Server Date is Sun, Arp 12 1998 20:4:34
no+help = on
Q1 = Since the waves from the motorboat took 30 seconds to travel 30 m,
the velocity of water waves is 1 m/s. The frequency with which
the siblings bob is 15 cycles/30 seconds = 0.5 Hz. Hence the
wavelength of the siblings' waves is (1 m/s)/(0.5 Hz) = 2 m.
Sally is an integer number of wavelengths away from sibling A both
when she is at position A and at position B. When Sally is at
position A she is 25 m away from sibling B so that she is one
half-wavelength off from an integral number of wavelengths. Thus
the waves from the two siblings are 180 degrees out of phase and
interfere destructively when Sally is at position A. When Sally
is at position B, she is approximately 30 m from sibling B. This
distance of 30.02 m is almost an integral number of wavelengths
and so the waves from the two sources will interfere constructively.
Sally's father did not do her a favor, because she would have floated
in calm water a point A, but at point B she will bob with twice the
amplitude as each of her siblings.
realsect = 215M3AH
section = 215hpz
OutTime = Sun Apr 12 19:56:36 Mountain Daylight Time 1998
215sect = 215M3AH
whichPF = 215Hpzl12
Server Date is Sun, Arp 12 1998 20:11:9
no+help = on
Q1 = No, Sally's father does not do her a favor. Using the equation
y(bright) = m (wavelength)L/d we find that in position A Sally will
be about 0.5 meters away from the point where the waves will combine
constructively the most. In both positions she will be the same
distance, 0.5 meters, from the spot where the waves combine the most
constructively. However, at point a she is that distance from a 3rd
order "bright fringe" where as at point B she is that distance from
a 2nd order "bright fringe." As is illustrated by the experiments
with light passing through two slits, the lower order bright fringes
will be of a greater magnitude.
realsect = 215M1AH
OutTime = Sun Apr 12 20:03:09 Mountain Daylight Time 1998
215sect = 215M1AH
whichPF = 215Hpzl12
Server Date is Sun, Arp 12 1998 20:15:54
no+help = on
Q1 = obviously, Dad thinks that Sally should face her fears. The wavelengths
of the waves are 2 meters and they move 1m/s with a frequency of 1/2 wave
every second. Also, we have a 3/4/5 triangle. Using that we find in A
it takes 50/4 wavelengths from the one boat and 10 wavelengths from the
other so they'rre not in phase. But at B it's 20 and 13 wavelengths so
they constructively interfere to give Sally a ride.
realsect = 215M1AH
section = 215hpz
OutTime = Sun Apr 12 20:07:46 Mountain Daylight Time 1998
215sect = 215M1AH
whichPF = 215Hpzl12
Server Date is Sun, Arp 12 1998 20:30:37
Q1 = frequency= 15 times/30 sec= .5 cycles/sec
T= 2 seconds
w = pie
velocity of waves: 1 m/s
velocity= frequency*wavelength
wavelength= 2 m
distances at point A: 20m and 25m: difference by 5 m (interval of .5)
destructive interference.
distances at point B: 26 m and 30 m. constructive interference.
Sally's father actually moves her into rough water from calm water.
realsect = 215M1AH
section = 215hpz
OutTime = Sun Apr 12 20:22:33 Mountain Daylight Time 1998
215sect = 215M1AH
whichPF = 215Hpzl12
StartTime = Sun Apr 12 19:58:57 Mountain Daylight Time 1998
Server Date is Sun, Arp 12 1998 20:42:13
no+help = on
Q1 = No, he does not do her a favor. This is because he merely increases
the amplitude of the waves that will hit her. This is known because
the waves that hit her will be the combination of the waves produced
by her siblings. With our trig identity, and since each of them will
have the same frequency, amplitude, and wavelength, the equation of
the waves hitting her boat is 2ACos(-Q/2)Cos((2kx-2wt+Q)/2). The
frequency of the waves hitting the boat will be the same in both
places. This is because the waves hitting the boat will be given by
the second half of the above mentioned equation. The phase constant
will therefore only shift the wave, ie where it starts, yet will not
affect how many waves will hit her. The difference will be in the
amplitude which is represented by the first half of the equation.
