A square metal frame in the vertical plane is hinged at O at its centre . a bug moves along the rod PN which is at a distance l from the hinge,such that the whole frame is always stationary ,even though the frame is free to rotate in the vertical plane about the hinge .then the motion of the bug will be simple harmonic , with time period??

To analyze the motion of the bug on the square metal frame, we need to consider the dynamics of the system. The frame is hinged at point O and can rotate freely in the vertical plane. The bug moves along the rod PN, which is at a distance l from the hinge. Given that the frame remains stationary while the bug moves, we can deduce that the bug's motion will indeed exhibit simple harmonic characteristics. Let's break this down step by step.

Understanding the System Dynamics

The square frame can be thought of as a rigid body that is pivoted at its center. When the bug moves along the rod PN, it creates a torque about the hinge point O. This torque is a result of the gravitational force acting on the bug, which tries to pull it downward. The key here is that the frame is stationary, meaning that any motion of the bug must be balanced by the frame's response.

Torque and Restoring Force

As the bug moves, it creates a torque given by the formula:

• Torque (τ) = r × F

Where:

• r is the distance from the hinge to the bug (which is l), and
• F is the gravitational force acting on the bug (F = mg, where m is the mass of the bug and g is the acceleration due to gravity).

This torque will create a restoring force that tries to bring the bug back to its equilibrium position. The restoring force is proportional to the displacement of the bug from its equilibrium position, which is a characteristic of simple harmonic motion (SHM).

Establishing the Motion's Characteristics

For the motion to be classified as simple harmonic, it must satisfy the condition that the acceleration of the bug is directly proportional to its displacement from the equilibrium position and directed towards that position. Mathematically, this can be expressed as:

• a = -ω²x

Where:

• a is the acceleration,
• ω is the angular frequency, and
• x is the displacement from the equilibrium position.

Calculating the Time Period

The time period (T) of simple harmonic motion is given by the formula:

• T = 2π/ω

To find ω, we can relate it to the physical parameters of the system. The angular frequency for a simple harmonic oscillator can be derived from the formula:

• ω = √(g/l)

Substituting this into the time period formula gives us:

• T = 2π√(l/g)

Final Thoughts

Thus, the time period of the bug's motion along the rod PN, while the square frame remains stationary, is given by:

• T = 2π√(l/g)

This relationship shows that the time period depends on the distance l from the hinge and the acceleration due to gravity g. The further the bug is from the hinge, the longer the time period, which aligns with our understanding of pendulum-like systems. This analysis illustrates the elegant nature of simple harmonic motion in a dynamic system.