acceleration question
jgi...@hgo.net
the plane of the surface. A thread is drawn between points
P and Q, where P=(x1,y1) and Q=(x2,y2) and the magnitude of
P > the magnitude of Q. A stone is dropped at O=(0,0),
creating a wavefront of constant speed s that reaches point
Q first, travels along line segment QP and reaches point P
last.
What is the equation of motion of the wavefront along line
segment QP? I need the equations of the instantaneous velocity
and acceleration of the circular wavefront as it intersects the
line QP.
Let
tq=time for wave to reach Q
tp=time for wave to reach P
s(tq)=|Q|
s(tp)=|P|
r(t)=radius of wavefront at any time t = st
c=|QP|
c'=(dc/dt)
R=position vector of the intersection of the
wavefront with line QP.
At R(tq), r(t)=s(tq)=|Q|
At R(tp), r(t)=s(tp)=|P|
At R(t), tq <= t <= tp, r(t)=?
Supposing I used the law of cosines
Let
|Q|=a, |P|=b, angle POQ = A. Then
2cc'=2aa'+2bb'-2(ab'+a'b)cosA
But a'=b'=s and
cc' = s(a + b)(1 - cosA)
c dc/dt - C = 0, C = s(a+b)(1-cosA)
c dc = C dt
(1/2)c^2 = s(a+b)(1-cosA)(tp-tq)
c' = (s/c)(a+b)(1-cosA)
Does anyone have any idea?
Please reply to: jgi...@hgo.net
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> On the surface of an infinite pond, the x and y axis are on
> the plane of the surface. A thread is drawn between points
> P and Q, where P=(x1,y1) and Q=(x2,y2) and the magnitude of
> P > the magnitude of Q. A stone is dropped at O=(0,0),
> creating a wavefront of constant speed s that reaches point
> Q first, travels along line segment QP and reaches point P
> last.
> What is the equation of motion of the wavefront along line
> segment QP? I need the equations of the instantaneous velocity
> and acceleration of the circular wavefront as it intersects the
> line QP.
> Let
> tq=time for wave to reach Q
> tp=time for wave to reach P
> s(tq)=|Q|
> s(tp)=|P|
> r(t)=radius of wavefront at any time t = st <----
> c=|QP|
> c'=(dc/dt)
> R=position vector of the intersection of the
> wavefront with line QP.
> At R(tq), r(t)=s(tq)=|Q|
> At R(tp), r(t)=s(tp)=|P|
> At R(t), tq <= t <= tp, r(t)=?
r(t)=st (you are given that!)
I suppose instead you want the motion along the line.
Let the equation of the line joining Q to P be x=x2+w(x1-x2),
y=y2+w(y1-y2) and we also have x^2+y^2=s^2*t^2 (s=radial velocity of wave,
t=time, w=parameter along the line from Q=(x2,y2) to P=(x1,y1): w=0 at Q,
w=1 at P: w is the percentage of the distance from Q to P where the point
(x,y) on the line segment lies).
Do you want w as a function of t? (w measures the location of the point
along the line as a percentage of the displacement from Q to P) If so...
simply substitute x=x2+w(x1-x2), y=y2+w(y2-y1) into x^2+y^2=s^2*t^2 and
solve for w (you will get Aw^2+Bw+C=s^2*t^2 where, for example,
A=(x1-x2)^2+(y2-y1)^2=the_square_of_the_PQ_Distance) (just multiply out)
and can differentiate to get 2Aw*w'+Bw'=2s^2*t (w'=dw/dt) (if the distance
between P and Q is "1" then w' is the velocity along the line... if the
distance is D, then wD is the velocity) or w'=(2s^2*t)/(2Aw+B)
(e.g. at w=0, at position Q, the motion along the line has speed
2s^2*t_q/B where t_q is the time at Q=|Q|/s). For the acceleration,
differentiate 2Aw*w'+Bw'=2s^2*t again to get: 2Aw'^2+2Aw*w"+Bw"=2s^2 (and
plug in w' that we just got and solve for w").
William L. Bahn
What if P = (3,0) and Q = (-2,0)?
In order to reach point Q first and THEN "travel" unidirectionally along QP
to point P it seems like the angle OQP would have to be obtuse.
jgi...@hgo.net wrote in message <6pecit$u2s$1...@nnrp1.dejanews.com>...
>On the surface of an infinite pond, the x and y axis are on
>the plane of the surface. A thread is drawn between points
>P and Q, where P=(x1,y1) and Q=(x2,y2) and the magnitude of
>P > the magnitude of Q. A stone is dropped at O=(0,0),
>creating a wavefront of constant speed s that reaches point
>Q first, travels along line segment QP and reaches point P
>last.
>
>What is the equation of motion of the wavefront along line
>segment QP? I need the equations of the instantaneous velocity
>and acceleration of the circular wavefront as it intersects the
>line QP.
Do you want the instantaneous velocity and acceleration of the circular
wavefront, or do you want the instantaneous velocity and acceleration of the
point of intersection?
