Theoretishe Physik A WS 2000/01
Prof. Dr. J.Kuhn / Dr. W.Kilian / I. Shwarze Blatt 3 3. 11.2000
Losungsvorshlage
1. Ameisenspaziergang(5 Punkte)
(a)
s(x)= Z
x
a dx
q
1+y 0
2
= Z
x
a dx
p
1+sinh 2
x= Z
x
a
dxoshx=sinhx sinha
(b)
x(s)=Arsinh(s+sinha)
y(s)=oshx(s)= q
1+sinh 2
[x(s)℄= q
1+(s+sinha) 2
()
sinh 0
t=osht ) Arsinh 0
t=
1
osh (Arsinht)
= 1
p
1+t 2
dx
ds (s)=
1
q
1+(s+sinha) 2
;
dy
ds (s)=
s+sinha
q
1+(s+sinha) 2
dx
ds
2
+
dy
ds
2
=
1
2 +
(s+sinha) 2
2
=1
(a)
a
x
(t)= a
h sin!t
a
y
(t)=a
h os!t
(+1) fur 0t 1
2 T
( 1) fur 1
2
T t T
a
z
(t)=a
v 8
<
:
(+1) fur0t 1
4 T
( 1) fur 1
4
T t 3
4 T
(+1) fur 3
4
T tT
(b)
v
x
(t)=v
0 a
h
!
(1 os!t)
v
y (t)=
a
h
!
sin!t
(+1) fur0t 1
2 T
( 1) fur 1
2
T tT
v
z
(t)=a
v 8
<
:
t fur 0t 1
4 T
1
2 T t
fur 1
4
T t 3
4 T
(t T) fur 3
4
T t T
()
x(t)=v
0 t
a
h
w
t 1
! sin!t
y(t)= a
h
w 2
(1 os!t)
(+1) fur0t 1
2 T
( 1) fur 1
2
T tT
z(t)= 1
2 a
v 8
>
<
>
:
t 2
fur 0t 1
4 T
h
1
8 T
2
1 t 1
2 T
2 i
fur 1
4
T t 3
4 T
(t T) 2
fur 3
4
T t T
(d)
z
1
2 T
= 1
16 a
v T
2
=z
max
) a
v
= 16z
max
T 2
x(T)=v
0 T
a
h T
!
=0 ) a
h
=v
0
!
x
1
T
= v
0
= 1
x
tot
) v
0
= 1
!x
tot
x y
y z
(f)
v 2
(t)=v 2
x (t)+v
2
y
(t)+v 2
z
(t)=v 2
0 os
2
!t+v 2
0 sin
2
!t+a 2
v t
2
=v 2
0 +a
2
v t
2
s(t)= Z
t
0 dt
0
v 2
(t 0
)= Z
t
0 dt
0 q
v 2
0 +a
2
v t
0 2
= v
2
0 a
2
v
2a 3
v
Arsinh a
2
v t
v
0 a
v +
a 2
v t
2a 2
v q
v 2
0 +a
2
v t
2
= v
2
0
2a
v
Arsinh a
v t
v
0 +
t
2 q
v 2
0 +a
2
v t
2
s(T)=4s
1
4 T
=
2
x 2
tot
2z
max
Arsinh 2z
max
x
tot +
q
2
x 2
tot +4z
2
max
=x
tot
1
Arsinh+ p
1+ 2
mit:=
2z
max
x
tot
= 1
x
tot
(0:984+1:049)128m
DiesesErgebnisisterstaunlihnaheamUmfangvonzweiKreisen, 2x
tot
126m.
Die Hohe von 10m bewirkt alsonur eine kleine Korrektur.
