Homework Problem Set 5 Solutions ( )2

4 a.) continued
11
Averaging the second pair of velocities yields
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( ) = 1 2 [ v x( t 3/ 2) + v x ( t 1/ 2 )]
= 1 [ 8189 m/s + 8400 m/s ]
2
( ) = 8295m/s.
v x t 1
v x t 1
Averaging the final pair of velocities yields
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( ) = 1 2 [ v x( t 5/ 2) + v x ( t 3/ 2 )]
= 1 [ 7773m/s + 8189 m/s ]
2
( ) = 7981m/s.
v x t 2
v x t 2
These results are summarized in the table below.
n
t n
(s)
x( t n )
(m)
v x ( t n )
(m/s)
0 0 0 8400
1 1×10 −15 8.40× 10 −12 8295
2 2× 10 −15 1.66 ×10 −11 7981
b.) Compare your results from part (a) to the exact solution obtained in problem 3.
The equations for the exact analytic solutions for the position and velocity are given in problem 3. The
position is
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x( t) = ( 0.531 Å)sin( 1.58 ×10 14 s −1 t) ,
with time in seconds and position in angstroms. The velocity is
v( t) = ( 8400 m/s) cos( 1.58 ×10 14 s −1 t) ,
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with the velocity in m/s and time in seconds. The results for the analytic solution, using the same time
points as in part (b), yields € the results presented in the table below.
n
t n
(s)
x( t n )
(m)
v x ( t n )
(m/s)
0 0 0 8400
1 1×10 −15 8.363×10 −12 8295
2 2× 10 −15 1.652 ×10 −11 7982
We can see from these results that the leapfrog algorithm predicts the velocities with little error compared to
the analytic solution. The percent errors for the n = 0, 1, and 2 time steps are 0%, 0%, and 0.01%,
respectively. The leapfrog algorithm also does a reasonable job in predicting the position of the particle
compared to the analytic solution. The errors are somewhat larger than those for the velocity, with errors of
0%, 0.4%, and 0.6%, respectively.