Question

A simple pendulum while oscillating rises to a maximum vertical height 5cm from its rest position when it reaches to its extreme position on one side.if mass of the Bob of the simple pendulum of 500g (g-10m/s2 find total energy of the simple pendulum at any instant while oscillating.find velocity of the Bob at its mean position

Answer 1

By using energy conservation,

(5cm=1/20m)

mghrel=1/2mv^2

g/20=v^2/2

v^2=1

v=1m/s

energy at any position is constant ,so energy of bob at mean position is=

1/2x1/2x1=1/4Joules

Answer 2

Answer:

Total energy of the simple pendulum at any instant while oscillating is equal to 0.25J. The velocity of the bob at its mean position will be equal to 1m/s.

Explanation:

We have given, the mass of the bob, m= 500g = 0.5Kg

The maximum vertical height of pendulum, h= 5cm = 0.05m

The acceleration due to gravity, g = 10m/s²

We assume that there is no loss of energy due to friction with air.

Total energy of simple pendulum = potential energy at extreme position

Total energy = P.E. = mgh

Total energy = (0.5)(10)(0.05)

Total energy = 0.25J

The velocity of the bob at its mean position :

At mean position, kinetic energy = potential energy

$\frac{1}{2}mv^2=mgh$

$\frac{1}{2}v^2=gh$

$v=\sqrt{2gh}$

Substitute the value of 'g' and 'h' in above equation:

$v=\sqrt{(2)(10ms^{-2})(0.05m)}$

$v=1ms^{-1}$

Therefore, the velocity of the bob at its mean position is equal to 1 m/s.