tell me about acceleration of free fall by means of simple pendulum?

Answer 1

I presume you are interested in PE = KE; that is, potential energy equals kinetic energy, which is representative of the conservation of energy.

When a pendulum weight is h height above the lowest point of the swing of the pendulum, and the weight is motionless, its potential energy is PE = mgh; where m = mass of the weight (assuming a massless pendulum arm), g = 9.81 m/sec^2, and h = that height above lowest point in meters.

When that weight is let go, the conservation of energy says all that PE must be converted into KE = 1/2 mv^2; where v = the tangential velocity of the weight at the bottom of the swing. Thus we have mgh = 1/2 mv^2; and gh = 1/2 v^2; so that v^2 = 2gh or v = sqrt(2gh) if you are solving for the tangential velocity.

If you are solving for acceleration (g) you have g = v^2/(2h), This is one way to verify that g ~ 9.81 m/sec^2, but you have to be able to measure v at the bottom of the swing in experimentation. Generally, one just assumes g = 9.81 m/sec^2 at Earth's surface because that's usually close enough. But, as g does vary inversely with the square of the distance from the center of the Earth, your elevation on Earth could make a difference if you are looking for precision.

Answer 2

I'm not sure I can type in the appropriate symbols, but the period of oscillation for a pendulum is proportional to: 1/(square root of) L/g --- where L is the length of the pendulum and g is the acceleration of gravity. Perhaps the term 2 pi precedes the value for length of pendulum. Length of pendulum should be in feet and "g" expressed as 32 feet/sec/sec for an answer in seconds. The usual pendulum in a wall clock is timed for 1/2 second, but large "grandfather" clocks may swing only once per second.