s11f1005

# s11f1005 - 2 = b π 2 = 1 ⇒ b = 2 π so that this path is...

This preview shows page 1. Sign up to view the full content.

HOME AND CLASS WORK – SET 1 Solution for assignment 5. 1. The Lagrangian is L = 1 2 ˙ x 2 - 1 2 x 2 . Thus, the Euler-Lagrange equation becomes 0 = ∂L ∂x - d dt ∂L ˙ x = - x - ¨ x and the general solution is given by x ( t ) = A sin( t ) + B cos( t ). The integration constants are determined by the boundary conditions x (0) = B = 0 and x ( π/ 2) = A = 1 so that the exact path is x ( t ) = sin( t ) . For this path the action is given by S [ x ( t )] = 1 2 Z π/ 2 0 dt h cos 2 ( t ) - sin 2 ( t ) i = 1 2 Z π/ 2 0 dt cos(2 t ) = 0 . 2. For a linear path x ( t ) = a + b t subject to the boundary conditions: x (0) = a = 0 and x ( π/
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2) = b π 2 = 1 ⇒ b = 2 π so that this path is x ( t ) = 2 π t . For this path the action is given by S [ x ( t )] = 1 2 Z π/ 2 dt " ± 2 π ² 2-± 2 π ² 2 t 2 # = 2 π 2 Z π/ 2 dt ³ 1-t 2 ´ = 2 π 2 " π 2-1 3 ± 2 π ² 3 # = 1 π µ 1-π 2 12 ! = 1 π-π 12 = 0 . 0565 . . . . 3. Assuming that the previous result is in J · s , we ﬁnd for the large number S/ ¯ h ≈ 5 . 37 × 10 32 ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online