a meter rule is suspended from a horizontal pin through the 20 cm mark and allowed to oscillate (with simple harmonic motion). In this position, the ruler completes 20 oscillations in 45.8 seconds.

derive an expression for the period T in terms of I20, M, g and d where I20 is the moment of inertia about an axis through 2o cm mark, g is the acceleration due to gravity and d is the distance between the point of suspension and the center of mass...

I AM AT A LOSS as even where to start

Well, they tell you what the period is

45.8/20 = 2.29 seconds
However that is not the point
d = pivot to center (CG) (I think it is 30 cm to the center of the meter stick but they do not say that)
force down at CG = weight = Mg
torque about pivot = Mg d sin theta
for small angle sin theta = theta in radians
so
torque = -M g d theta
torque = I alpha
if theta = A sin w t
then alpha =-Aw^2 sin wt
then
Torque = -I A w^2 sin w t
where w = 2 pi/T
M g d A sin w t = I w^2 A sin w t
M G d = I w^2
w = sqrt(M g d/I)
2 pi/T = sqrt (M g d/I)
T = 2 pi sqrt (I/Mgd)
I am calling your I20 simply I

By the way, you know d = 30 from the geometry of the meter stick

You can compute I20 assuming the meter stick is uniform in mass distribution along its length.
That I20 will have M in it, but it cancels the M in the period equation
Use g = 9.8
Then you can actually compute T in seconds.

To derive an expression for the period of the oscillations, we need to analyze the physical principles and equations governing the system. We can use the concepts of torque and rotational motion to establish the relationship.

1. Start by considering the equation for the period of a simple pendulum:
T = 2π√(l / g)
where T is the period, l is the length of the pendulum, and g is the acceleration due to gravity.

2. The meter rule is acting as a physical pendulum and oscillating about the pin attached at the 20 cm mark. Therefore, the effective length of the pendulum is the distance from the point of suspension to the center of mass, which is given by (d + l/2), where d is the distance between the point of suspension and the center of mass, and l is the length of the pendulum (1 meter).

3. Next, we need to relate the moment of inertia (I20) about the axis through the 20 cm mark to the length of the pendulum and the distance (d). The moment of inertia of the meter rule can be calculated as:
I20 = (1/3) * M * l^2
where M is the mass of the meter rule and l is the length of the pendulum.

4. The moment of inertia (I20) is related to the effective length (d + l/2) by the parallel axis theorem:
I20 = Icm + Md^2
where Icm is the moment of inertia about an axis passing through the center of mass and d is the distance between the point of suspension and the center of mass.

5. Rearranging the equation from step 4, we can solve for Icm:
Icm = I20 - Md^2

6. Finally, substitute this expression for Icm back into the equation for the period of a simple pendulum (step 1), using the effective length (d + l/2) as the length of the pendulum:
T = 2π√((d + l/2) / g)

Putting it all together, the expression for the period T in terms of I20, M, g, and d is:
T = 2π√((d + l/2) / g)
T = 2π√((d + 1/2) / g)
T = 2π√((d + 0.5) / g)