- #1

castrodisastro

- 82

- 0

## Homework Statement

A uniform rod of mass

*M*, and length

*L*swings as a pendulum with two horizontal springs of negligible mass and constants

*k*

_{1}and

*k*

_{2}at the bottom end as shown in the figure. Both springs are relaxed when the when the rod is vertical. What is the period

*T*of small oscillations?

## Homework Equations

*T*= 2π/ω

*I*

_{uniform rod}= ⅓

*ml*

^{2}= ⅓

*ML*

^{2}

ω = √[mgh

**/**I] = √[Mg(½

*L*)

**/ ⅓**

*ML*

^{2}]

Potential Energy

*U*= mgh =

*M*g(½

*L*)

Kinetic Energy

*T*= ½kx

^{2}= ½(

*k*_{1}+*k*_{2})L^{2}

## The Attempt at a Solution

In order to find the period

*T*, I need to find the angular velocity, ω.

To find the angular velocity I need to know the moment of Inertia for the uniform rod, and we need to know every component that would affect the motion of the rod when disturbed from equilibrium.

The moment of Inertia of the rod is given by ⅓

*ML*

^{2}

and the restoring forces are

*M*g(½

*L*) and ½(

*k*

_{1}+

*k*

_{2})

*L*

^{2}

Solving for

*T*:

= 2π √[(⅓

*ML*

^{2})

**/**(

*M*g(½

*L)+*½(

*k*

_{1}+

*k*

_{2})

*L*

^{2}]

Factoring out an

*L*from the numerator and denominator, and factoring out a ⅓ from the numerator, and a ½ from the denominator...

= 2π √[((⅓

*L)ML*)

**/**((½

*L*)(

*M*g

*+*(

*k*

_{1}+

*k*

_{2})

*L*))]

The

*L*'s cancel and rearranging the ⅓ and the ½ I get...

= 2π √[(2

*ML*))

**/**(3

*M*g

*+*3(

*k*

_{1}+

*k*

_{2})

*L*)]

as my final answer.

The answer in the back of the book however, is

= 2π √[(2

*ML*))

**/**(3

*M*g

*+*6(

*k*

_{1}+

*k*

_{2})

*L*)]

and I have it on good authority that this book is NOT wrong.

Is my reasoning incorrect. Any help is appreciated.

Thank you