Final answer:
When the elevator is accelerating downward at 4.22 m/s², the effective acceleration due to gravity is 5.58 m/s², and thus the period of the 4.99 m long pendulum is approximately 5.94 seconds.
Step-by-step explanation:
To find the period of simple harmonic motion for a pendulum in an elevator accelerating downward, we can modify the standard equation for the period of a pendulum based on the fact that the effective acceleration due to gravity on the pendulum is reduced by the acceleration of the elevator. The standard period of a simple pendulum where g is the local acceleration due to gravity is given by:
T = 2π√(L/g)
Where L is the length of the pendulum and T is the period. In the elevator accelerating downward at 4.22 m/s², the effective acceleration due to gravity, g', would be:
g' = g - a = 9.80 m/s² - 4.22 m/s² = 5.58 m/s²
The modified equation for the period T' is then:
T' = 2π√(L/g')
Plugging in our values:
T' = 2π√(4.99 m / 5.58 m/s²)
Calculating this yields:
T' ≈ 2π√(0.8946) ≈ 2π(0.946) ≈ 5.94 s
Therefore, the period of the pendulum in the accelerating elevator to the nearest hundredth is approximately 5.94 seconds.