A closed circular wire hung on a nail in awall undergoes small oscillations 2 degree & the time period 2sec.Find (a) the radius of the circular wire (b)the speed of the particle farthest away from the point of suspension as it goes through the mean position (c) the acceleration of this particle as it goes through the mean positon &(d)acceleration of this particle when it is at the exterme position.Take g = pi^2m/sec^2

This is referred to as a "physical pendulum" problem. The equation you need, to relate the period P to the geometry, can be found at

http://hyperphysics.phy-astr.gsu.edu/HBASE/pendp.html

In this case, it becomes

P = 2 pi sqrt[I/(M g R)]

where I is the moment of inertia for oscillating about the nail.

Use the parallel axis therorm to derive the relation

I = 2 MR^2

M cancels out and you are left with

P = 2 pi sqrt(2R/g)

From P you can calculate R.

Use conservation of energy to get the maximum oscillation speed, going through the equilibrium position.
You will need to equate the potential energy rise at maximum swing angle to the kinetic energy (1/2) I w^2 at the equilibrium position.

(c) At the equilibrium position, the acceleration of a particle is centripetal. Use the maximum velocity to compoute it.

(d) At the extreme position, acceleration is tangential. Maximum angular acceleration equals w times maximum angular velocity

This is a rather long problem and I will leave you go through the computational steps and verfy my logic.

To solve this problem, we need to start with the relationship between the time period and the radius of the circular wire.

(a) To find the radius of the circular wire, we use the formula for the time period of a simple pendulum:

T = 2π√(L/g)

Where:
T = Time period
L = Length of the pendulum (in this case, the radius of the circular wire)
g = Acceleration due to gravity (given as pi^2 m/sec^2)

We are given that the time period T = 2 seconds. Therefore, we can rearrange the formula to solve for the radius L:

L = (T^2 * g) / (4π^2)

Substituting the given values, we have:

L = (2^2 * pi^2 m/sec^2) / (4π^2)

Simplifying:

L = pi^2 / 2 m

So the radius of the circular wire is pi^2 / 2 meters.

(b) The speed of the particle farthest away from the point of suspension as it goes through the mean position can be calculated using the formula:

v = ω * r

Where:
v = Velocity
ω = Angular frequency = 2π / T (T is the time period)
r = Radius of the circular wire

Substituting the given values:

ω = 2π / 2 = π rad/sec
r = pi^2 / 2

v = π * (pi^2 / 2) = π^3 / 2 m/sec

So the speed of the particle farthest away from the point of suspension is π^3 / 2 meters per second.

(c) The acceleration of this particle as it goes through the mean position can be calculated using the formula:

a = ω^2 * r

Substituting the given values:

ω = π rad/sec
r = pi^2 / 2

a = π^2 * (pi^2 / 2) = π^4 / 2 m/sec^2

So the acceleration of this particle as it goes through the mean position is π^4 / 2 meters per second squared.

(d) The acceleration of this particle when it is at the extreme position can be calculated by considering the vertical component of the gravitational force acting on it:

a = g * sin(θ)

Where:
a = Acceleration
g = Acceleration due to gravity (given as pi^2 m/sec^2)
θ = Angle of displacement from the mean position (which is 2 degrees, or π/90 radians)

Substituting the given values:

θ = π/90
g = pi^2

a = pi^2 * sin(pi/90) m/sec^2

So the acceleration of this particle at the extreme position is pi^2 * sin(pi/90) meters per second squared.

I hope this helps! Let me know if you have any further questions.