GOAL Apply the concept of time dilation. PROBLEM The period of a pendulum is mea
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Question
GOAL Apply the concept of time dilation. PROBLEM The period of a pendulum is measured to be 3.00 s in the inertial frame of the pendulum at Earth's surface. What is the period as measured by an observer moving at a speed of 0.950c with respect to the pendulum? STRATEGY Here, we're given the period of the clock as measured by an observer in the rest frame of the clock, so that's a proper time interval delta tp. We want to know how much time passes as measured by an observer in a frame moving relative to the clock, which is delta t. Substitution into the time dilation equation then solves the problem. SOLUTION Substitute the proper time and relative speed into the time dilation equation. LEARN MORE REMARKS The moving observer considers the pendulum to be moving, and moving clocks are observed to run more slowly: while the pendulum oscillates once in 3 s for an observer in the rest frame of the clock, it takes nearly 10 s to oscillate once according the moving observer. QUESTION Suppose a mass-spring system with the same period as the pendulum is placed in the observer's spaceship. When the spaceship is traveling at a speed of 0.95c relative to an observer on Earth, what is the period of the mass-spring system as measured by the Earth observer? PRACTICE IT Use the worked example above to help you solve this problem. The period of a pendulum is measured to be 3.86 s in the inertial frame of the pendulum. What is the period as measured by an observer moving at a speed of 0.945c with respect to the pendulum? s EXERCISE Use the values from PRACTICE IT to help you work this exercise. What is the period of the pendulum as measured by a third observer moving at 0.867c? delta t = SExplanation / Answer
As we know that
t' = t/(1 - v^2/c^2)
Therefore
In MASS SPRING System Case
t' = 3/sqrt(1 - 0.95^2)
= 9.607 sec
Similarlily in Pendulum Case
t' = t/(1 - v^2/c^2)
t' = 3.86/sqrt(1 - 0.945^2)
= 11.80 sec
t' = t/(1 - v^2/c^2)
t' = 3.86/sqrt(1 - 0.867^2)
= 7.746 sec
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