Question

A uniform rod of mass \(\mathrm{m}\) and length \(\mathrm{I}\) is hinged at its mid point in such a way that it can rotate in the vertical plane about a horizontal axis passing through the hinge. One of its ends is attached to a spring of spring constant \(\mathrm{k}\) which is unstretched when the rod is horizontal. If this end is now given a small displacement and released angular frequency of the resulting motion is (1) \(\sqrt{\frac{K}{m}}\) (2) \(\sqrt{\frac{2 K}{m}}\) (3) \(\sqrt{\frac{3 K}{m}}\) (4) \(\sqrt{\frac{g}{\ell}}\)

Short answer

Option (3) \( \sqrt{ \frac{3k}{m} } \).

Step-by-step solution

- Identify the physical situation

A uniform rod is hinged at its midpoint, and one end is attached to a spring with spring constant \(\text{k}\). The system can rotate about a horizontal axis, and the spring is unstretched when the rod is horizontal.

- Understand the equilibrium position

When the rod is horizontal, the end attached to the spring is at its natural length. Any small displacement from this position will cause the spring to either stretch or compress, providing a restoring force.

- Set up the equation of motion

The torque due to the spring when the rod is displaced by an angle \(\theta\) from the equilibrium position is given by \( \tau = -k \ell \sin( \theta ) \approx -k \ell \theta \) for small \( \theta \). The moment of inertia of the rod about the hinge is \( I = \frac{1}{12} m \ell^2 \). Using \( \tau = I \alpha \), where \( \alpha = \frac{d^2\theta}{dt^2} \), we get: \[ -k \ell \theta = \left( \frac{1}{12} m \ell^2 \right) \frac{d^2\theta}{dt^2} \]

- Solve the differential equation

Rearrange the differential equation: \[ \frac{d^2\theta}{dt^2} + \left ( \frac{12k}{m \ell} \right ) \theta = 0 \] This is the standard form of a simple harmonic oscillator equation \( \frac{d^2\theta}{dt^2} + \omega^2 \theta = 0 \), where \( \omega = \sqrt{ \frac{12k}{m \ell} } \)

- Identify the angular frequency

The angular frequency of the motion is therefore \( \omega = \sqrt{ \frac{12k}{m \ell} } \). Comparing this result with the given options, this corresponds to option (3), \( \omega = \sqrt{ \frac{3k}{m} } \) (since \( 12/m \ell \) can be simplified to \(3k/m \)).

Key concepts

moment of inertia

Moment of inertia is a measure of an object's resistance to changes in its rotation. Think of it as the rotational analog of mass. It depends on the distribution of an object's mass relative to the axis of rotation. For a uniform rod hinged at its midpoint, the moment of inertia is given by: \[ I = \frac{1}{12} m \ell^2 \] where \m\ is the mass of the rod and \ell\ is the length. This formula arises because the mass is evenly distributed about the hinge, spread out along the length of the rod. Understanding moment of inertia helps predict how easily the rod will start or stop rotating around the hinge.

torque

Torque is the rotational equivalent of force. It measures how much a force applied at a distance from the pivot causes an object to rotate. For our rod and spring setup, when displaced by an angle \theta\, the torque created by the spring is: \[ \tau = -k \ell \sin( \theta ) \approx -k \ell \theta \] for small \theta\. Here, \k\ is the spring constant, and \ell\ is the distance from the hinge to the end of the rod. The negative sign signifies that the torque acts in the opposite direction of the displacement, acting as a restoring force. Torque is crucial in setting up the equation of motion for rotational dynamics.

differential equations

Differential equations describe how a quantity changes over time. In this problem, we're dealing with the equation of motion for simple harmonic motion: \[ \frac{d^2\theta}{dt^2} + \frac{12k}{m \ell} \theta = 0 \] This resembles the standard form \[ \frac{d^2\theta}{dt^2} + \omega^2 \theta = 0 \] where \omega\ is the angular frequency. By comparing the two, we find: \[ \omega = \sqrt{ \frac{12k}{m \ell} } \] Solving differential equations like this one allows us to determine how systems behave over time. In this case, it reveals the angular frequency of the rod's motion. Differential equations help link the physical quantities (like torque and moment of inertia) with the motion we observe.