Harmonic Motion (

III

)

•Simple and Physical Pendulum •SHM and uniform circular motion Physics 1D03 - Lecture 34 1

Simple Pendulum

Gravity is the “restoring force” taking the place of the “spring” in our block/spring system.

Instead of x, measure the displacement as the arc length s along the circular path.

Write down the tangential component of F=ma:

θ L T Restoring force 

mg

sin 

m d

But 2

dt

2

s s

 

ma t L

 

d

2 

dt

2    

mg

sin(  )

g L

sin 

mg s

mg sin θ Physics 1D03 - Lecture 34 2

SHM: Simple pendulum:

d

2

x



dt

2

d dt

2  2     2

x g L

sin  The pendulum is

not

a simple harmonic oscillator!

However, take

small

oscillations: sin    (radians) if  is small.

Then

d dt

2  2  

g L

sin   

g L

 Physics 1D03 - Lecture 34 3

For small  : This looks like

d dt

2  2  

g L



d dt

2 2

x

   2

x

, with angle  instead of

x

.

The pendulum oscillates in SHM with an

angular frequency

 

g L

and the position is given by  (

t

)   o cos( 

t

  )

amplitude (2

p

/ period) phase constant

Physics 1D03 - Lecture 34 4

• a simple harmonic oscillator is a mathematical ‘approximation’ to the full problem • for large amplitudes, the solution that the SHO gives us start to deviate from what they actually should be :

T

 2 p

l g

   1  1 4 sin 2    

o

2         Unlike the SHO, the actual solution depends on the amplitude!

Physics 1D03 - Lecture 34 5

Quiz Pendulum clocks (“grandfather clocks”) often have a swinging arm with an adjustable weight. Suppose the arm oscillates with T=1.05sec and you want to adjust it to 1.00sec. Which way do you move the weight?

A) Up B) Down ?

Physics 1D03 - Lecture 34 6

Question:

A simple pendulum hangs from the ceiling of an elevator. If the elevator accelerates upwards, the period of the pendulum: a) Gets shorter b) Gets larger c) Stays the same

Question:

What happens to the period of a simple pendulum if the mass

m

is doubled?

Physics 1D03 - Lecture 34 7

Question:

A geologist is camped on top of a large deposit of nickel ore, in a location where the gravitational field is 0.01% stronger than normal. the period of his pendulum will be a) longer b) shorter (and by how much, in percent?)

Question:

How high is the ceiling?

Physics 1D03 - Lecture 34 8

Simple pendulum

: A particle on a massless string.

“

Physical” pendulum

: any rigid body, pivoted at P, and free to swing back and forth.

To find the period: 1) Consider the

torque

due to gravity 2) Write t

(



)



I

a 

I (d 2



/ dt 2 )

3) SHM if t is proportional to  

P



CM mg

Physics 1D03 - Lecture 34 9

Example: a metre stick, pivoted at one end. What is its period of oscillation?

“Uniform thin rod, pivot at end”: I =

1 / 3 ML 2



Mg L Calculate torque about the end:

t 

I

a 

Mg

sin  

L

2  1 3

ML

2  a and so a 

d

2 

dt

2   3

g

2

L

sin 

Note, this does not describe SHM!

Physics 1D03 - Lecture 34 10

But,

for

small

oscillations, sin    so

d

2 

dt

2   3

g

2

L

    2  This looks like

d dt

2 2

x

   2

x

, with angle  instead of

x

.

The

angular frequency

is and the period is 

T

  2 p  3

g

2

L

 2 p 2

L

3

g

This is like a simple pendulum of length 2 / 3

L

.

Physics 1D03 - Lecture 34 11

Any swinging object can be analysed in a similar way; we just need to know its moment of inertia,

I ,

about the pivot point.



Mgd

sin  

I d

2 

dt

2 d  

Mgd

in general

θ I Mg For a simple pendulum, I = Ml 2 and

 

g l

 (

t

)   o cos( 

t

  )

amplitude (2

p

/ period) phase constant

Physics 1D03 - Lecture 34 12

Quiz:

What happens to the period of the metre stick when the pivot is moved closer to the centre?

A) The period gets longer.

B) The period gets shorter.

C) The period stays the same.

D) It’s rather difficult to tell.

mg P

Physics 1D03 - Lecture 34 13

 

Mgd I

, or

T

 2 p   2 p

I Mgd I

is the moment of inertia

about the pivot

. From the parallel-axis theorem:

I



I CM



Md

2  1 12

ML

2 

Md

2 (for a uniform thin rod).

So

T

 2 p 1 12

L

2

gd



d g

(for a uniform thin rod).

Physics 1D03 - Lecture 34 14

SHM and Circular Motion

Uniform circular motion about in the xy plane, radius A , speed v , and angular velocity  = v/A : 

(t) =

 0

+



t and so x y

 

A

cos 

A

sin   

A

cos(  0

A

sin(  0  

t

)  

t

)

A



Real particle moves on the x axis “Imaginary” particle moves in a circle.

Physics 1D03 - Lecture 34 15

x y

 

A

cos  

A

sin  

A

cos( 

o A

sin( 

o

  

t



t

) ) Compare with our expression for 1-D SHM.

x



A

cos( 

t

  ) Result: SHM is the 1-D projection of uniform circular motion.

An interesting side problem: (try this out on your own) Start with 

r



A

cos 

i



A

sin 

j

Differentiate twice, and then show

a

r  

v

2

r

Physics 1D03 - Lecture 34 16

Summary

The projection of uniform circular motion onto an axis is SHM in 1-D.

The oscillation of a simple pendulum is

approximately

SHM, if the amplitude is small, with angular frequency  

g L

Physics 1D03 - Lecture 34 17

Practice problems, Chapter 15

3, 5, 11, 19, 23, 31, 67 (6 th ed – Chapter 13) 1, 3, 5, 9, 19, 23, 29, 67 Physics 1D03 - Lecture 34 18