Asimple pendulum consists of a point mass, m, attached to the end of a massless string of length l. it is pulled out of its straight-down equilibrium position by a small angle theta and released so that it oscillates about the equilibrium position in simple harmonic motion with frequency f. what will happen to the frequency if the pendulum is pulled from equilibrium by 2 times theta instead?

The frequency will decrease by a factor of square root of 2 (ω = √(2 (g / L)).

Explanation:

A simple pendulum consists of a mass m hanging from a string of length L and fixed at a pivot point P. When displaced to an initial angle and released, the pendulum will swing back and forth with periodic motion. By applying Newton's second law for rotational systems, the equation of motion for the pendulum may be obtained

τ = I α ⇒ - mg sin(θ) L = mL² (d²θ/dt²)

where

The above expression can be rearranged as

(d²θ/dt²) + g / L (sin(θ)) = 0

If the amplitude of angular displacement is small enough that the small angle approximation () holds true, then the equation of motion reduces to the equation of simple harmonic motion

(d²θ/dt²) + g / L (θ) = 0

The simple harmonic solution is

θ(t) = θ₀ cos(ωt + Ф)

where

The frequency is expressed as

ω = √(g / L)

If the pendulum is pulled from equilibrium by 2 times theta, The simple harmonic solution will be

θ(t) = θ₀ cos(2 ωt + Ф)

and therefore, the frequency will be

ω = √(2 (g / L))

answer: (431,2)

step-by-step explanation: