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A large, 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.)

User Sriram C G
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1 Answer

6 votes

Answer:

t = 1,144 s

Step-by-step 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

User Munyengm
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