(S.h.m & waves).(reflection & refrection).(interferenc) prove s & important rules
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Transcript of (S.h.m & waves).(reflection & refrection).(interferenc) prove s & important rules
All Physics Proves & Summary Of Rules
Edited By Eng : Mustafa Kashif
S.H.M & WaveS
1-prove that, the Total energy of particle is moving in a simple harmonic motion is a Constant….
The differential equation of any particle moving in S.H.M is
Y=A sin (wt+α) y’=A w cos(wt+α)y’’=Aw2 sin(wt+α)y’’=-w2y
K.E=0.5 m y’2=0.5 m w2 A2 cos2(wt+α)1
P.E= =
P.E= =0.5 m w2 y2 P.E= 0.5 m w2 A2 sin2 (wt+ α)
ToTal Energy=Etot= K.E+P.E
E=0.5 m w2 A2[cos2(wt+ α)+sin2(wt+ α)] = 0.5 m w2 A2 Constant
/ / 2y = -w y
y
0- fdy ''my dy
y
0-
y 2
0+ mw ydy
S.H.M & Waves
Block of mass m attached to a spring the block free to move on horizontal, frictionless surface .if the spring constant is K …Prove that the Block Moves in S.H.M and Determined Angle speed ..
Spring Force F= KX .The eqn of motion MX’’=-KX X’’= -K/M *X The deffrential eq of S.H.M is X’’=-W2X The Block Moves in S.H.M with angular speed W W2=K/MW=(K/M)0.5 T=2Л/W =2Л(K/M)0.5 To Prove that the particle moves in S.H.M ..Prove the realtion between the
angular speed and the distanceAcceleration = -W2(Displacement)
S.H.M & WaveS
Summary of chapter 1 rules:-
1-General eq of simple Harmonic motion X=A cos(wt+α)
Y=A Sin(wt+α)
W=ө/t =2Л/T
Total energy = K.E+P.E =0.5m W2A2
V=y’ , a =Y’’=-W2 Y
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When Summing 2 or more S.H.M needS some Rules
N N
2 2t i i i i
i=1 i=1
A = [ A Cosθ ] +[ A Sinθ ]
N
i ii=1
t N
i ii=1
A sinθtan(θ ) =
A cosθ
S.H.M & WaveS
When Summing (n) of S.H.M we use some of rules:-
t 1
αα = (N -1) +α
2
t t tY = A sin(wt +α )
t
nαsin( )
2A =α
sin( )2
S.H.M & WaveS
To Prove that the wave is Traveling wave Prove that Y is
F(x+vt) or F(x-vt).
-General Eqn of Harmonic wave : Y(x,t) = A Sin(wt-kx+α)
W=2ЛF ,,,, (K) wave num =2Л/ג
(Intensity)
Energy EI = =
Area* time S*t
2 2tautspring
1Power = μw A v
2
2 21Power = μρSw A v
2
S.H.M & WaveS
Standing waves 1-Position of nodes amplitude=0
x=0.5 m ג
2- position of AntinodeS
x=0.25 (2m+1)ג- To Solve Standing waveS question:-- 1-put the 2 eqn in the form of Y=Asin( ) - 2- sum the 2 eqn with Rule- A[sin(a)+Sin(b)]=A[2cos0.5(a-b)Sin0.5(a+b)]- 3-the term tha don’t have t is the amlitude for standing
wave and be the fn of X only
4-let the amlitude =0 & amplitude =2A
For Getting Antinodes and nodes placeS
5- draw the standing wave form
6-Put Y axsis Perpendicular on X axsis and in sutable place depen on the places of nodes and antinodes .
S.H.M & WaveS
Doppler effectThe movement in same direction:
The movement in opposite direction
rr s
s
v + vf = f
v + v
rr s
s
v + vf = f
v v
Reflection & Refraction
Prove(1) Define the Critical angle өc is and Polarized angle өp and drive the relation between them
The critical angle өc is the incident angle over which there is total internal reflection (no transmission to the second medium )
If n1>n2
n =n2/n1<1
As өi increase , өt increase
If өt =90 өi = өc
n1sin өc =n2 sin 90
Sin өc = n2/n1=n
Reflection & Refraction
Prove(2)-Derive the equation for the amplitude reflection and transmitted coefficient in TM mode (rtanseverse magnetic mode parallel electric mode).
