A pendulum is released from rest at a displacement ...?

Question

A pendulum is released from rest at a displacement of .46 meters from its equilibrium position. It is stopped abruptly and uniformly at its equilibrium position and it is observed that a loose bit of metal slides without resistance off the pendulum and falls to the floor .86 meters below.
•If the projectile started off with a velocity in just its horizontal direction, and if travels .5 meters in the horizontal direction during its fall, what was the velocity of the pendulum at equilibrium?
•What is the period of motion of the pendulum?

Answer 1

Ok, here you need to make a graphic of the pendule, at the first moment and at the moment it reached it's equilibrium position.

You will find out, that it's falling from certain altitude "H" and when it reached the equilibrium position, you need to graph the movement of the bit of metal that slides off the pendulum.

So here we go :

The bit of metal has a horizontal velocity and no vertical velocity. That bit of metal falls 0.86 meters bellow and move 0.5 meters to the right.

0.5 = V*t >>> V = horizontal velocity ( velocity of the pendulum)

0.86 = g*t^2 / 2 >>> no vertical velocity

g = 9.8 m/s^2

t = 0.41 s

0.5 = V*t, >>> V = 1.22 m/s >>> this is the velocity of the pendulum.

Now, we need to use the conservation of energy, when the pendulum was falling until it reached it's equilibrium :

m*g*H = m*v^2 / 2

9.8*H = 1.22^2 / 2

H = 0.076 meters

Now, you will find this, a rectangle triangle with :

L = lenght of the pendulum

L - H = another lenght

0.3 >>> distance of the pendulum from the point it was falling to the equilibrium :

Let's use Pitagoras :

L^2 = 0.3^2 + (L - 0.076)^2

L = 0.63 (m)

the period of a pendulum : 2*pi*sqrt(L / g)

period = 2*3.1416*sqrt(0.63 / 9.8)

period = 1.59 (s)