Question 26.1: Pendulum Periods GOAL Apply the concept of time dilation. PR...
Pendulum Periods
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 t_{p}. We want to know how much time passes as measured by an observer in a frame moving relative to the clock, which is Δt. Substitution into Equation 26.2 then solves the problem.
{\Delta t={\cfrac{\Delta t_{p}}{\sqrt{1-v^{2}/c^{2}}}}=\gamma~\Delta t_{p}} [26.2]
Substitute the proper time and relative speed into Equation 26.2:
\Delta t={\frac{\Delta t_{p}}{\sqrt{1~-~v^{2}/c^{2}}}}={\frac{3.00\,{\mathrm{s}}}{\sqrt{1~-~{\frac{(0.950~c)^{2}}{c^{2}}}}}}=~9.61~s
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.