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1.1: Wave Basics
a) f=1/T  ::  T is the period from one peak to the next
b) λ=v/f
c) λ = distance from one peak to the next
d) f=v/λ


1.2: Traveling Wave
a) A = the peak of a wave
b) f=1/T  ::  T is the period from one peak to the next
c) λ=v/f
d) choose: x=0 and t=0, then A=A and y=A.  These cancel and you get
   1=sin(φ).  Solve for φ.


1.3: Wave Equation
*if wave is in form f(ax-bt), then v=b/a.
*if wave is in form f1(a1x-b1t) + f2(a2x-b2t), then v must equal b1/a1 = b2/a2
a) v=b/a
b) v=b/a (+/- does not matter)
c) v=b/a
d) v=b/a
e) v=0 because this is a product, not a sum
   

1.4: Phasor Practice
a) Phasor 1:
    A1 = given amplitude
    angle in radians α = plug in values given for t and T
    angle in degrees α = (angle in radians)*(180/π)
b) Phasor 2:
    A2 = given amplitude
    angle in radians β = plug in values given for t and T
    angle in degrees β = (angle in radians)*(180/π)
c) Phasor Sum:
    Ax=[A1cos(α)+A2cos(β)]2 ; Ay=[A1sin(α)+A2sin(β)]2 ; Amplitude = sqrt(Ax+Ay)
    angle in radians θ = (π/180)*(angle in degrees)
    angle in degrees θ = (β-α)/2


1.5: Two and Three Slit Interference Pattern
a) A phase difference of 0 = 5 (two straight arrows)
b) A phase difference of π/2 radians = 4 (arrow pointing right, arrow pointing up)
c) A phase difference of π radians = 2 (two opposite arrows)
d) A phase difference of 3π/2 radians = 6 (arrow pointing right, arrow pointing down)
d) A phase difference of 2π radians = 5 (two straight arrows)
f) φ=(2πd/λ)sinθ   ::  plug in values of φ d λ and solve for θ
g) y=L*tanθ

h) A phase difference between adjacent slits of 0 radians = 6 (three straight arrows)
i) A phase difference between adjacent slits of π/3 radians = 3 ('open' pointing up)
j) A phase difference between adjacent slits of 2π/3 radians = 1 (triangle pointing up)
k) A phase difference between adjacent slits of π radians = 2 (three opposite arrows)
l) A phase difference between adjacent slits of 4π/3 radians = 8 (triangle pointing down)
m) A phase difference between adjacent slits of 5π/3 radians = 7 ('open' pointing down)
n) A phase difference between adjacent slits of 2π radians = 6 (three straight arrows)
o) φ=&pi/2
p) φ=(2πd/λ)sinθ   ::  plug in values of φ d λ and solve for θ
   distance=L*tanθ


1.6: Two Speakers
a) δ=r2-r1  (use pythagorean to find r2)
   I1 is the intensity the observer hears when either speaker is sounded alone
   I is the intensity the observer hears when both speakers are on
   I=4I1cos2(φ/2)   ::  solve for φ and find the three lowest values for φ
   λ = δ*2π / φ
   f=v/λ
b) I=0 ; same steps as (a)
c) I=8 ; same steps as (a)
d) I=4 ; same steps as (a)


NOTE ABOUT HOMEWORK B: THESE ANSWERS MAY NOT BE CORRECT. USE THEM AT YOUR OWN RISK.

HomeworkB 01
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You are setting up a stereo system in your apartment. You put two speakers in front of your couch, arranged as in the picture:

Using a test disc, you play a variety of frequencies. The speakers emit sound in phase. The speed of sound is 330 m/s. For a 130 Hz tone, when one speaker at a time is turned on, you measure (at the left end of the couch, directly in front of the left speaker) 6 Watts/m2. Now, turn both speakers on.

1) What is the phase difference, φ, between the waves from the speakers measured at the left end of the couch? 
 φ=2πδ/λ &there4 φ=[2π(r2-r1)f]/v
2) If you vary the frequency while keeping the power output of the individual speakers constant, the maximum sound intensity you can get from both speakers at once is 
 Imax when cos(φ)=1 &there4 Imax=4I1
3) When the frequency is adjusted so that the phase difference φ is 2.5 rad, what sound intensity do you measure at the left end of the couch? 
 I=4I1cos2(φ/2)
4) If the frequency is then increased by a small amount, the intensity will 
 decrease
Hint: One way to understand this is to draw a phasor diagram and note the frequency dependence of φ.