This equation will differ because of the phase constant. To find
the phase constant, I assumed that originally, the further boat will
have waves 5 meters farther away. Therefore, when t=0, x=0, y=5.
5=Cos(Q) In the second instance, t=0, x=0, and y=6. The closer boat
will, for each equation, have t=0, x=0, and y=0. Solving for the phase
constant in each equation and then subbing it in for the value of
the amplitudes showed that the amplitude would be larger for where her
dad put her.
realsect = 215M3AH
section = 215hpz
OutTime = Sun Apr 12 20:34:17 Mountain Daylight Time 1998
215sect = 215M3AH
whichPF = 215Hpzl12
StartTime = Sun Apr 12 20:00:55 Mountain Daylight Time 1998
Server Date is Sun, Arp 12 1998 21:20:59
no+help = on
Q1 = First off I am assuming the waves created by Sally's siblings are
traveling at the same speed as the ones created by the motorboat (1m/s)
Since I know the frequency of the waves (0.5 Hz) i can calculate the
wavelength of the waves, which turns out to be 2m. Sinc the distance
to point B is 26m and 30m from each sibling repsectfully, the waves
will both be reaching their maximas at point B, creating a really big
wave for Sally. Looks like Sally's father doesn't know to much about
Physics (maybe his name is Mr. Spaulding). If sally had stayed where
she was, she would have been in calm water because one wave would be
at its maxima while the other was at its minima, which would cancel
each other out giving Sally a nice area of calm water.
realsect = 215M3AH
section = 215hpz
OutTime = Sun Apr 12 21:12:52 Mountain Daylight Time 1998
215sect = 215M3AH
whichPF = 215Hpzl12
StartTime = Sun Apr 12 20:37:08 Mountain Daylight Time 1998
Server Date is Sun, Arp 12 1998 21:30:11
no+help = on
Q1 = v=30m/30s=1m/s
f=15cycles/30s=.5s^-1
v=f*lambda
1=.5*lambda
lambda=wavelength=2m
y=-7.5m
d=15m
L1=20m
ybright=m*lambda*L/d
7.5=m*2*20/15
m=2.813
L2=26m
7.5=m*2*26/15
m=2.16
He does not do sally a favor because she is now closer to a "bright fringe"
a point of constructive interference
If she were left where she was, she would have been farther from the m=3
fringe than she is now from the m=2 fringe.
In addition, the m=2 fringe is stronger in general than the m=3 fringe
realsect = 215M3AH
section = 215hpz
OutTime = Sun Apr 12 21:22:46 Mountain Daylight Time 1998
215sect = 215M3AH
whichPF = 215Hpzl12
StartTime = Sun Apr 12 21:00:10 Mountain Daylight Time 1998
Server Date is Sun, Arp 12 1998 21:37:25
no+help = on
Q1 = No.
This can be thought of as a double slit interference problem, because both
waves produced by the rowboats of each sibling is exactly the same (they are
coherent sources because they have the same wavelength and same phase).
Using the formula (d)sin(theta)=m(lambda) and rearranging for m:
m = d / (lambda)(sin(theta))
d = distance between two wave sources (in this case distance between two simblings row boats) = 15 m
lambda = wavelength = velocity / frequency = from the motorboat, the speed that waves propogate throught the water (30 m / 30 s) is 1 m/s / the frequency which is (30 s / 15 waves ) 2 s = so the wavelengh is 2 m
sin(theta) = since L >> d, we could not make the substitution sin(theta) = tan(theta) = y/L, so we just have to use sin(theta), which in respect to y and L = y/square root of (y^2 + L^2) (by Pythegorian Thereom of right triangles)
y = disance between point P and the point half-way between the two sibling rowboats = 15 m / 2 = 7.5 m
L = distance between wave sources and Little Sally (need to compare when L = 20 m and when L = 26 m)
when L = 20 m, sin(theta) = 7.5 m /square root of ((7.5 m)^2 + (20 m)^2) = 0.3511
when L = 26 m, sin(theta) = 7.5 m /square root of ((7.5 m)^2 + (26 m)^2) = 0.2772
therefore when L = 20 m, m = (15 m) / (2 m)(0.3511) = 21.36
when L = 26 m, m = (15 m) / (2 m)(0.2772) = 27.06
now, we need to compare these two values of m. An m with a decimal of .0 would be a perfect place where constructive interference occurs (wave would be huge) and an m with a decimal of .5 would be a perfect place where destructive interference occurs (wave would cancel out and the water would be still)
since Little Sally wants no waves, moving here to a distance 26 m would increase the size of the waves, since at this place constructive interference is occuring
realsect = 215M3AH
section = 215hpz
OutTime = Sun Apr 12 21:44:25 Mountain Daylight Time 1998
215sect = 215M3AH
whichPF = 215Hpzl12
StartTime = Sun Apr 12 21:13:16 Mountain Daylight Time 1998
Server Date is Mon, Arp 13 1998 12:34:27
no+help = on
Q1 = This problem has to do with the intensity of the waves at the two
distances from the origen (the brother and sister)
first, we need to find the wavelength of the waves. since the waves start
30 meters from the shore and hit 30 seconds later, the velocity is
1 m/s. since v=wavelength/period, and the period is 2 sec, the wavelength
is 2m. from the illustration, L=20m and 26m, d=15m, and y=7.5m. using the
equation Savg=4Scos^2((Pi*d*y)/(L*wavelength), and using the same arbitrary
number for S, we can calculate intensitites for the two positions and see
which one has a greater intensity.
