Assignments for Physics 5103 - 2019 Reading in Classical Mechanics with a BANG! and Lectures

9/5 Assignment Set 3 - Read Unit 1 Ch. 3 thru Ch.7 Due Wed. 9/18/19 Name_____________

Pseudo-Rotations for Independent Bounce ModelExercise 1 Estrangian plot in Fig. 5.2 (Details on p.51-59 of Lecture 3) has mass ratio " and has nearly periodic path plot. (Experiment using BounceIt link in Lect.3 p.74-77. https://modphys.hosted.uark.edu/markup/BounceItWeb.php?scenario=1014) (Let small-mass be " here.) Changing to " gives more nearly periodic symmetry paths seen below.

" " (a) What order N=___ of CN or DN polygonal symmetry is appearing here? (b) Give a closed formula for value of " (to 7 figures) that approaches exactly periodic behavior. Simplest formula should relate the tangent of a desired Estrangian rotation half-angle " to mass " . Plot an Estrangian velocity path on the protractor graph-paper attached assuming initial velocity V1=1 and V2=0. (c) Ceiling height (It is " for cases above) may eventually affect or destroy periodicity. Use BounceIt to show cases that are affected. (Many have chaotic behavior.)

KE becomes PE Exercise 2 A mass m1=1kg ball is trapped (like Fig. 6.3) between two smaller mass m2=1gm balls of high speed (v2(0)=1000m/s for x=0). Suppose this affects m1 with an effective force law F(x) of isothermal approximation (6.11). Assume m1 motion is small and slow around x=0. (“Balls” idealize as point masses here.)

(a) A further approximation is the one-Dimensional Harmonic Oscillator (1D-HO) force and PE in (6.12). If each mass m2 start in an interval Y0=1m, derive approximate 1D-HO frequency and period for mass m1.

(b) What if the adiabatic approximation is used instead? Does the frequency decrease, increase, or just become anharmonic? Compare isothermal and adiabatic quantitative results for m1=1kg ball being hit by two m2=1gm balls each having speed of v2(0)=1000m/s as each starts bouncing in a space of Y0=1m on either side of the equilibrium point x=0 for the 1kg ball.

(c) How does the frequency decrease or increase in isothermal case versus the adiabatic case if we shorten the run interval Y0=1m to one-quarter meter?…What if we reduce the mass ratio m1/ m2 by one-quarter?(d) Derive the adiabatic frequency and period for the case M=50kg in adiabatic force of two m=0.1kg masses of initial speed v0=20m/s and range Y0=3m. Compare with Fig. 1.6.3c.

M1 /m2 = 49/1

m2=1 M1= 48.3..

M1= 48.3...θ/2 M1

ymax = 7.0

Assignments for Physics 5103 - 2019 Reading in Classical Mechanics with a BANG! and Lectures

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Assignments for Physics 5103 - 2019 Reading in Classical Mechanics with a BANG! and Lectures

Assignment 3- Solutions to Ex. 1 Solution to Pseudo-Rotations for Independent Bounce ModelEx.1(a) Symmetry is that of 22-point polygon (22-agon), a cyclic group C22 or dihedral group D22.(b) Solving the mass ratio equation (5.10b) for " in terms of angle θ that simplifies to:

Given angle or and m2=1 predicts

Assignment 3 Solutions to Ex. 2. Ex.2 (a) A mass m1=1kg ball is trapped (like Fig. 6.3) between two smaller mass m2=1gm balls of high speed (v2(0)=1000m/s for x=0). Suppose this affects m1 as effective force law F(x) of isothermal approximation (6.11).

For isothermal we set: " to give: "

For Y0=10cm there is an isothermal spring constant due to left (Y0+x) AND right (Y0-x) walls.

…

If Y0=1m, m2=1gm=10-3kg has speed v2=1000m/s so k= =2(10-3)(1000)2=2000.

Then frequency of m1= 1 kg is ω=√(k/m1)=√2000=44.7 rad/sec. Then υ=√2000/2π=√500/π=7.12Hz with period π=0.14sec.(b) What if the adiabatic approximation is used instead? Does the frequency decrease, increase, or just become anharmonic? Compare isothermal and adiabatic quantitative results for Ex. 2(a).For adiabatic v2 is not constant but inversely dependent on Y so we set:

, ~f (for:Y~1).

"

So 1kg mass angular frequency is ω=√2000√3=77.5 rad/sec. (Increases by factor of √3 over the isothermal values.) υ=√2000√3/2π=12.33 Hz. 1/υ=τ=1/12.33=0.0833 sec (c) How does the frequency decrease or increase in isothermal case and in the adiabatic case if we shorten the run interval Y0=1m to one-quarter meter?……..if we reduce the mass ratio m1/ m2 by one-quarter?

(Both frequencies increase by factor of 4.) (Both frequencies increase by factor of 2.) (d) Derive the adiabatic frequency and period for the case M=50kg in adiabatic force of two m2=0.1kg masses of initial speed v0=20m/s and range Y0=3m. Compare with Fig. 1.6.3c.

" (Close to Fig. 6.3)

m1 /m2

cosθ ≡ m1 −m2

M⎛⎝⎜

⎞⎠⎟ and : sinθ ≡

2 m1m2

M⎛

⎝⎜⎞

⎠⎟cosθ = m1 −m2

m1 +m2

=

m1

m2

−1

m1

m2

+1 and : m1

m2

= 1+ cosθ1− cosθ

=cos2 θ

2sin2 θ

2

= cot2 θ2

or : cotθ2= m1

m2

θ = 2π /22 = π /11 θ /2 = π /22 m1 = cot2 π22

= 48.3741500787

v2 = const. = v2IN = v2(0)

F isoth (1wall Y )=±m2v2

2

Y= ±

m2v22 (0)

Y=± const.

Y=± f

Y

k ≡ 2 f = 2·m2v2

2 (0)Y0

2

Fisoth = fY0 + x

− fY0 − x

= fY01− xY0

+ ...⎡

⎣⎢

⎤

⎦⎥ −

fY01+ xY0

+ ...⎡

⎣⎢

⎤

⎦⎥ ≈ −2 f · x

Y02 = −2m2v2

2(0)· xY02 = −k·x

2m2v22 (0)

v2 =const.Y

=v2INY0Y

F adibad (1 wall)=±m2v2

2

Y!± m2

v2INY0( )2Y 3

= ± gY 3

where: g=m2 v2INY0( )2

F2−walladibad = g

(Y0 + x)3 −

g

(Y0 − x)3 =

g

Y 30

g

(1+ xY0)3− g

(1− xY0)3

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= g

Y 301− 3 x

Y0+ ...

⎡

⎣⎢

⎤

⎦⎥ −

g

Y 301+ 3 x

Y0− ...

⎡

⎣⎢

⎤

⎦⎥ = − 6g

Y 40·x + ...

= −6·m2v2

2(0)Y 20Y 40

·x + ...= −6·m2v2

2(0)

Y 20·x + ... (Effective spring constant increases by factor of 3)

υ = ω2π

= 12π

kM

= 12π

6·m2v22(0)

M ⋅Y 20= 12π

6·0.10·202

50·32= 12π

2445

= .732π

So period is τ = 1υ= 2π 15

8= 8.6sec.

Assignments for Physics 5103 - 2019 Reading in Classical Mechanics with a BANG! and Lectures

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π/22=8.18°

π/11=16.36°

-π/11=-16.36°