A massless rod is pivoted at point 0. A string carrying a mass m at one end is attached to the point A on the rod.The period ofsmall vertical oscillation of mass m aroundits equilibrium position is :

To determine the period of small vertical oscillations of a mass \( m \) attached to a massless rod pivoted at point \( O \), we can analyze the system using principles of physics, particularly those related to simple harmonic motion (SHM). The setup resembles a pendulum, where the mass oscillates around an equilibrium position due to gravitational forces.

Understanding the System

In this scenario, the mass \( m \) is attached to a point \( A \) on the rod, which is pivoted at point \( O \). When the mass is displaced slightly from its equilibrium position and released, it will experience a restoring force due to gravity, causing it to oscillate vertically.

Key Concepts

• Restoring Force: When the mass is displaced, gravity acts on it, creating a torque about the pivot point \( O \).
• Small Angle Approximation: For small displacements, we can assume that the angle \( \theta \) (the angle between the rod and the vertical) is small enough that \( \sin(\theta) \approx \theta \).
• Moment of Inertia: The moment of inertia of the mass about the pivot point is crucial for determining the dynamics of the oscillation.

Deriving the Period of Oscillation

To find the period \( T \) of the oscillation, we can use the formula for the period of a physical pendulum:

Step-by-Step Calculation

1. Torque Calculation: The torque \( \tau \) acting on the mass when displaced by an angle \( \theta \) is given by:

   \( \tau = -m g L \sin(\theta) \approx -m g L \theta \)

   where \( L \) is the distance from the pivot \( O \) to the mass \( m \) and \( g \) is the acceleration due to gravity.
2. Angular Acceleration: Using Newton's second law for rotation, we have:

   \( \tau = I \alpha \)

   where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration. For a point mass, \( I = m L^2 \), so:

   \( -m g L \theta = m L^2 \alpha \)

   Since \( \alpha = \frac{d^2\theta}{dt^2} \), we can rewrite this as:

   \( -g L \theta = L \frac{d^2\theta}{dt^2} \)

3. Forming the Equation of Motion: Rearranging gives us:

   \( \frac{d^2\theta}{dt^2} + \frac{g}{L} \theta = 0 \)

   This is a standard form of the simple harmonic motion equation.
4. Finding the Period: The general solution for the period \( T \) of a simple harmonic oscillator is given by:

   \( T = 2\pi \sqrt{\frac{I}{m g L}} \)

   Substituting \( I = m L^2 \) into the equation gives:

   \( T = 2\pi \sqrt{\frac{m L^2}{m g L}} = 2\pi \sqrt{\frac{L}{g}} \)

Final Result

The period of small vertical oscillations of the mass \( m \) attached to the rod is therefore:

T = 2π√(L/g)

This result indicates that the period of oscillation depends only on the length of the rod \( L \) and the acceleration due to gravity \( g \), not on the mass \( m \) itself. This is a fascinating aspect of simple harmonic motion, showing that the mass does not influence the period in this specific setup.