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Answer :
The period of small oscillations for the simple pendulum in the elevator accelerating upward at 5.00 m/s² is approximately 6.40 seconds.
To find the period of small oscillations for a simple pendulum located in an elevator accelerating upward at 5.00 m/s², we can use the equation for the period of a simple pendulum:
T = 2π * √(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, the length of the pendulum (L) is given as 5.00 m.
The effective acceleration due to gravity in the elevator can be calculated by subtracting the acceleration of the elevator from the actual acceleration due to gravity. Since the elevator is accelerating upward, the effective acceleration due to gravity will be reduced.
g_effective = g - a
where g is the acceleration due to gravity (approximately 9.8 m/s²) and a is the acceleration of the elevator (5.00 m/s²).
Substituting the values into the equation for the period of a simple pendulum:
T = 2π * √(L/g_effective)
T = 2π * √(5.00 / (9.8 - 5.00))
Calculating the period T:
T ≈ 2π * √(5.00 / 4.8)
T ≈ 2π * √(1.0417)
T ≈ 2π * 1.020
Using the approximation π ≈ 3.14:
T ≈ 6.40 s
Therefore, the period of small oscillations for the simple pendulum in the elevator accelerating upward at 5.00 m/s² is approximately 6.40 seconds.
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