# The Oscillation Period of A Rod Pivoted at One End

• Beginner@Phys
In summary, the problem at hand involves a 230 g uniform rod pivoted at one end and attached to a horizontal spring with a spring constant of 3.0N/m. The length of the rod is 0.20m and the question asks for the rod's oscillation period, assuming the rod's angle from vertical is always small. The relevant equations for solving this problem include restorative force, torque, moment of inertia for a rod pivoted about one end, and angular frequency. However, the previous attempt at a solution did not consider the stiffness added by the spring, resulting in incorrect calculations. Therefore, a fresh start is needed to accurately solve this problem.
Beginner@Phys

## Homework Statement

The figure shows a 230 g uniform rod pivoted at one end. The other end is attached to a horizontal spring. The spring is neither stretched nor compressed when the rod hangs straight down. K=3.0N/m and the length of the rod is 0.20m.

What is the rod's oscillation period? You can assume that the rod's angle from vertical is always small.

## Homework Equations

Restorative force=F=-kΔx
Torque=Fd=force * length of lever arm
Moment of inertia for a rod pivoted about one end: I=(1/3)mL2
Angular frequency(w)=2π/T

## The Attempt at a Solution

-kΔx=I*-w2Θ
<=>
-kr2Θ=I*-w2Θ
<=>
-kr2Θ/I=-w2Θ
Substitute I into the equation, L=r in this case.
-3K/M*Θ=-w2Θ

Therfore w=sqt(3K/M)
<=>
2π/T=sqt(3K/M)
<=>
T=2π/sqt(M/3K)=1.00s

But, I don't really know if my calculations are correct.

#### Attachments

• Fig 14.jpg
4.4 KB · Views: 695
Sorry, I don't know how to do this problem. But check
-kΔx=I*-w2Θ
Looks like kg m/s^2 on the left and kg m^2/s^2 on the right.

If the spring were not there, you would have a simple pendulum. It does not look to me like you have taken this aspect of the problem into your formulation at all. What you have in this problem is a simple pendulum with additional stiffness added. I suggest that you take a second look at the problem (start over).

Dr.D said:
If the spring were not there, you would have a simple pendulum. It does not look to me like you have taken this aspect of the problem into your formulation at all. What you have in this problem is a simple pendulum with additional stiffness added. I suggest that you take a second look at the problem (start over).

Are you sure? It rotates around an axle..that's why I thought it was a physical pendulum. See: http://session.masteringphysics.com/problemAsset/1070632/3/14.CP80.jpg

OK, pardon the error in termnology. It is a physical pendulum, you are correct. That said, it is just a physical pendulum with a spring added to increase stiffness. Now, shall we start from the beginning?

## 1. What is the oscillation period of a rod pivoted at one end?

The oscillation period of a rod pivoted at one end is the time it takes for the rod to complete one full swing or rotation.

## 2. What factors affect the oscillation period of a rod?

The oscillation period of a rod can be affected by several factors including the length, weight, and material of the rod, as well as the angle at which it is pivoted and the force of gravity.

## 3. How is the oscillation period of a rod calculated?

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

## 4. Does the oscillation period of a rod change if the pivot point is moved?

Yes, the oscillation period of a rod will change if the pivot point is moved. The closer the pivot point is to the center of mass, the shorter the period will be.

## 5. How does the oscillation period of a rod change with different materials?

The oscillation period of a rod can vary depending on the material it is made of. For example, a lighter material will have a shorter period than a heavier material due to differences in mass and moment of inertia.

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