Problem The period of a pendulum is measured to be 3.00 s in theinertial frame o

Question

Problem The period of a pendulum is measured to be 3.00 s in theinertial frame of the pendulum at Earth's surface. What is theperiod as measured by an observer moving at a speed of0.950c with respect to the pendulum?

Strategy Here, we're given the period of the clock as measured byan observer in the rest frame of the clock, so that's a proper timeinterval tp. We want to know how muchtime passes as measured by an observer in a frame moving relativeto the clock, which is t. Substitution into thetime dilation equation then solves the problem. SOLUTION Substitute the proper time and relative speedinto the time dilation equation. t = tp = 3.00 s = 9.61 s 1 -v2/c2 1 - (0.950c)2/c2 LEARN MORE Remarks The moving observer considers thependulum to be moving, and moving clocks are observed torun more slowly: while the pendulum oscillates once in 3 s for anobserver in the rest frame of the clock, it takes nearly 10 s tooscillate once according the moving observer.

Question Suppose a mass-spring system with the same period as thependulum is placed in the observer's spaceship. When the spaceshipis traveling at a speed of 0.95c relative to an observeron Earth, what is the period of the mass-spring system as measuredby the Earth observer?
SOLUTION Substitute the proper time and relative speedinto the time dilation equation. t = tp = 3.00 s = 9.61 s 1 -v2/c2 1 - (0.950c)2/c2 SOLUTION Substitute the proper time and relative speedinto the time dilation equation. t = tp = 3.00 s = 9.61 s 1 -v2/c2 1 - (0.950c)2/c2 Substitute the proper time and relative speedinto the time dilation equation. t = tp = 3.00 s = 9.61 s 1 -v2/c2 1 - (0.950c)2/c2 Substitute the proper time and relative speedinto the time dilation equation. t = tp = 3.00 s = 9.61 s 1 -v2/c2 1 - (0.950c)2/c2 t = tp = 3.00 s = 9.61 s 1 -v2/c2 1 - (0.950c)2/c2 LEARN MORE Remarks The moving observer considers thependulum to be moving, and moving clocks are observed torun more slowly: while the pendulum oscillates once in 3 s for anobserver in the rest frame of the clock, it takes nearly 10 s tooscillate once according the moving observer.

Question Suppose a mass-spring system with the same period as thependulum is placed in the observer's spaceship. When the spaceshipis traveling at a speed of 0.95c relative to an observeron Earth, what is the period of the mass-spring system as measuredby the Earth observer?
LEARN MORE Remarks The moving observer considers thependulum to be moving, and moving clocks are observed torun more slowly: while the pendulum oscillates once in 3 s for anobserver in the rest frame of the clock, it takes nearly 10 s tooscillate once according the moving observer.

Question Suppose a mass-spring system with the same period as thependulum is placed in the observer's spaceship. When the spaceshipis traveling at a speed of 0.95c relative to an observeron Earth, what is the period of the mass-spring system as measuredby the Earth observer?
t = tp = 3.00 s = 9.61 s 1 -v2/c2 1 - (0.950c)2/c2

Explanation / Answer

the spaceship is traveling at a speed of v = 0.95crelative to an observer on Earth the period of the mass-spring system as measured by the Earthobserver is t = [to/(1 -v2/c2)1/2] or to= t * (1 -v2/c2)1/2 t = 3.00 s or to= 3.00 * (1 -(0.95c)2/c2)1/2) or to= 0.93 s

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