You are at a furniture store and notice that a Grandfather clock has its time regulated by a physical pendulum that consists of a rod with a movable weight on it. When the weight is moved downward, the pendulum slows down; when it is moved upward, the pendulum swings faster. If the rod has a mass of 1.23 kg and a length of 1.25 m and the weight has a mass of 1.92 kg, where should the mass be placed to give the pendulum a period of 2.00 seconds? Measure the distance in meters from the top of the pendulum.

P= 2PI sqrt(L/g)

2.00=2PI sqrt(l/9.8) solve for L

but wouldn't the mass and the Length of the rod come in play?

Wouldn't the formula be T=2Pisqrt(I/mgd) where you solve for d, but my question iss do I add the two masses or am I over thinking this one

To determine the position where the mass should be placed to give the pendulum a period of 2.00 seconds, we need to use the formula for the period of a simple pendulum:

T = 2π√(L/g)

where T is the period in seconds, L is the length of the pendulum in meters, and g is the acceleration due to gravity which is approximately 9.8 m/s² on Earth.

Given:
L = 1.25 m
T = 2.00 s
g = 9.8 m/s²

We can rearrange the formula to solve for the square of the period:

T² = (4π²/g) * L

Now, let's substitute the given values and solve for the unknown position of the mass:

(2.00 s)² = (4π²/9.8 m/s²) * 1.25 m

4.00 s² = (4π²/9.8 m/s²) * 1.25 m

Multiplying both sides by (9.8 m/s²):

(9.8 m/s²) * 4.00 s² = 4π² * 1.25 m

39.2 m/s² * s² = 4π² * 1.25 m

Dividing both sides by (4π²):

(39.2 m/s² * s²) / (4π²) = 1.25 m

9.9 m = 1.25 m

To find the distance from the top of the pendulum, we subtract 1.25 m from the length of the pendulum:

Distance = 1.25 m - 1.25 m
Distance = 0 m

Therefore, the mass must be placed at the top of the pendulum (0 meters from the top) to give a period of 2.00 seconds.