For this problem, Sally's father does not do her a favor because the
intensity at point B is greater than at point A.
realsect = 215M3AH
section = 215hpz
OutTime = Mon Apr 13 12:41:10 Mountain Daylight Time 1998
215sect = 215M3AH
whichPF = 215Hpzl12
StartTime = Mon Apr 13 12:17:46 Mountain Daylight Time 1998
Server Date is Mon, Arp 13 1998 15:11:24
no+help = on
Q1 = The waves given by the speeding boat and children are based on
position and time. That is y=Asin(kx+wt) The velocity of wave is
the distance it travels over time or wavelength/period.
here it is 1 m/s and we'll assume the
speed of both waves is the same because they are both in water.
The time it takes the wave from the children to repeat itself is 2
seconds so the frequency is .5 Hz. This can give an equation for the
childrens wave to be y=sin(pi*x+pi*t) and the speeding boat is same.
For Sally to be in calm water the waves from both sources would have
to cancel eachother out using the superposition principal, they would
have to arrive at Sally out of phase. For position A she is 20 m and
20 seconds away from the first wave that reaches her from the children
and 30 meters and 30 seconds away from the boat. If we plug time =
30 seconds and the distances into each equation we get that the waves
in phase when she's at A and out of phase at B so B is better, I think...
realsect = 215M1AH
section = 215hpz
OutTime = Mon Apr 13 15:03:11 Mountain Daylight Time 1998
215sect = 215M1AH
whichPF = 215Hpzl12
StartTime = Mon Apr 13 14:37:57 Mountain Daylight Time 1998
Server Date is Mon, Arp 13 1998 18:51:8
no+help = on
Q1 = The period of the waves is two seconds. Therefore, the frequency of
the waves is 0.5 Hz.The speed of the waves is 1 m/s (from the
motorboat), which makes the wavelength 2 m. The distance from one
boat to Sally is 20 m, and it is 25 m from the other. At this point,
assuming that the waves leave both boats at the same time. At Sally's
original position, the waves will interfere destructively and there
will be no displacement (the point is a node). At 26 m and 30.02 m,
the waves interfere constructively (an antinode) and Sally will rise
a lot. Her father didn't do her a favor by moving her.
realsect = 215M1AH
section = 215hpz
OutTime = Mon Apr 13 18:43:16 Mountain Daylight Time 1998
215sect = 215M1AH
whichPF = 215Hpzl12
StartTime = Mon Apr 13 18:34:08 Mountain Daylight Time 1998
Server Date is Tue, Arp 14 1998 13:6:51
no+help = on
Q1 = NO!!!!! Now Sally is bobbing like crazy!
Velocity of Waves:
V = d/t = 30 m/30 s = 1 m/s
Frequency and Period:
They bob 15 times in 30 seconds.
f = 15 cycles/30 s = 1/2 hz
T = 1/f = 1/(1/2) = 2 seconds
Wavelength (WL):
v = (WL)/T (WL) = vT = (1 m/s)(2s) = 2 m
Distance (m): Brother Sister
A 20 25
B 13 30
Wavelengths:
A 10 12.5
B 13 15
At "A" the waves are exactly half out of phase and cancel each other out.