From maxwell’s eqn
Applying continuity of B-compnents
Bi+Br=Bt
Applying Conuity of E-components
EiCos(өi)-ErCos (өi)=EtCos (өt) 2
Using E/B=v=c/n
B=n*E/c substitute in (1)
Reflection & Refraction
1
2 t1 i 2 r
2i r t
n En E n E+ =
c c cn
E +E = En
2
1
nlet = n
n
i r tE +E = nE
Substitute from 3 in 2
3
i i r i i r t
i i t r i t
i trp
i i t
1E cosθ -E cosθ = (E +E )cosθ
nE [nCosθ - cosθ ] = E [ncosθ + cosθ ]
ncosθ - cosθEr = =
E ncosθ + cosθ
Divide the eqn 3 by Ei
tr
i i
pp
EE1+ = n
E E
1+r= t
n
tp
i
Et =
E
Reflection & Refraction
Prove (3)Derive the equations for the amplitude reflection and transmitted coefficent in T E mode (normal elctric field Mode)
-applying continuity of E- components
Ei+Er=Et 1
Applying continuity of B-components
-BiCosөi+Brcosөi=-Btcosөt 2
By using E/B=v=c/n
B=nE/c substitute in 2
Reflection & Refraction
2 t1 i 1 ri i t
2i i r i t t
1
i i r i t t
n E-n E n Ecosθ + cosθ = - cosθ
c c cn
-E cosθ +E cosθ = - E cosθn
-E cosθ +E cosθ = -nE cosθ 3
tr
i i
tn n
i
EE1+ =
E E
E1+r = t =
E
Substitute from 1 in 3
Dividing eqn 1 By Ei
i i r i i r t
i i r i i r t
i i t r i t
i trn TE
i i t
-E cosθ +E cosθ = -n[E +E ]cosθ
X(-1)
E cosθ -E cosθ = n[E +E ]cosθ
E [cosθ -ncosθ ] = E [cosθ +ncosθ ]
[cosθ -ncosθ ]E= r r =
E [cosθ +ncosθ ]
Reflection & Refraction
Prove(4) Draw the relation between amplitude reflection coefficient and incident angle for both external and internal reflection.
For external reflection
At өi=0 normal incdence
Өt=0 from snell’s law
+ve value
i tn
i t
[cosθ -ncosθ ]r =
[cosθ +ncosθ ] n
1-nr =
1+n
i trp
i i t
ncosθ - cosθEr = =
E ncosθ + cosθ
p
n -1r =
n +1
Reflection & Refraction
At өi=90
rp =-1
rn =-1
At ө= өp
rp=0
Reflection & Refraction
For Internal reflection
N=n2/n1<1
At ө =0 өt=0
At өt=90
өi= өc
Reflection & Refraction
Important RuleSn1=c/v1n2=c/v2n=n2/n1Өi= өrSnell law- n1sinөi=n2sinөt_____________________External reflectionn=n2/n1=nH/nL>1The Beam tranmitted near from the verticalөt> өin2>n1
Reflection & Refraction
Internal reflection
Өt> өi
n1>n2
n<1
When өt=90 өi is called ө(c) crtical angle
The beam transmitted far from vertical
Є= Є(r)* Єo
Intensity=0.5 Єo n c Eo2
Reflection & Refraction
In Normal mode (TEM) in Parallel Mode(TMM)
i trn TE
i i t
[cosθ -ncosθ ]E= r r =
E [cosθ +ncosθ ]
tr
i i
tn n
i
EE1+ =
E E
E1+r = t =
E
i trp
i i t
ncosθ - cosθEr = =
E ncosθ + cosθ
pp
1+ r= t
n
tp
i
Et =
E
Interference