Time Dilation Practice Problems

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Time Dilation Practice Problems

Lecture number:
37
Pages:
2
Type:
Lecture Note
School:
The University of Vermont
Course:
Phys 012 - Elementary Physics
Edition:
1
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Unformatted text preview:

Lecture 37 Outline of Last Lecture I. Special Relativity a. Newtonian mechanics fails to describe the motion of very fast moving objects (close to the speed of light). b. Event: a physical happening that occurs at a certain place at a certain time i. An observer need three spatial and one temporal dimension (x, y, z, t) c. For special relativity, we need to observe from inertial frames of reference (observational reference frames in which Newton’s Laws of motion are valid). i. Accelerated frame of reference is not inertial. d. Postulates of Special Relativity i. Relativity postulate: the laws of physics are the same in every inertial reference frame ii. Speed of light postulate: the speed of light in a vacuum “c” as measured in any inertial frame of reference always has the same value regardless of how fast the source of light is moving e. Interpretation i. Any inertial reference frame is as good as any other one. There is no absolute velocity. ii. In relativistic mechanics, there is no such thing as absolute length or an absolute time interval. iii. Two events that are observed as simultaneous in one frame of reference are, in general, not simultaneous in another frame moving relative to the first. II. Time Dilation a. Ex) person in a moving train (at velocity v) holding a flashlight jumping up and down to a height of h i. Time for light to go up and down for person on train = proper time = Δtp = 2d/c ii. Time for light to go up and down for stationary person watching train = Δt = (2d)/(c√(1-[v2/c2])) = Δtp/√(1-[v2/c2]) Outline of Current Lecture III. Time Dilation a. Problem: A pendulum has a period of 3 seconds in the reference frame of the pendulum. What is the period of the pendulum within the reference frame of an observer moving at a constant velocity of 0.95 c? i. γ = 1/√(1-[(0.95c)2/c2]) = 1/√(1-[0.952]) = 3.2 ii. Δt = Δt0γ = (3)(3.2) = 9.6 seconds Physics 012 1st Edition

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