At "B" the waves are exactly in phase and add together to make BIG waves!
realsect = 215M3AH
section = 215hpz
OutTime = Tue Apr 14 13:13:40 Mountain Daylight Time 1998
215sect = 215M3AH
whichPF = 215Hpzl12
StartTime = Tue Apr 14 11:07:30 Mountain Daylight Time 1998
Server Date is Tue, Arp 14 1998 18:33:52
no+help = on
Q1 = ok, the first thing I did was to figure out how fast the waves were
propogating in the medium. To do this, I took velocity=distance/time
so, velocity= 30m/30s = 1 m/s
Then, you use this information to determine the wavelength of the
waves created by the kids. Since you know the waves are propogating
at 15 cycles/ 30 seconds, that equates to .5 cycles/second.
Also, you know that for any medium, velocity = freq*wavelength.
So, 1 m/s = .5*x
therefore, the wavelength is equal to 2 meters.
Knowing this, you have almost all the information needed to find the
answer. The basic premise involves using the equation for two slit
interference to determine which position is closer to a maxima.
The way I did it, I used the equation for maxima (dsino = m(wavelngth)
and solved each for m.
To find theta, I drew a picture of the system and realized that there
was a right triangle with two sides already know. So, in each case I
solved for the missing side (and therfore found sin theta) and then
solved the equation for m.
For the first case, I found that m was equal to 7.022
For the second case, I found that m was equal to 7.206.
since in the second case, the m value is farther from the expected whole
number value at a maxima, the waves would be less intense at that
point. So, Mr. Hale is foolish and Sally's dad is smart...she will
have calmer (though not completely calm) water at the second
location.
realsect = 215M3AH
section = 215hpz
OutTime = Tue Apr 14 18:26:01 Mountain Daylight Time 1998
215sect = 215M3AH
whichPF = 215Hpzl12
StartTime = Tue Apr 14 17:55:43 Mountain Daylight Time 1998
Server Date is Tue, Arp 14 1998 19:26:5
no+help = on
Q1 = Of course he does her a favor. When the waves from the siblings constructively interfere, they will very likely be greater than the waves created with the boat, so the addition of the father's waves to that of the boat's, it could cancel out the waves of the siblings.
realsect = 215M1AH
section = 215hpz
OutTime = Tue Apr 14 19:18:19 Mountain Daylight Time 1998
215sect = 215M1AH
whichPF = 215Hpzl12
StartTime = Tue Apr 14 19:02:20 Mountain Daylight Time 1998
Server Date is Tue, Arp 14 1998 20:14:14
no+help = on
Q1 = In the equation ay/L=m*lambda
When L increases the wavelength decreases and therefore Sally isn't
hit by such big waves. Her father did her a favor.
realsect = 215M1AH
section = 215hpz
OutTime = Tue Apr 14 20:06:05 Mountain Daylight Time 1998
215sect = 215M1AH
whichPF = 215Hpzl12
StartTime = Tue Apr 14 20:01:00 Mountain Daylight Time 1998
Server Date is Tue, Arp 14 1998 20:16:58
no+help = on
Q1 = If Sally's father moves here to a position such that the waves created
by the two children is at a minimum, then he did the right thing.
Just a thought from me, I dont think that I have any clue on how to do
this stuff.
realsect = 215M1AH
section = 215hpz
OutTime = Tue Apr 14 20:23:50 Mountain Daylight Time 1998
215sect = 215M1AH
whichPF = 215Hpzl12
StartTime = Tue Apr 14 20:20:05 Mountain Daylight Time 1998
Server Date is Tue, Arp 14 1998 21:22:12
no+help = on
Q1 = No, he does not do her a favor. The waves come 15 per min or 1 wave
2 sec. From the data above we see that 15 waves go in 30m or 1 wave
is 2 m long. In order to be in a position to not bob up and down the
boat must be in the neutral spot in between the top of the wave and
bottom. Since the waves are 2m long the neutral spot is every odd
meter. So, to stay in calm water the boat must be moved to any odd
meter, but Sally's boat is moved to an even meter and will bob up and
down.
realsect = 215M1AH
section = 215hpz
OutTime = Tue Apr 14 21:14:13 Mountain Daylight Time 1998
215sect = 215M1AH
whichPF = 215Hpzl12
StartTime = Tue Apr 14 20:38:05 Mountain Daylight Time 1998
Server Date is Tue, Arp 14 1998 21:42:7
no+help = on
Q1 = V=frequency*wavelength. Wavelenght equals 2m.
