Answer:The period of a pendulum, as measured by an observer moving at a relativistic velocity with respect to the pendulum, can be calculated using the concept of time dilation from special relativity.
The formula for time dilation in special relativity is given by:
Δt' = Δt / sqrt(1 - (v^2 / c^2))
where:
Δt' is the time interval measured by the moving observer
Δt is the time interval measured in the rest frame of the pendulum
v is the relative velocity between the pendulum and the moving observer
c is the speed of light in vacuum
In this case, the relative velocity between the pendulum and the moving observer is 0.95c, where c is the speed of light in vacuum (approximately 3 x 10^8 meters per second). Let's assume the period of the pendulum as measured in its rest frame is 3 seconds.
Plugging in the values into the formula:
Δt' = 3 / sqrt(1 - (0.95c)^2 / c^2)
Simplifying the expression:
Δt' = 3 / sqrt(1 - 0.95^2)
Using a calculator to evaluate the square root and simplify further, we get:
Δt' = 3 / sqrt(0.0975)
Δt' = 3 / 0.3125
Δt' ≈ 9.6 seconds
So, the period of the pendulum, as measured by an observer moving at a speed of 0.95c (95% of the speed of light) with respect to the pendulum, would be approximately 9.6 seconds. This demonstrates the concept of time dilation in special relativity, where the observed time interval changes due to relative motion at relativistic velocities.
Step-by-step explanation: