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

(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

answer; i believe the correct answer is (40,000 j) good luck

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explanation: