# What is the equation for the period of oscillation for a physical pendulum?

• brendan3eb
In summary, the conversation discusses a problem involving a physical pendulum with a meter stick pivoted at a distance d from the 50 cm mark. The period of oscillation is given as 2.5 s and the goal is to find the value of d. Relevant equations involving the center of mass, the parallel axis theorem, and the moment of inertia are discussed. The correct equation for the period is derived, which involves the variables d, m, g, and I. The final solution is found by correctly incorporating the center of mass and using the approximation of sin(theta) ~ theta for small theta.
brendan3eb

## Homework Statement

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50 cm mark. The period of oscillation is 2.5 s. Find d

## Homework Equations

T=2PI*sqrt(I/mgh)
I(com)=(1/2)mL^2
parallel axis theorem
I=I(com)+m(L/2-d)^2
L=length of stick
h=L/2=center of mass

## The Attempt at a Solution

so T=2.5 s
center of mass = .50 m
I=(1/2)mL^2 + m*(L/2-d)^2
then plug T, I, and h in for the equation
T=2PI*sqrt(I/mgh)
2.5 s=2PI*sqrt((3L^2/4-2Ld+d^2)/(gL/2))
however when I solve for d I do not get 5.6 cm..anyone see my mistake?

I just spent another 30 minutes working on the problem...and no luck :(

I think your original 'relevant equations' are wrong.
The moment of inertia of a stick through its center of mass of total length L is: (mL^2)/12
Combining this with the parallel axis theorem, you should get through the point d cm from its center of mass:

I = (mL^2)/12 + md^2

Note that h in the period equation is actually d. It represents the distance from the center of mass. So if you actually swing it through the center of mass, guess what? It swings forever and if you divide by h approaching 0, the answer is a period approaching infinity.

Hope that helps.

ah! that helps a lot. Thank you so much. When I changed it to (1/12)mL^2 and used d instead it worked out great. I was forgetting about where it was rotating. Once again, thanks so much!

You're welcome ^_^

Ya u got it, but in case u wonder how 2 get the equation for the period, here is my quick derivation:

$$\Sigma\tau=I*\ddot{\theta}=r\timesF=$$
$$-d*m*g*sin(\theta)=-d*m*g*\theta$$ we can approximate sin(theta) ~ theta for small theta

=>$$-d*m*g*\theta - I*\ddot{\theta} = 0$$

Solving Differential Equations

w = sqrt(d*m*g / I)

T = 2*pi/lamba =2*pi*sqrt(I / d*m*g)

## 1. What is a physical pendulum?

A physical pendulum is a rigid body that is suspended from a fixed point and is free to oscillate. It consists of a mass attached to a pivot point by a rigid rod or arm.

## 2. How does a physical pendulum oscillate?

A physical pendulum oscillates due to the force of gravity acting on the suspended mass. As the pendulum swings back and forth, it alternates between potential energy and kinetic energy, resulting in a periodic motion.

## 3. What factors affect the oscillation of a physical pendulum?

The oscillation of a physical pendulum is affected by its length, mass, and the force of gravity. The material and shape of the pendulum can also have an impact on its oscillation.

## 4. What is the equation for the period of a physical pendulum?

The period of a physical pendulum can be calculated using the equation T = 2π√(I/mgd), where T is the period, I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and d is the distance from the pivot point to the center of mass.

## 5. How is a physical pendulum different from a simple pendulum?

A physical pendulum is different from a simple pendulum in that it has a larger mass and can have a varying length and shape. This results in a more complex motion and a different equation for calculating the period.

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