Simple Harmonic Motion

Equations

f = 1/t

x(t) = A cos (wt)

v(t) = -wAsin(wt)

a(t) = -w² A cos(wt)

w = 2 pi f = 2 pi / t

Tsp = 2 pi root (m / k)

Tp = 2 pi root (l / g)

Restoring force (and acceleration) always points towards equilibrium position

Simple Harmonic Motion needs a linear restoring force present

Simple harmonic motion is a sinusoidal function

The highest or lowest point occurs at ½ a period

position time graph = cosine function

velocity time graph = negative sine function

velocity is at a max when displacement is at 0 (equilibrium), aka 0 spring potential

velocity is 0 at A

acceleration time graph = negative cosine function

acceleration is at max when velocity is at 0

amplitude - maximum displacement from equilibrium position

NEVER USE KINEMATICS to solve velocity, acceleration, energy are always changing

Solving Spring Motion - use spring energy and kinetic energy, no gravitational potential energy

Solving Pendulum Motion - started with f = mgsin0, if the angle is small enough you can change it to mg0, 0 = S/L (arclength or displacement / radius), leading to the equation mg S/L

Comparing the equation F = kx and the equation F = mg S/L, you can see that k = mg / L (all those values are constant) and x = S

in the equations about (wt) stays constant

A = max displacement

VMAX = -wA

AMAX = -w²A

When looking at graphs always pay attention to the initial value (either positive negative or zero) to see if it makes sense for the situation

mass increases period

Energy in SHM

Kinetic energy is at maximum at equilibrium

Potential energy is at a maximum at the amplitudes

Kinetic and potential are equal towards the amplitude NOT AT A/2

Kmax = Umax = Etotal

to double Etotal double either quantity directly proportional to Kinetic energy (mass) or potential energy (spring constant)

deriving T using energy on a spring

Kmax = Umax

½ mv² = ½ k x²

½ m ((2 pi A)/T)² = ½ k A² (using velocity equation from UCM and A as the displacement)

½ and A cancel out

Solve for T

T = 2 pi root (m / k)

A does not affect period on a system with a mass on a spring

deriving T using forces on a pendulum

F = mg sin 0 (x component of weight)

F = mg 0 (small angle)

0 = S/L

F = mg S / L

Tp = 2 pi root (m / k)

k = mg / l

Tp = 2 pi root (m l / mg)

Tp = 2 pi root (l / g)

Tension at equilibrium mirrors uniform circular motion, where the acceleration points upwards making it greater then then weight force

To find L at an angle us L - L cos theta

Dont always assume newtons 3rd laws when asked about forces and acceleration