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Unformatted text preview: itial equilibrium point) is x, as usual. This displacement
motion is considered sinusoidal: η= η0cosωt. Discussion about the forces on the bob (gravity and tension) follows,
concluding that for small motions in which the string remains nearly vertical, the tension is just mg, and the restoring
force in the x direction (which arises entirely from tension) is mgsinθ. The only difference between this and what
you should have seen in an introductory mechanics course is that θ is toward a moving, rather than fixed, suspension
point so that sin x l sinθ, mx bx mg x
l . Then, Newton’s Law gives mx bx mg sin , or by putting in the definition of mg 0 cost . Dividing by m and using ω02=g/l and γ=b/m, the DE becomes
l 2
2 x 0 x 00 cos t , which is exactly analogous to our previous driven motion DE, with ω02η0 replacing x F0/m (38:00). If we make this substitution in the solutions, the solutions...
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This note was uploaded on 02/18/2011 for the course PHYS 320 taught by Professor Martinconner during the Spring '10 term at Open Uni..
 Spring '10
 martinconner
 Force, Mass

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