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Key Equations

Relationship between frequency and period f=1T
Position in SHM withϕ=0.00 x(t) =Acos(ωt )
General position in SHM x(t) =Acos(ωt+ϕ )
General velocity in SHM v(t) =Aωsin(ωt+ϕ )
General acceleration in SHM a(t) =Aω2cos(ωt+ϕ )
Maximum displacement (amplitude) of SHM xmax=A
Maximum velocity of SHM |vmax | =Aω
Maximum acceleration of SHM |amax | =Aω2
Angular frequency of a mass-spring system in SHM ω=km
Period of a mass-spring system in SHM T=2πmk
Frequency of a mass-spring system in SHM f=12π km
Energy in a mass-spring system in SHM ETotal=12 kx2+12 mv2=12 kA2
The velocity of the mass in a spring-mass
system in SHM
v=±km (A2x2 )
The x-component of the radius of a rotating disk x(t) =Acos(ωt+ϕ )
The x-component of the velocity of the edge of a rotating disk v(t) =vmaxsin(ωt+ϕ )
The x-component of the acceleration of the
edge of a rotating disk
a(t) =amaxcos(ωt+ϕ )
Force equation for a simple pendulum d2θ dt2 =gL θ
Angular frequency for a simple pendulum ω=gL
Period of a simple pendulum T=2πLg
Angular frequency of a physical pendulum ω=mgL I
Period of a physical pendulum T=2πImgL
Period of a torsional pendulum T=2πIκ
Newton’s second law for harmonic motion md2x dt2 +bdx dt +kx=0
Solution for underdamped harmonic motion x(t) =A0eb2m tcos(ωt+ϕ )
Natural angular frequency of a
mass-spring system
ω0=km
Angular frequency of underdamped
harmonic motion
ω=ω02(b2m ) 2
Newton’s second law for forced,
damped oscillation
kxbdx dt +Fosin(ωt ) =md2x dt2
Solution to Newton’s second law for forced,
damped oscillations
x(t) =Acos(ωt+ϕ )
Amplitude of system undergoing forced,
damped oscillations
A=Fo m2(ω2ωo2 ) 2+b2ω2
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