Conclusion: Equation of SHM d-x +02x=0 dt Solution:!x=Acos(t+φ)
Conclusion: Equation of SHM x 0 dt d x 2 2 2 + = Solution: x = Acos(t + )
Example: A particle dropped down a hole that extends from one side of the earth, through its center, to the other side. Prove that the motion is shm and find the period. Solution F= GmM Fr、R g 、mP公 M 4 r≤R,M M R M GmM d r r= R dt d-x dr /GM +02x=0 r=0 dt R
Solution: 2 ' g r GmM F = − r R, M R r r 3 4 R 3 4 M M 3 3 3 3 ' = − = − R O Fg r M r R GmM Fg 3 = − 2 2 dt d r = m r 0 R GM dt d r 2 3 2 + = 2 Example: A particle dropped down a hole that extends from one side of the earth, through its center, to the other side. Prove that the motion is SHM and find the period. x 0 dt d x 2 2 2 + =
EXample: An astronaut on a body mass measuring device(BMMD), designed for use on orbiting space vehicles, its purpose is to allow astronauts to measure their mass in the 'weight-less' condition in earth orbit The BMMD is a spring mounted chair. if M is mass of astronaut and m effective mass of the bMd, which also oscillate, show that M=(4n2)-m
M = k T − m 2 2 ) 4 ( Example: An astronaut on a body mass measuring device (BMMD), designed for use on orbiting space vehicles, its purpose is to allow astronauts to measure their mass in the ‘weight-less’ condition in earth orbit. The BMMD is a spring mounted chair. if M is mass of astronaut and m effective mass of the BMMD, which also oscillate, show that
+M Ea16-5(u=2x/T) k k M
Example: the system is as follow, prove the block will oscillate in SHM Solution: mg-T=ma 1) (Ti-12)R=la(2) 72=k (3) a=aR We have dy ky dt m+<0
0 2 2 2 = + + R m I ky dt We have d y Example: the system is as follow, prove the block will oscillate in SHM (4) (3) ( ) (2) (1) 2 1 2 1 a R T k x T T R I m g T m a = = − = − = Solution: o y