Solutions to Week 11 Assignment
-
From a proportion with the intensities treating the jet airplane as an isotropic
source.
Ij/IE = (P/4prj2)/(P/4prE2)
= rE2/rj2
Ij
= (rE/rj)2IE =
(1/5.2)2(1400W/m2) = 52W/m2. -
(a) v = Dx/Dt =
(1.80m - 1.50m)/0.20s = 1.5m/s
x = x0
+ vDt = 1.80m + (1.5m/s)(3.00s - 0.20s)
= 6.0m.
(b) t = (x -
x0)/v + t0 = (4.00m - 1.80m)/(1.5m/s) +
0.02s = 1.7s. -
f = v/l = (120m/s)/(30.0
x 10-2m) = 400Hz.
-
A = 0.120m, l = 0.300m, v =
6.40m/s, and y(x, t) = Asin(w
t + kx)
w
=2pv/l
= 2p(6.40m/s)/(0.300m) = 134s-1
k
= 2p/l =
2p/(0.300m) = 20.9m-1
So, y(x, t) =
(0.120m)sin[(134s-1)t + (20.9m-1)x] -
(a) ymax = 2.6cm, so A = 2.6cm.
(b)
l = Dx = 16m
- 2m = 14m.
(c) v = Dx/Dt =
(7.5m - 5.5m)/(0.10s) = 20m/s.
(d)
f = v/l = (2.0 x10-1m/s)/(14m)
= 1.4Hz.
(e) T = 1/f
= 1/(1.4Hz) = 0.70s. -
Let y1 = Asin(w t
+ kx) and y2 = Asin(w t
+ kx - f) and use the trigonometric identy
sina
+ sinb = 2sin[(a
+ b)/2]cos[(a
- b)//2]. Use the principle of superposition.
y
= y1+ y2= Asin(w t
+ kx) + Asin(w t
+ kx - f) = 2Asin(w t
+ kx - f/2)cos(f//2)
= A'sin(w t
+ kx - f/2)
where A' = 2Acos(f//2)
= A.
Find f
2Acos(f//2)
= A
2cos(f//2)
= 1
cos(f//2) =
1/2
f//2
= cos-1(1/2)
f/ = 120° -
f = v/l and the frequency is
the same in both mediums.
va/la
= vw/lw
lw = (vw/va)la
=0.750(0.500 x 10-6m) = 375nm. -
Intensity is proportional to the amplitude squared. Find A1/A2.
A1/A2
= Ö(I1/I2) = Ö(25/15) =
Ö(5.0/3.0)
For constructive interference, the
resultant amplitude is the sum of the original amplitudes.
A = A1
+ A2 = A2Ö(5.0/3.0)
+ A2 = A2[1 + Ö(5.0/3.0)]
Ö(I/I2)
= A/A2 = 1 + Ö(5.0/3.0)
I
= [1 + Ö(5.0/3.0)]2I2
= [1 + Ö(5.0/3.0)]2(15mw/m2)
= 79mW/m2. -
(a) fn = nv/2L and v = ÖT/m
f1 = v/2L =
1/2LÖT/m
= 1/2(1.5m)Ö12N/1.2 x10-3kg/m = 33Hz.
(b) f3 = 3v/2L = 3/2LÖT/m
4L2f32/9
= T/m
T = 4mL2f32/9
= 4(1.2 x 10-3kg/m)(1.5m)2(0.50 x10Hz)2/9
= 300N.