Predicting Vibration Frequencies of the Diamond Wafer

Preview:

DESCRIPTION

Predicting Vibration Frequencies of the Diamond Wafer. Presented by: Austin Antoniou Site: Nuclear Physics Mentor: Dr. Richard Jones. Modes of oscillation. Oscillation on the Z axis Rotation About X Axis Rotation About Y Axis Rotation About Z Axis. y. x. z. Model of the diamond. - PowerPoint PPT Presentation

Citation preview

1

Predicting Vibration Frequencies of the Diamond Wafer

Presented by: Austin AntoniouSite: Nuclear PhysicsMentor: Dr. Richard Jones

2

Modes of oscillation

x

y

z

Oscillation on the Z axis

Rotation About X Axis

Rotation About Y Axis

Rotation About Z Axis

3

Model of the diamond

Wire length=w

Diamond length=aDistance between wire joints=s Mass of

diamond=mVelocity of a wave traveling on wire=v

Tension Forces

4

Oscillation on the Z axis

5

Diagram of Z Axis Oscillation

6

Equation Used for Z axis Oscillation(continued)

z(y,t)=A+sin(ky-ωt)+A-sin(ky-ωt)

z(0,t)=A+sin(-ωt)+A-sin(ωt)

z(0,t)=(-A++A-)sin(ωt)z(y,t)=A[sin(ky)cos(ωt)+cos(ky)sin(ωt)+

sin(ky)cos(ωt)-cos(ky)sin]z(y,t)=2Asin(ky)cos(ωt)Where 2Asin(kw)=z0, A=z0/(2sin(kw))

z’(w,t)=z0kcot(kw)cos(ωt)z’(w,t)=kcot(kw)

z’(w,t)=kcot(kw)

7

Equation Used for Z Axis Oscillation

(Continued)

z’(w,t)=kcot(kw)

This equation can be related to the restoring force:

FR= -kFTcot(kw)z

FR=(d2z/dt2)m=-kFTcot(kw)z(d2z/dbt2)+[-(kFTcot(kw)/m]z=0

The generic equation used for ω is ω2=(K/m)

In this case, K is equal to

Therefore,

cot(kw)z-kFmF Tdtzd

R 2

2

0zmcot(kw)kF

dtzd T2

2

mK2

w)cot(FT vv

m

w)cot(F vT2

8

Graph of ω2= ωcot(ω)

ω (rad/s)

f(ω

) (

Rad

2/s

2)

9

Rotation about the X axis

10

Diagram of X Axis Rotation

11

Rotation About the Y Axis

12

Diagram of Y Axis Rotation

13

Rotation about the Z axis

14

Diagram of Z Axis Rotation

15

Predicted frequencies of the diamond wafer

Oscillation on Z Axis

Rotation About X Axis

Rotation About Y Axis

Rotation About Z Axis

45.1 78 55.5 2560

5060 5060 506010100 10100 10100

Frequencies(Hz) (Hz)(Hz)(Hz)

16

ω2=ωcot(ω)

ω (rad/s)

f(ω

) (

Rad

2/s

2)

17

What should we make of these calculations?

The higher-frequency oscillations are nearly identicalMathematically, this is because the slopes of the graphs of the graphs of ω only change slightly from mode to modePhysically, this is because the system has natural frequencies due to properties such as the length of the wires

18

Questions?

Recommended