kT^2}{\pi^2}$	Calculation error in the lever arm squared $(1/2 \times 1/2 = 1/4)$.
(D)	$3 \frac{kT^2}{\pi^2}$	Incorrectly used full length $L$ instead of $L/2$ for torque.

Final Answer

The mass is determined by the equation: $\boxed{M = \frac{3}{4} \frac{kT^2}{\pi^2}}$

Common Mistakes

• Lever Arm Squared: Students often forget that the distance $L/2$ appears twice: once in the force calculation ($F = k \Delta x$) and once in the torque calculation ($\tau = rF$). This results in a factor of $L^2/4$, not $L^2/2$.
• Inertia Reference Point: Ensure the $I$ value matches the pivot point. If the rod was pivoted at the end, $I$ would be $\frac{1}{3}ML^2$, changing the final coefficient significantly.

Related Topics: Torsional Oscillators, Parallel Axis Theorem, Small Angle Approximation in Pendulums.