A ground observer measures the period of a pendulum moving as a part of an inertial frame of reference to be 2.30 s as the inertial frame moves by at a velocity of 0.800c.What would the observed period be of the same pendulum if its inertial frame were at rest with respect to the observer?
A) 3.03 s
B) 1.38 s
C) 1.84 s
D) 4.25 s

Question

Quizplus · Accepted Answer

According to time dilation, the time interval between two events in a moving frame of reference is longer than that in a rest frame of reference. The observed period in the ground observer's frame will be longer than the actual period in the pendulum's rest frame. Using the time dilation equation:

Δt' = Δt/γ

where Δt is the time interval in the rest frame, Δt' is the time interval in the moving frame, and γ is the Lorentz factor:

γ = 1/√(1 - v^2/c^2)

where v is the velocity of the moving frame and c is the speed of light.

In this problem, Δt' = 2.30 s and v = 0.800c. Solving for γ:

γ = 1/√(1 - (0.800c)^2/c^2) = 1.67

Substituting into the time dilation equation:

Δt = Δt'/γ = 2.30 s/1.67 = 1.38 s

Therefore, the observed period of the pendulum in the rest frame of reference is 1.38 s, which corresponds to choice B.

A Ground Observer Measures the Period of a Pendulum Moving