Simple+Harmonic+Motion+equations

=Mass on a spring=

Image from [|www.physics.usyd.edu.au]
 * F = ma, F= -kx

ma = - kx

m x" + kx = 0 { as a = x" }

x" +( k/m )x = 0 || x = x 0 cos ωt v = x' = −ωx 0 sin ωt a = v' = - ω 2 x 0 cos ωt = - ω 2 x { as x = x 0 cos ωt }

- ω 2 x + ( k/m )x = 0 ω = root (k/m )

T=2π/ω, T = 2π root( m/k ) || where, ω is the angular frequency of oscillation ( It is given by 2π / T or 2π f)

T is the period of oscillation (s)


 * //x// ** is the displacement (m)

= = || Mass - Spring: T = 2π root(m/k) ||  || Ex)x = 0, t = 0, v = -3, k = 10, m = 0.1
 * //x// ** 0 is the maximum displacement (m) ||
 * x" +( k/m )x = - ω 2 x + ( k/m )x = 0

ω = root( k/m ) = 10 rad/s T = 2π/ω = 0.628 s f = 1.6 Hz

X 0 cos ωt = 0 ωt = π/2, 3π/2, .. { as ωt =0 }

-3 = x 0 sin ωt x 0 =


 * Work done in stretching a spring = area of the triangle of F - x graph(Hooke's law) ||
 * W = 1/2 k ( x 0 2 − x 2 )W = E = E K = 1/2 mω 2 (x 0 2 − x 2 ) { as ω = root( k/m ), m ω 2 = k ) } ||
 * E T = 1/2 k x 0 2 = 1/2 mω 2 x 0 2 ||

Pendulum