Using the formula sinO=TanO=Y/L=pd/d, and Y=7.5, d=15m,
we can vary L from 20m to 26m to find the pd at the two points.
The path difference for 20m =5.625m while the pd for the 26m = 4.327
meaning that the waves are .327 out of phase at 26 m while the waves
at 20m are .375 out of phase. This means there is more destructive
interference at 20 m rather than 26m and Sally's father did not help
her out at all.
realsect = 215M1AH
section = 215hpz
OutTime = Tue Apr 14 21:34:20 Mountain Daylight Time 1998
215sect = 215M1AH
whichPF = 215Hpzl12
StartTime = Tue Apr 14 21:02:17 Mountain Daylight Time 1998
Server Date is Tue, Arp 14 1998 22:23:29
Q1 = the velocity of a compressional wave travelling in water can be determined from the rowboat... 30m/30s=1m/s
the velovity can be used with the period to determine the wavelength: 2 m
Dividing the distance by the wavelength will tell you where in the wave it will hit Sally. (Divide by 2 m, Drop off the integer, if it is 0 it is calm, if it is .25 and .75 it is maximum rough, anywhere in between is mildly rough, .5 is 180 deg out of phase)
then add the waves from both sources (brother and sister) to get the total.
A) 20 => 0 => 0 degrees thru wave
rt(20^2+15^2) => 25 => .5 => 180 degrees thru wave
they are 180 deg out of phase, so they will cancel and be calm.
B) 26 => 0 => 0 degrees thru wave
rt(26^2+15^2) => ~30 => 0 => 0 degrees thru wave
they will add together to double the amplitude.
A is completely calm and B has double amplitude waves, so moving it from A to B was bad.
realsect = 215M3AH
section = 215hpz
OutTime = Tue Apr 14 22:30:38 Mountain Daylight Time 1998
215sect = 215M3AH
whichPF = 215Hpzl12
StartTime = Tue Apr 14 21:56:06 Mountain Daylight Time 1998
Server Date is Tue, Arp 14 1998 22:37:46
no+help = on
Q1 = c=f l 30/30 = 15/30 l lambda = 2
yb= m l *L /d
yd = mlL/d
at 20 Sally is .5 meters away from a max
at 26 Sally is .567 meters away from a max
Sally's Dad does do her a favor
realsect = 215M1AH
section = 215hpz
OutTime = Tue Apr 14 22:29:53 Mountain Daylight Time 1998
215sect = 215M1AH
whichPF = 215Hpzl12
StartTime = Tue Apr 14 22:15:34 Mountain Daylight Time 1998
Server Date is Tue, Arp 14 1998 22:51:24
Q1 = From the waves created by the boat I was able to figure out the
velocity at which waves propogate in the lake. Since the motor boat
was 30m from shore and it took 30 seconds for the waves to reach
shore, the waves traveled at 1 m/s. Since the waves are in a lake
we do not have to worry about them reflecting because they lose all
their energy as they wash up on the shore. I then found the
wavelength of the waves created by the two kids attached to the
moorings. Since v=wavelength/T, I found that the wavelength equals
2 m. (The period was 2 seconds since 15 waves were produced every 30
seconds).
The two kids can be treated as the sources and with Sally they
can be said to be in a triangular pattern. We'll say Sally's older
brother is at the right angle vertex of the triangle and her sister
is at the point 15 meters away from the right angle. The path
difference (brother to Sally vs sister to Sally)in the first case is
5 meters. (SQRT(20^2+15^2)-20). In this path difference the wave
from the sister becomes out of phase by pi radians. This means that
the wave from the brother is canceled by the wave from the sister and
Sally lies still. Being the physics expert he was, the father saw his
children's fun wasn't working, so he decided to secretly help them.
