Sorry I’ll try to explain it better. We can write the spatial part of the solution as ##\varphi (x) = A\sin{nx}## and the temporal part as ##\psi (t) = B\cos{nt} + C\sin{nt}##, with ##n \in \mathbb{Z}##. Part of the full solution then is $$ u_{n}(x,t) = \varphi (x)\psi (t) = (AB\cos{nt} +...
Okay thank you. I’m still a bit confused though… In the book I’m following m is restricted further and only m = 1,2,3,… is considered. If m = 0 the solution vanishes. I’m not understanding why the case in which m<0 resolves to that of m>0 when the coefficients for cosine differs by a -ve sign...
Is it necessary to write ##\sqrt{(-m^2)}=a+ib##? Can I not just write ##e^{imx} = e^{-imx}##
This is what I got writing the exponent out as a + ib:
$$ e^{(a+ib)\pi} = e^{-(a+ib)\pi} \rightarrow e^{2\pi(a+ib)} = 1 \rightarrow e^{2\pi a}(\cos{2\pi b} + i\sin{2\pi b}) = 1 $$
Then I guess a has...
I am trying to follow through a derivation of the solution to the wave equation governing a string with fixed ends via separation of variables but I am stuck at the step which concludes ‘m’ must be a natural number 1,2,3,… etc. as opposed to an integer. After analysing each case: m > 0, m = 0...
Ahhh okay, thank you. For some reason it didn’t click that I could solve for V using ## T = \frac{2a}{R}\left(\frac{1-\frac{b}{V}}{b}\right) ## and then sub that into the other equation 😬 I’ve got it now
Hey. Thanks for the response. Do you have any idea how I would go about obtaining the expression my lecturer has provided? Namely, $$ P_{inv} = \left[\frac{2a}{b^2} - \frac{RT}{b}\right]e^{\frac{1}{2}-\frac{a}{RTb}}$$ If I am to rewrite ## T = \frac{2a}{R}\left(\frac{1-\frac{b}{V}}{b}\right) ##...
The notes my lecturer has provided state that the maximum temperature can be found taking p = 0 in the inversion curve formula, given as:
I’m not sure how to obtain this??
These are the formulas:
This is my attempt at a solution :
Not sure if this approach is right?
Ohhh okay, that makes everything a lot clearer... So dp is the difference in pressure between the top and bottom portions of the slap, therefore p_top = p_bottom + dp
So the force from the pressure at the top of slab acts downwards and is given by (p+dp)A
I think its mainly the (p+dp)A force I'm not understanding. Specifically the dp part, I'm not sure exactly what that is supposed to represent or why its acting downwards. I thought there would be more pressure at the bottom?
dw is the weight of the slab of fluid and the negative sign implies that it acts downwards.
(p+dp)A is the force on the slab due to the fluid above it, it also acts downwards, hence the negative sign.
pA is the force acting upwards on the slab and so its positive...
The net force acting on...