No book handy so will derive pendulum
let theta be angle from vertical then
(1/2) m v^2 max = m g h = m g L (1- (1-theta^2/2)) maximum
(1/2) v^2 = g L(theta^2/2)
but v = L d theta/dt
L^2 d (theta/dt)^2 = g L theta^2
let Theta = A sin 2 pi t/T
max = A
then d theta/dt = A (2 pi/T) cos 2pit/T
max = A(2 pi /T)
so
L (2 pi/T)^2 = g
2 pi/T = sqrt(g/L)
T = 2 pi sqrt(L/g)
so now the problem
T =85/25 = 3.4 seconds per swing
2 pi/T = 1.85
T^2 = 3.42 = L/9.81
L = 33.5 m
A science museum has asked you to design a simple pendulum that will make 25.0 complete swings in 85.0 {\rm s}.
How long should the pendulum be? in meters
3 answers
T =85/25 = 3.4 seconds per swing
T/2pi = .541 = sqrt(L/9.8)
L/9.8 = .293
L = 2.87 m
T/2pi = .541 = sqrt(L/9.8)
L/9.8 = .293
L = 2.87 m
A science museum has asked you to design a simple pendulum that will make 54 complete swings in 73 s. What length should you specify for this pendulum?