Given a simple pendulum of length r and mass m, determine the oscillation period T if the perturbation angle θ is very small (i.e. θ < 10degrees).
How can I derive an equation for period using just that info?
newterra.chemeketa.edu/faculty/ejensen/docs/Pendulum.doc - Similar pages - Note this
The equation of motion can be written
d(theta)/dt + (g/r) theta = 0
if the approximation sin theta = theta
is made. g is the acceleration of gravity and r is the length of the pendulum. The mass cancels out in the derivation.
The general solution of this "simple harmonic motion" differential equation is
theta = A sin wt + B cos wt
where w = sqrt (g/r)
The period is 2 pi/w
To derive an equation for the period of a simple pendulum using the given information, you can use the principles of circular motion and the concept of restoring force for small angles.
Let's start by considering a simple pendulum with a mass 'm' attached to a string of length 'r'. When the pendulum is displaced from its equilibrium position by a small angle θ, it experiences a restoring force that tries to bring it back to its original position.
The restoring force acting on the mass can be approximated as a linear function of the angle for small angles. It follows Hooke's law, which states that the force is proportional to the displacement. Therefore, the restoring force can be given as:
F = -mg sin(θ)
where 'g' represents the acceleration due to gravity.
According to Newton's second law of motion, the net force acting on an object is equal to the product of its mass and acceleration. Here, the net force is the restoring force:
-mg sin(θ) = m a
We can represent angular acceleration as the second derivative of the angle with respect to time:
-mg sin(θ) = m r α
where α is the angular acceleration.
The angular acceleration, α, can be written as the second derivative of the angle, θ, with respect to time, t:
-mg sin(θ) = m r d²θ/dt²
Since the perturbation angle, θ, is very small, we can make the approximation sin(θ) ≈ θ (in radians). This approximation is valid because for small angles, the sine function and the angle become almost equal.
After substituting the approximation into the equation, we get:
-mg θ = m r d²θ/dt²
Now, rearranging the terms:
(g/r) dt² = -d²θ/θ
To solve this differential equation, we need to make the assumption that the solution has a form of θ = A sin(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase constant.
Taking the second derivative of θ with respect to time, we get:
d²θ/dt² = Aω² sin(ωt + φ)
We can substitute these values back into our differential equation:
(g/r) dt² = -(Aω² sin(ωt + φ))/ (A sin(ωt + φ))
We can cancel out the common terms:
(g/r) dt² = -ω²
Now, we can integrate both sides of the equation:
∫(g/r) dt² = ∫ -ω² dt
Integrating gives:
(g/r) t = -ωt + C
where C is the constant of integration.
Rearranging the terms:
ωt = -(g/r)t + C
Using the initial condition that at t = 0, θ = 0, we find that C = 0.
Therefore, the equation becomes:
ωt = -(g/r)t
From here, we can determine the angular frequency, ω, as:
ω = √(g/r)
The period, T, is the reciprocal of the angular frequency:
T = 2π/ω = 2π√(r/g)
So, the final equation for the period of a simple pendulum, for small angles, is:
T = 2π√(r/g)
That's how you can derive the equation for the period of a simple pendulum using the given information.