Homework Problem Set 5 Solutions ( )2

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4 a.) continued 10 For the next time step, with n = 1, the velocity equation becomes v x ( t 3/ 2 ) = v x ( t 1/2 ) − hk m x ( t 1). In order to evaluate this equation, it is helpful to convert the force constant into SI units, € v x t 3/ 2 k = ( 760 kcalmol −1 Å −2 ⎛⎛ 4.184 kJ ⎞⎞ ) ⎜⎜ ⎟⎟ 1000 J 2 ⎛⎛ ⎞⎞ ⎛⎛ 1Å ⎞⎞ ⎜⎜ ⎟⎟ ⎜⎜ ⎝⎝ 1kcal ⎠⎠ ⎝⎝ 1kJ ⎜⎜ ⎠⎠ ⎝⎝ 10 −10 ⎟⎟ ⎟⎟ m⎠⎠ k = 3.01×10 26 J mol −1 m −2 . Substituting the force constant into the velocity equation yields € ( ) = v x ( t 1/ 2 ) − hk ( ) m x t 1 = 8400 m/s − 10−15 s v x ( t 3/ 2 ) = 8189 m/s. For the time step with n = 1 the position equation becomes ( )( 3.01× 10 26 J mol −1 m −2 )( 8.40 × 10 −12 m) ( 0.012 kg mol −1 ) x( t 2 ) = x( t 1 ) + h v x ( t 3/ 2 ) . Substituting, € x t 2 ( ) ( ) = 8.40 ×10 −12 m + 10 −15 s ( 8189 m/s ) x( t 2 ) = 1.66 ×10 −11 m. Finally, for the final time step, with n = 2, the velocity equation becomes € € v x ( t 5/ 2 ) = v x ( t 3/ 2 ) − hk m x ( t 2). Using the force constant in the velocity equation yields € v x ( t 5/ 2 ) = 8189 m/s − v x ( t 5/ 2 ) = 7773 m/s. Averaging the first pair of velocities, the equations are v x t 0 ( 10 −15 s) 3.01×1026 J mol −1 m −2 ( )( 1.66 ×10−11 m) ( 0.012 kg mol −1 ) ( ) = 1 2 [ v x( t 1/ 2) + v x ( t −1/ 2 )] = 1 [ 8400 m/s + 8400 m/s ] 2 v x ( t 0 ) = 8400 m/s. €
4 a.) continued 11 Averaging the second pair of velocities yields € ( ) = 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 € ( ) = 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 € 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) , € € 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.
Page 1 and 2: Chemistry 380.37 Dr. Jean M. Standa
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Page 15 and 16: 15 6. The equipartition theorem sta

Homework Problem Set 5 Solutions ( )2

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