What is the frequency of a simple pendulum 2.7 m long in the following situations?

(a) in a room that is stationary Hz

(b) in an elevator accelerating upward at a rate of 1.1 m/s2 Hz

(c) in free fall Hz

Answer: (a) 0.35Hz (b) 0.37Hz (c) 0

Explanation: frequency of a simple pendulum is expressed as;

f = 1/2π√ (a/L)

(a) in a room that is stationary, a, is the acceleration due to gravity = 9.8m/s²

f = 1/2π√ (g/L)

f = 1/2π (9.8/2)

f = 0.35Hz

(b) If the pendulum is in an elevator accelerating at a rate of 1.1m/s²

The total force will surely to both a and g

Therefore apparent acceleration = a+g = 1.1 + 9.8=10.9 m/s²

f = 1/2π √ (10.9/2)

f = 0.37Hz

(c) For a free fall, apparent acceleration is 0

f = 1/2π √ (0/L)

f = 0