"In Young's double slit experiment, assuming the distance between the slits is 0.07mm and the wavelength of light used is 600 nm, when the screen is 70 cm away, what kind of interference is there."
I'm not sure what I'm being asked to determine here.
Plugging in the figures into the...
The question calls for using Cauchy's integral formula to compute the integral for Int.c z/[(z-1)(z-3i)] dz, assuming C is the loop |z-1|=3.
Taking z = 1 and f(z) = z/(z-3i), I came up with (2pi*i)/(1-3i), which seems like it could be simplified, but I'm not sure how.
Alright then, so z = (z + i)A + (z - 2i)B, then expanding gives
z = z(A + B) + i(A - 2B)
Since the LHS has no i, then A - 2B = 0, and likewise, A + B = 1
But...this is going nowhere. Where am I slipping up?
I need to find the partial fraction expansion of the integrand z/[(z-2i)(z+i)]
Just doing 1/(z-2i) + 1/(z+i) results in (2z-i)/(z-2i)(z+i).
It seems easy, but I can't figure out what to multiply by to get the correct numerator.
As part of finding the integral of z/(z^2 -1), I'm stuck on getting the partial fraction for it. 1/2 [(1/(z-1) - 1/(z+1)] gives 1/(z^2-1). What should I do to get the z in the numerator. Any hints welcome.
Regards
Cauchy integral question
The question calls for finding the integral of dz/((z-i)(z+1)) (C:|z-i|=1)
I can't figure out how to do this for (C:|z-i|=1). How does this differ from, say, (C: |z|=2)
Regards
What exactly does a path integral measure? Is it area between the ends/bounds of the line? Or is it the length of the line? Just started complex analysis and am comletely confused by this.
I get it. Since 20 degrees = pi/9, moving LHS A to RHS gives
sin (-120pi*t + pi/9) = 1
so, -120pi*t + pi/9 = pi/2
and finally t = 7/2160 seconds
Many thanks
"There are two sine waves having a phase difference of 20 degrees. After one reaches its maximum value, how much time will pass until the other reaches its maximum, assuming a frequency of 60 Hz."
Should I go about this by assuming...
sin(120pi*t) = sin(120pi(t + x) - 20)
Any hints...
The problem is to find the definite integral of e^2x/(e^x + e^-x) with the limits of 0 and log 2. Finding the integral was easy enough, e^x - log (1 + e^2x) , but how do I plug log 2 into this. Any hints appreciated.
Just finished working out the integral of 1/(1 + cosx)^2 as
(1/2)(tan(x/2)) + (1/6)(tan (x/2))^3 + C
I'm just wondering whether it's possible to use any substitutions to simplify this further.
regards
Would very much appreciate a hint on how to start this off.
So far I have cos theta = (1-t^2)/(1+t^2). After squaring this and doing various substitutions, I get (cos theta)^3, which can't be right.
I've struggled with this since this morning with hardly any progress.
"There is a level pipe filled with water flowing through it. Branching off from this pipe are two pipes going perpendicularly upward, with a cross sectional area Sa, Sb, respectively, and there being a difference in water...
Many thanks
I think I get it. Based on your equations, the ratio of the young's moduluses should equal the ratio of the individual increases, that is, 13/10 = L1/L2
Therefore, 10L1 = 13L2
Since, L1 = L2 = 0.5, then L1 = 0.5 - L2. Plugging this into the above gives 10(0.5 - L2) = 13L2, so...
Young's modulus problem -- need a hint
There are two wires, one brass the other copper, both 50 cm long and 1.0 mm diameter. They are somehow connected to form a 1m length. A force is applied to both ends, resulting in a total length change of 0.5 mm. Given the respective young's moduluses of...
Rereading my question, i noticed that i typed the wrong sign in the displacement equation. I meant to say x=(2 m) cos (6pi(t) - pi/2) ["minus" before pi/2]. This now gives 2 m with t = 0. Since the object is first moving to the right (positive), shouldn't the phase angle be -p/2?
"An object is moving along the x-axis in simple harmonic motion. It starts from its equilibrium position which is at the origin at t=0 and is moving to the right. The amplitude of its motion is 2 m and its frequency is 3 hz. (1) Determine the expression for the objects displacement. (2) Where is...
"An elevator is moving upward at a constant velocity of 2.5 m/s. A ball is launched straight upward from the floor of the elevator with an initial velocity of 4.9 m/s (relative to the floor). (1) Determine the time the ball takes to reach its highest point from the standpoint of a person not...
"In an elevator that is moving with constant acceleration, a person of 50 kg mass is standing on a bathroom scale that shows "55 kg". Determine the accerlation of the elevator."
I'm pretty sure my answer is right. I'm just wondering if there's another way.
So, 55 kg is 1.1 times 50 kg...
Yes, i got the same answer. Since this is a conservation of energy problem, obviously your approach is the more elegant of the two. Makes sense: the PE that has been consumed has been converted to speed.
Thanks
Ball, spring and curved incline -- is this right?
Picture a smooth, level, frictionless surface (A) that leads into a curved incline (B-C). The top of the incline (C) is 0.4 meter above the flat surface. At the start of A is a spring with a spring coefficient of 10, with one end fixed and a...
Just realized I forgot about the board. So since, as Gza said, momentum is concerved here, p(initial) = p(final)
so, (0.5)3 = (1 + 0.5)V'
Solving for v' gives 1 m/s (the velocity of the clay/board system)
Plugging this into E(initial) = E(final) gives
1/2mv^2 = 1/2kx^2
x = 0.1 meter...
Did some more reading.
Since this is a conservation of energy problem, E(initial) = E(final)
Now, E(initial) is the kinetic energy of the clay 1/2mv^2 = (1/2)(.5kg)(3m/s)^2 = 2.25 J
E(final) is the potential energy stored in the compressed spring 1/2kx^2 = 1/2(150)x^2
Therefore... 2.25 =...
So, if the momentum of the clay is mv= 0.5kg(3 m/s)=1.5 kg/ms, do you mean that all this energy will be conserved by the spring?
What exactly is the work energy theorem?
Thanks
"On a frictionless, horizontal surface there is a spring with one end fixed and the other attached to a board of mass 1 kg. The spring coefficient is 150 N/m. 500g of clay moving at 3.0 m/s on this horizontal surface collides squarely with the board. As a result, the board and clay stick...