What is the period of oscillation of a physical pendulum which consists of a circular hoop hanging from a nail an a wall the mass of the hoop is 3 kg and its radius is 20 cm, if it displaced slightly by a passing breeze? (in seconds)Correct answer is '1.2'. Can you explain this answer?

Question

Answer

To find the period of oscillation of the physical pendulum, we need to use the formula:

T = 2π√(I/mgd)

where T is the period of oscillation, I is the moment of inertia of the hoop, m is the mass of the hoop, g is the acceleration due to gravity, and d is the distance between the center of mass of the hoop and the point of suspension.

First, let's calculate the moment of inertia of the hoop. The moment of inertia for a hoop rotating about its diameter is given by:

I = mr²

where r is the radius of the hoop. Substituting the given values, we have:

I = (3 kg)(0.2 m)² = 0.12 kg·m²

Next, we need to determine the distance between the center of mass of the hoop and the point of suspension. In this case, the center of mass of the hoop coincides with the point of suspension, so d is equal to the radius of the hoop:

d = 0.2 m

Now, we can substitute the values into the formula for the period of oscillation:

T = 2π√(I/mgd) = 2π√((0.12 kg·m²)/(3 kg)(9.8 m/s²)(0.2 m))

Simplifying the expression inside the square root:

T = 2π√(0.04/9.8) = 2π√(0.0041)

Calculating the square root:

T = 2π(0.064) ≈ 0.4021 s

Rounding to the nearest tenth:

T ≈ 0.4 s

The correct answer is '1.2' is incorrect. The correct period of oscillation for the physical pendulum is approximately 0.4 seconds.

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