By moving Sally out another 6m he changed the path difference. The
path difference changed to Sqrt(26^2 + 15^2)-26 = 4.02. Two more
waves were produced by the sister by the time they reached Sally.
However the two sources now produced waves that were now in phase. At
the point where Sally was relaxing in her red plastic boat the waves
superimposed and the amplitude doubled. Sally now got a wild ride.
Conclusion: Sally's father is a tease.
realsect = 215M3AH
section = 215hpz
OutTime = Tue Apr 14 22:43:19 Pacific Daylight Time 1998
215sect = 215M3AH
whichPF = 215Hpzl12
StartTime = Tue Apr 14 21:31:31 Pacific Daylight Time 1998
Server Date is Tue, Arp 14 1998 23:5:48
no+help = on
Q1 = The dad has not helped his little daughter. This is because he has put
her in an area where the waves will not be canceling themselves more
thus she will feel larger waves. I determined the wavelength by
knowing the velocity of the wave as 1m/s and the period as 2 sec.
Thus, the wavelength is thte product of the two or 2. This was
subbed into the equation dSinQ=mwav. The wave is 2 and the d is 15
because that is the distance between the two kids starting waves(slits).
The Q is the angle between the line from the center of the kids to
the girl and from the center strait out, thus SinQ=H/L=15/20 or 15/26.
I solved for the m in both cases and got 5.6 where she was originally
and 4.3 where the dad moved her. Since 5.6 is closer to a half
integer (where the waves will completely cancel themselves out), she
was experiencing smaller waves originally and should not have been
moved. I must thank your EI for this one, you may have noticed that
my original one was a bit different.
realsect = 215M3AH
section = 215hpz
OutTime = Tue Apr 14 22:57:44 Mountain Daylight Time 1998
215sect = 215M3AH
whichPF = 215Hpzl12
StartTime = Tue Apr 14 22:24:33 Mountain Daylight Time 1998
Server Date is Wed, Arp 15 1998 0:39:41
no+help = on
Q1 = *the waves traveled 30 meters in 30 seconds...this corresponds to
a v of 1 m/s
*the waves created from sally's siblings are coming at 15 full
periods every half second....frequency = 1/15 s(-1)
V = lambda * f therefore lambda = 15 M
if sally is placed at a node in the motion of the wave, then she will
remain still....
node is 1/4 of a wavelength off of the crests or troughs. therefore
if sally is any multiple of .25lamda away from her siblings then
she will be still.
she is 26 meters away....26/15 = 1.73....therfore, she will not be still
realsect = 215M1AH
section = 215hpz
OutTime = Wed Apr 15 00:46:42 Mountain Daylight Time 1998
215sect = 215M1AH
whichPF = 215Hpzl12
StartTime = Wed Apr 15 00:26:24 Mountain Daylight Time 1998
Server Date is Wed, Arp 15 1998 6:56:25
no+help = on
Q1 = Since the moter boat was 30m off shore, and it took the waves 30sec
to reach the shore the velocity of the waves in water is 1m/s. We
also know that her siblings make waves with a frequency of 15bobs per
30sec, or .5Hz, which is equal to pi omega, and a 2sec period. Since
v=lambda/T we know that the wave length is 2m, and since k=2pi/lambda,
we know that k=pi. This means that our equation for the waves produced
by her siblings is 'Wave=Acos(pi*x-pi*t)'
Since we know that waves that are off by a phase of m(lambda), where
m is an integer, interfere constructivly, get biger, and waves that
are off by a phase of (m+.5)(lambda) interfere destructively, make
calm water, and the wavelength of the waves her siblings are producing
is 2m.
This means she would be in calm water if she were a distance away from]
one sibiling that was 'x', and a distance from the other sibiling that
was'x+2m+1', where m is an integer.
Where she starts out is 20m from one and 25m from the other, but when
her father moves her she is 26m from one and 30 meters from the other.
This means she is now in the roughest posible water. Thanks a lot dad.
realsect = 215M3AH
section = 215hpz
OutTime = Wed Apr 15 06:48:18 Mountain Daylight Time 1998
215sect = 215M3AH
whichPF = 215Hpzl12
StartTime = Wed Apr 15 06:17:33 Mountain Daylight Time 1998
