Derive the equation T = 2pi(l/g)^1/2.
That's period of motion.
hang mass m (will not matter) from string of length L
deflect angle A from up and down
Tension T in string
m g = T cos A
horizontal restoring force = -T sin A
so
F =-( m g/cosA) sin A = - m g tan A
if A is small
F = - m g A
distance x from centered
x = L sin A
for small angle
x = L A
if x = s sin (2 pi t/T)
v = s ( 2 pi/T) cos (2 pi t/T)
a = - s (2 pi/T)^2 sin (2 pi t/T)
or
a = - (2 pi /T)^2 x
now F = m a
-m g A = -m(2 pi/T)^2 L A
(2 pi/T)^2 = g/L
2 pi/T = (g/L)^.5
T = 2 pi (L/g)^.5
To derive the equation for the period of motion, T, we will start by considering the motion of an object undergoing simple harmonic motion (SHM). Simple harmonic motion refers to the motion of an object back and forth along a straight line, where its acceleration is directly proportional to its displacement and is always directed towards its equilibrium position.
Let's consider a mass, m, attached to a spring with a spring constant, k. When the mass is displaced a distance x from the equilibrium position, the force exerted by the spring follows Hooke's Law: F = -kx. Here, the negative sign indicates that the force is always directed towards the equilibrium position.
According to Newton's second law, F = ma, where a is the acceleration of the object. Therefore, we can rewrite Hooke's Law as ma = -kx.
We can rearrange this equation to obtain the acceleration of the object as a = (-k/m)x.
Now, let's consider the equation for simple harmonic motion:
a = (-k/m)x
Since acceleration is the second derivative of displacement with respect to time, we can rewrite the equation as:
d^2x/dt^2 = (-k/m)x
To solve this differential equation, let's assume that the solution is of the form:
x = A*cos(ωt)
where A is the amplitude of the motion and ω is the angular frequency.
Now, let's differentiate x twice with respect to time:
dx/dt = -A*ω*sin(ωt)
d^2x/dt^2 = -A*ω^2*cos(ωt)
Substituting these expressions into our differential equation, we have:
-A*ω^2*cos(ωt) = (-k/m)(A*cos(ωt))
The cos(ωt) terms cancel out, leaving us with:
ω^2 = k/m
Rearranging this equation, we get:
ω = sqrt(k/m)
The period of the motion, T, is the time it takes for one complete oscillation. It is given by T = 2π/ω.
Substituting ω = sqrt(k/m) into the equation for T, we have:
T = 2π/ sqrt(k/m)
Rationalizing the denominator, we obtain the final equation for the period of motion:
T = 2π * sqrt(m/k) = 2π * sqrt(l/g)
where l is the length of a pendulum and g is the acceleration due to gravity.
So, the derived equation for the period of motion is T = 2π(l/g)^(1/2).