Question

Show that a simple pendulum bob which has been pulled aside from its equilibrium position through an angle q and then released will pass through the equilibrium position with speed v= root [2gL(1-cos q)] where L is the length of the pendulum.

Answer 1

length of pendulum = L
velocity = v
height at equilibrium position= h = L- Lcosq = L(1-cosq)

at equilibrium, KE= PE
1/2 mv² = mg L(1-cosq)
v² = 2gL (1-cosq)
v= √2gL (1-cosq)

Answer 2

v = √2gL (1 - cosq)

Explanation:

When the bob of a simple pendulum is pulled to one side from its equilibrium position at an angle q and then released, the pendulum will start swinging. The bob will pass through the equilibrium position at a speed and the length of the pendulum is a constant.

Given length of pendulum is L and the velocity of the bob is v.

At equilibrium position, the height of the bob can be found as h = L - Lcosq = L (1 - cosq)

Also at equilibrium, the kinetic energy will be equal to the potential energy.

KE = PE

1/2 mv² = mg L(1-cosq) where m is mass and g is gravity.

Since m is on both sides of the equation, we can cancel m.

So, we get v² = 2gL (1 - cosq)

                v = √2gL (1 - cosq)