Alarge, simple pendulum is on display in the lobby of the united nations building. if the pendulum is 18.5 m in length, what is the least amount of time it takes for the bob to swing from a position of maximum displacement to the equilibrium position of the pendulum? (assume that the acceleration of gravity is g = 9.81 m/s2 at the un building.)

     t = 1,144 s

Explanation:

The simple pendulum consists of an inextensible string with a mass at the tip, the angular velocity of this is

     w = √( L / g)

The angular velocity is related to the frequency and period

     w = 2π f

      f = 1 / T

     w = 2π / T

Let's replace

     2π / T = √ (L / g)

     T = 2π √ (g / L)

Let's calculate

     T = 2π √ (9.81 / 18.5)

     T = 4,576 s

The definition of period in the time it takes the ball to come and go to a given point (a revolution) in our case we go from the end to the middle point that is a quarter of the path

     t = T / 4

     t = 4,576 / 4

     t = 1,144 s

show the diagram then i will answer

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer