Question

the period of a pendulum is measured to be 3.00s in the refrence frame of the pendulum.what is the period when measured by an observer moving at a speed of 0.960c relative to the pendulum?

Answer 1

Δt = Δτ / (1 - β^2)1/2

Δt is the time observed

Δτ is the proper time (3.00 s)

β = v / c = .96

Δt = 3.00 / (1 - .96^2)^1/2 = 10.7 sec