Archive for January, 2010

Physics Friday 106

January 29, 2010

Consider two parallel rings of radius R, separated by a perpendicular distance 2d, with a soap bubble (in equilibrium) stretching between them. What is the shape of the soap bubble?

Solution:

Monday Math 105

January 25, 2010

The Laplace Transform
Part 6: Frequency shifting, time shifting, and scaling properties

Expanding upon last time’s work on the Laplace transform of the exponential, we can prove quickly the “frequency shifting” property of the Laplace transform. Let real function f(t), t≥0, have Laplace transform . Then, for real number a, we see that
,
and thus multiplication by an exponential in the time domain equals a translation in the s domain.
This can be combined with our results for sine and cosine to get

.
And our results for powers of t to get
.
Note that for all of these, the shift in the “frequency” domain entails an equal shift in the region of convergence; if for , then for .

Next, let’s consider what happens when we scale our time variable. Again, we have f(t), t≥0, with Laplace transform . Now, let us find the Laplace transform for f(at), with a>0.
.
Making u-substitution u=at, then
.
As with our other result, the region of convergence will be transformed along with the scaling.

We found what happens in the time domain when the transform is shifted in the “frequency” domain, but what happens when we shift in the time domain? Recall that our function f(t) is defined for t≥0; when we shift our function forward, we must set those portions that are at t<0 before our shift to zero, so that our translated function is f(tτ)H(tτ), where H(t) is the Heaviside step function, and τ>0 is our time delay.
Integrating as we did for the transform of the Heaviside step function,
.
Now, making u-substitution u=tτ, we find
.
Thus, a time shift corresponds to multiplication by an exponential in the “frequency” domain.

Physics Friday 105

January 22, 2010

Suppose you have 30 identical 1 Ω resistors connected to form the edges of a regular icosahedron? What, then, is the effective resistance between adjacent vertices? What about the effective resistance between adjacent vertices of a regular dodecahedron formed of 30 identical 1 Ω resistors?
Solution:

Monday Math 104

January 18, 2010

The Laplace Transform
Part 5: Exponential and trigonometric functions

Next ia a very simple Laplace transform: that of exponential functions. If , we use the definition of the Laplace transform to find that
,
with for convergence.
From this, and the linearity of the Laplace transform, we see
,
and
,
both with .

Now, in the integral we used for the exponential, we see that the integration still holds, for , if we replace a with ıω; we get
.
Thus, we use and to get
,
and
,
both with .

Phyiscs Friday 104

January 15, 2010

Consider a massive rope, with length L and mass per unit length of λ. Let this rope be held so that it hangs vertically, with a thin sheet of paper just below the bottom end of the rope. This sheet of paper can support without breaking a maximum weight equal to twice the weight of the rope. If one releases the top of the rope, will the paper stay unbroken until the rope is entirely atop it? If not, what will the height of the top of the rope (relative to the paper) be when the paper breaks?
Solution:

Monday Math 103

January 11, 2010

The Laplace Transform

Part 4: Powers

Let us now find the Laplace transform for powers of t; letting f(t)=tr, we use the definition of the Laplace transform to find:
,
with as the region of convergence.
Using the u-substitution u=st,
.
You should recognise this last integral as the gamma function of r+1, and so
,
and so we see that it is necessary that r not be a negative integer. If r is a non-negative integer n, then we have
.
Note also that r=0 gives a transform of 1/s, which is what we previously found for the Laplace transform of a constant.
Setting , and using , we see
.
Similarly, since ,
.

Physics Friday 103

January 8, 2010

Suppose we have a long, uniform plank of thickness T balanced on a cylindrical support of radius R.

Now, let the plank tilt slightly, with no slipping or sliding along the pivot (sufficient friction). There are two possibilities: first, the plank could continue to tilt until it falls off the support (unstable); or, secondly, the plank could tip back, in a small rocking oscillation on the pivot (stable); what is the condition needed for the latter situation?
Solution:

Monday Math 102

January 4, 2010

The Laplace Transform
Part 3: Step and Delta Functions

The Laplace transform of the constant function f(t)=c, t≥0, can be computed easily from the definition of the transform; with (necessary for convergence of the integral),
.

Similarly, one can also find easily the Laplace transform of f(t)=H(tτ), where H(t) is the Heaviside step function, and τ a positive constant. [If τ≤0, then for t>0, f(t)=1, and the integral is identical to that above for the constant, and .]
Taking the integral, again with ,
.

Next, consider an impulse function f(t)=δ(tτ), where δ(t) is the Dirac delta function, and τ≥0.*.
Then we have
,
and letting τ=0, we see
.
Unlike the previous two Laplace transforms, this converges for all .

*The formal definition of the Laplace transform for acceptable distribution functions, such as the Dirac delta, is chosen so that any “point mass” at t=0 is entirely included in the transform; this is often written , where the lower limit 0 is shorthand for the limit as zero is approached from below:
.
Thus, the case of τ=0 above is valid.

Physics Friday 102

January 1, 2010

Torricelli’s law, mentioned in the previous Physics Friday, is derived from Bernoulli’s Principle, via a few assumptions. Let us have a small hole at the base of a large reservoir, with fluid depth h. Assuming an incompressible fluid and inviscid flow, Bernoulli’s principle says that is a constant along all points on a streamline, where v is the flow speed, z is the elevation, P is pressure, and ρ is the fluid density. For our problem, our streamlines all run from the upper surface of the fluid to the hole. Thus, choosing zero elevation at the hole, we have
,
where vs is the flow speed at the surface of the fluid, vh the speed at the hole, Ps the pressure at the surface, and Ph the pressure at the hole. However, we should note that Ps and Ph are both just atmospheric pressure, so that Ps=Ph, and the pressure terms cancel, giving us
.
If the hole is sufficiently small compared to the reservoir, then the flow rate out is small enough that the motion of the fluid surface is negligible; setting vs=0, and solving for vh, one gets Torricelli’s Law: .

Now, suppose we relax the last assumption, and do not neglect the speed at which the fluid drops in the reservoir. Assuming a hole of area a, and that the surface of the fluid in the reservoir has area A, then we see that the volume of fluid lost out through the hole is avh; this must equal the volume from the lowering of the surface, and so . Solving for the surface velocity, and plugging into our result from Bernoulli’s principle,
.
This is larger than the speed given by Torricelli’s law by a factor of . For , this factor is approximated by , and thus it is reasonable to neglect it, sticking to just Torricelli’s law.

2010

January 1, 2010

Happy New Year!