>
>Let
>tq=time for wave to reach Q
>tp=time for wave to reach P
>s(tq)=|Q|
>s(tp)=|P|
>r(t)=radius of wavefront at any time t = st
>c=|QP|
>c'=(dc/dt)
>R=position vector of the intersection of the
>wavefront with line QP.
>
>At R(tq), r(t)=s(tq)=|Q|
>At R(tp), r(t)=s(tp)=|P|
>At R(t), tq <= t <= tp, r(t)=?
>
David Ziskind
"derived wave" problem, which with changes in notations, is in substance the
following:
a) in 2 dim space, a straight line (the "line of derived motion") is
identified passing from point p to q. p=(p1,p2); etc.
b) function Z: t->Z(t), determines a derived motion. Two relations hold
for Z(t) (a 2 dim point):
1) Z(t) lies on the circumference of a circle with radius s*t, ie, Z(t)
lies on the wavefront of a radially traveling wave which started at
the origin at time 0 and has speed s;
2) Z(t) also lies on the line of derived motion;
3) in particular, Z(t0) = p.
c) Z is twice differentiable.
CHANGES IN NOTATION:
1. In the base post; point p was called (x1,y1), point q was (x2,y2).
2. In this post, the following symbols are NOT used: x1, y1, x2, y2.
3. The object here represented by Z(t) more usually might be represented by
X(t) or P(t), neither of which are used to avoid possible confusion with
other usages in the statement portion of the base post.
4. The development below is self-contained; the usages in the base post
solution section are not considered, do not apply here.
SOLUTION:
In the following deduction: A is a function; A(t) is a scalar; upper case
denotes a function; lower case an evaluated function, a fixed number, or a
fixed vector; * denotes scalar multiplication or vector inner product, and
prime (') denotes derivative. +-1 means: plus-or-minus-one. As implied by
b.3 above, "t0" is a distinguished value of t.
1. Assert as premise, "a", "b.2", and "b.3" of the problem statement. Thus:
Z(t)-p = A(t)*(q-p) and
A(t0) = 0
2. Assert as premise, "b.1" of the problem statment. Thus:
|Z(t)| = s*t
3. Introduce the following definitions for economy:
h = q-p
n = the angle between the vectors p, and q-p. (There is no implication
as to integer values.)
4. Square and differentiate 2.
5. Differentiate 1.a.
6. Substitute into Result-4 using 1.a and Result-5. Obtain a nonlinear
differential equation in A. Transform by using the identity
A' = 1/((Inv A)')(A). The resultant diff eq in Inv A is separable.
Integrate between limits, use 1.b. Evaluate result at arg=A(t), obtain a
quadratic in A(t), and solve. Obtain:
7. A(t) = |h|^(-2)* {-p*h +- [(p*h)^2+|h|^2*s^2*(t^2-t0^2)]^(1/2)}
8. Differentiate. By continuity, obtain +-1 = constant with respect to t.
Obtain for A'(t):
A'(t) = +-1*s^2*t / [(p*h)^2+|h|^2*s^2*(t^2-t0^2)]^(1/2)
9. Combining 1.a and 7 gives the general equation of motion for Z. The
primary objective however is to investigate the norms of the velocity
Z', and the acceleration Z''. We proceed as follows:
10. From 1.a and 3 we have:
|Z'(t)| = +-|h|*A'(t) and
|Z''(t)| = +-|h|*A''(t)
The calculation for |Z''| will be omitted since it is minor extension of
preceding results. For |Z'|, combine 10.a and 8, and use 3.b and obtain
after some calculation (and see below):
11. |Z'(t)| = s / denom; with
denom = [1 + (|p|^2*(cos n)^2 - s^2*t0^2)/(s^2*t^2)]^(1/2)
12. A significant result is: as t->inf, |Z'(t)|->s .
13. For t equal t0, we have the simple and important result:
|Z'(t0)| = s/|cos n|
Note that time t0 is the time with the property that: Z(t0) and the
line of derived motion form an angle of value n.
COMMENTS:
The motion Z is a "derived motion" of the type that can have arbitrarily
large velocity. As illustrated by 13, if angle n is 90 deg, then the
velocity is infinite at time t0. 11 gives the general circumstance for this
to occur.
The problem statment given here is similar to the searchlight problem that
is discussed in various special relativity treatments. Although the
analysis is different, the two common features are: 1) the wave that is
considered is "derived" from some other wave; 2) arbitrarily large
velocities are possible. The coordinate transformations in SR/GR of course
work so that arbitrarily large velocities are never generated. It may be a
simple matter to characterize the salient difference between SR/GR versus
the searchlight problem and its clones, but I would at this point suggest
that other writers be consulted.
Signs Multiplying Radical: In 7 and 8, we know most of the solution but the
analysis technique leaves it uncertain as to whether the sign multiplying
the radical is plus or minus. So which is it? In 11, denom has to be equal
to a positive quantity divided by a positive quantity and is thus positive,
and the radical is positive so the sign multiplying it must be positive.
For the general equation of motion, the sign of A(t) is determinable from
1.a, and the form of 7 plus a continuity argument determines the sign
multiplying the radical.
David Ziskind
zis...@ntplx.net