Question Detail
Question
A pendulum has a length of 50 cm. What is its period of oscillation if it is suspended in an elevator rising at a constant acceleration of 2 m/s²?
Correct Answer
Option C is the correct answer.
Detailed Explanation
From Worked Example 17 (v) (page 33), when an elevator is rising with constant acceleration 'a', the effective gravitational acceleration becomes (g+a). The period formula for a simple pendulum is T = 2π√(L/(g+a)). Given L = 0.50 m, g = 9.81 m/s² (assuming standard g as not specified otherwise for 17(v), though the example uses 9.8 for other parts), and a = 2 m/s². So, T = 2π√(0.50 m / (9.81 m/s² + 2 m/s²)) = 2π√(0.50 / 11.81) ≈ 1.29 s. The example uses 9.8m/s^2 for 'g' in 17(v).
Hint
When an elevator accelerates upwards, the effective gravitational acceleration (g_eff) experienced by a pendulum changes. Use the modified period formula: T = 2π√(L/g_eff).