Question

a. Assume that all of the pendulum’s mass M is contained in the “bob” at the end of the pendulum rod. Write Newton’s 2nd Law for the force balance along the circular arc traced out by the pendulum bob as it swings back and forth. [Hint: You can relate the position of the pendulum bob along its arc to the angle ! formed by the pendulum rod and the vertical line when the pendulum bob is at the “base” of the arc. It may help to draw this.]

b. Using the “small angle approximation” sin(theta) ≈ theta(t) and assuming that the pendulum bob is released from rest at an angle theta(zero), show that theta(zero) = (theta(zero))cos (omega t) solves the force balance equation in (a) for a particular value of omega. What value is it?

c.  The period of the pendulum corresponds to the time it takes the pendulum to oscillate through one cycle, i.e., from its initial position and back. Use the result in b to find the period. You should find the period to be a numerical factor times our estimate of the period from dimensional considerations.

Answer 1

Newton's second law states that the acceleration is deirectly propotional to the net force,or

F=ma,

where m=mass, a constant

Consider the equilibrium position is given by the angle So the force on the mass m along the horizontal axis is equal to ma at this position. The angle   is determined by the equations.

T sin   = ma, T cos   = mg

When the pendulum is displaced by a small amount , it will perform simple harmonic motion around the equilibrium position. Its equation of motion is

m x = –T sin(\theta + )

where x is the distance from the vertical OA.

For small angle \theta, sin(\theta + ) ≈ \theta cos + sin = θ mg /T+ ma/ T

Thus

m = – ma – mg\theta.

\theta and l are related geometrically as

x = l sin(\\theta + \phi) ≈ l \theta cos \phi + l sin \phi

or = l cos \\phi \ddot{\theta}

Hence l cos \phi \ddotθ = –a –gθ or

\ddotθ = – g /(l a cos \phi )*(\theta+a/g)

If we make the following substitution ψ = θ + a/ g

we get \ddotψ = – g /l cos φ ψ = – ω²ψ

with time period of oscillation T = 2\pi /ω = 2 \pi(lcos φ /g)^3/2

. Now, cos φ = mg /T = mg/\sqrt( m²a² +m²g²) + = g/ (a² g²)

So, T = 2\pi*(l/(\sqrt a²+g ²)^3/2