Topic: Oscillations

Question

Consider a simple pendulum of mass 0.703 kg and length 213 cm. If the pendulum is released from rest at an angle of 17.5^°, what is the tangential velocity of the mass at t = 4.43 seconds?

Answer 1

Let

m = 0.703 kg be the mass of the pendulum,

L = 213 cm = 2.13 m be the length of the pendulum,

g = 9.81 m/s² be gravitational acceleration,

T = 2π√(L/g) be the period of the pendulum,

θ = 17.5° be the angle at the time of release,

v(τ) be the tangential velocity at time τ since release,

V (unknown) be the maximum tangential velocity.

The tangential velocity undergoes harmonic oscillation with amplitude V

and period T.

That means

v(τ) = Vsin(2πτ/T) = Vsin(τ√(g/L))

In order to compute the desired v(t) we need to compute V.

We can do that from conservation of energy.

The pendulum has maximum velocity at its lowest point, having kinetic energy

mV²/2

It obtained that energy from the pendulum's potential energy at the point of release.

When released, the pendulum is higher than its lowest point by the amount

L(1-cosθ)

Therefore its potential energy is

mgL(1-cosθ)

By conservation of energy

mV²/2 = mgL(1-cosθ)

V = √(2gL(1-cosθ))

Thus the tangential velocity is

v(τ) = √(2gL(1-cosθ)) sin(τ√(g/L))

Substituting actual numbers

v(4.43) = √(2×9.81×2.13×(1-cos(17.5°)) × sin(4.43×√(9.81/2.13)) = 0.23 m/s

Notice that the result is independent of the mass of the pendulum.

That is a general property of pendula -- their behavior is independent of their mass.