- #1

Beginner@Phys

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## 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)mL

^{2}

Angular frequency(w)=2

**π**/T

## The Attempt at a Solution

-kΔx=I*-w

^{2}Θ

<=>

-kr

^{2}Θ=I*-w

^{2}Θ

<=>

-kr

^{2}Θ/I=-w

^{2}Θ

Substitute I into the equation, L=r in this case.

-3K/M*Θ=-w

^{2}Θ

Therfore w=sqt(3K/M)

<=>

2

**π**/T=sqt(3K/M)

<=>

T=2

**π**/sqt(M/3K)=1.00s

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