a – x

a = – 2x

Negative sign:Direction of a is opposite to x

O-xoxo

a a

200grams

VibratingTuning fork

A weight ona spring

A boy on a swing

F – x

F = – kx

O-xoxo

Point of equilibrium:a = 0F = 0

a

x

Measured from the point of equilibrium

Amplitude := xo

=Maximum distance

F = - kx

ma = - kx

a = km

- x a = - 2xcompare

2 =km

= angular frequency

= 2f = 2 T

F

x0 xo-xo

a

x0 xo-xo

2xo

-2xo

F = -kx a = - 2x

a = - 2x

d2xdt2 = - 2x

x = xo sin (t + )General solution

dxdt

= xo cos (t + )v =

d2xdt2

= -2xo sin (t + )a =

= - 2x

x = xo sin (t + )

phase

Phase constant

x = xo sin (t + ) depends on the initial condition

when t = 0

O-xoxo

At t = 0, x = 0

0 = xo sin ((0) + )

sin (0 + ) = 0

sin = 0

= 0

Therefore :

x = xo sin t

v = xo cos t

a = - 2xo sin t

= - 2x

x

t0

x = xo sin txo

-xoT 2T

v

T0-xo

xov = xo cos t

t2T

a

t0

a = -2xo sin t2xo

-2xoT 2T

t=0,v=xo

t=0,a=0

phase = t + /2

phase = t

x

t0

xo

-xoT 2T

v

T0-xo

xo

t2T

a

t0

2xo

-2xoT 2T

x = xo sin t

v = xo cos t

To compare the phase :They must all be expressed

as same function (either as sine

or cosine function)

Phase difference :=(t+ ) - t

2 =

2rad

= xo sin (t + /2)

v leading x

phase = t +

x

t0

xo

-xoT 2T

v

T0-xo

xo

t2T

a

t0

2xo

-2xoT 2T

x = xo sin tphase = t

v = xo cos t phase = t + /2

a = -2xo sin t = 2xo sin (t + )

Phase difference :=(t+) - t = rad

a leading xa and x are antiphase

x

t0

xo

-xoT 2T

v

T0-xo

xo

t2T

a

t0

2xo

-2xoT 2T

-xo xo0

t=0

x

x = xo sin (t + )General solution

Substitute :t=0 , x = xo

to find xo = xo sin ((0) + )

sin = 1

=2

Therefore , equation of x is: x = xo sin (t + /2)

t

= xo cos t

The motion of a body in simple harmonic motion is described by the equation :

x = 4.0 cos ( 2t + )3

Where x is in metres, t is in seconds and (2t + ) is in radians.

3a)What is

i) the displacement ii) the velocity iii) the acceleration and iv) the phaseWhen t = 2.0 s ?

b) Find i) the frequency ii) the periodof the motion.

x = 4.0 cos ( 2t + )3a) i)

Displacement = ?

t = 2.0 s ; x = 4.0 cos ( 2(2) + )3

= 2.0 m

ii) velocity = ?

v = dxdt

= -2(4.0) sin ( 2t + )3

v = -8.0 sin (4 + )t = 2.0 s ;3

= -21.8 ms-1

iii) acceleration = ?

a = dvdt

= -16.02 cos ( 2t + )3

t = 2.0 s ; a = -16.02 cos ( 2(2) + )3

= -79.0 ms-2

iv) Phase = ?

x = 4.0 cos ( 2t + )3

Phase = 2t + 3

t = 2.0 s ; Phase = 2(2) + 3

= 13.6 rad

The motion of a body in simple harmonic motion is described by the equation :

x = 4.0 cos ( 2t + )3

Where x is in metres, t is in seconds and (2t + ) is in radians.

3a)What is

i) the displacement ii) the velocity iii) the acceleration and iv) the phaseWhen t = 2.0 s ?

b) Find i) the frequency ii) the periodof the motion.

x = 4.0 cos ( 2t + )3

b) i) frequency = ?

x = xo cos (t + )

General solution :

= 2

2f = 2

f = 1.0 Hz

ii) Period = ?

T = 1 f

= 1.0 s

Figures (1) and (2) are the displacement-time graph and acceleration-time graph respectively of a body in simple harmonic motion.

x,m3

-3

t,sT0

(1) a,ms-2

12

-12

t,sT0

(2)

What is the frequency of the motion? Write an expression to represent the variation of displacement x with time t.

x,m3

-3

t,sT0

(1) a,ms-2

12

-12

t,sT0

(2)

frequency = ?

xo = 3 2xo = 12

2(3) = 12substitute

= 2 rad s-1

2f = 2f = 0.318 Hz

expression = ?

x = xo sin (t + )General solution

xo = 3 , = 2 ; x = 3 sin (2t + )

At t = 0, x = 3 3 = 3 sin (2(0) + )

sin = 1

= 2

rad

Therefore, the expression is

x = 3 sin (2t + )2

x = 3 cos 2t

a = dv = -2x dt

dv x dx = -2xdx dt

v dvdx

= -2x

v dv = -2x dx

v2

2= -2x2

2+ C

When x = xo, v = 0

0 = -2xo2

2+ C

C = 2xo2

2Hence ;

v2

2= -2x2

2 2xo

2 2

+

v2 = -2x2 + 2xo2

v = xo2 – x2

v

x0

-xo O xo

xo

-xo

xo-xo

v = xo2 – x2

A particle performs simple harmonic motion with an amplitude of 0.50 cm and a period of 3.0 s. Calculate

a)The angular frequency of the SHMb)The velocities of the particle when its

displacement is 0.20 cm. Explain why there are two velocities.

a) = 2f

= 2 T

= 2 3

= 2.09 rad s-1

b) v = xo2 – x2

= (2.09) 0.52 – 0.22

= 0.958 cm s-1

v = +0.958 cms-1 when the particle is moving in the positive direction and v = –0.958 cms-1 when the particle passes through the same point in the opposite direction on the return journey.

Total energy = E = K + U

At point, x = 0 :

K = ½ mv2

= ½ m2(xo2 – x2)

K = ½ m2(xo2 – 0)

= ½ m2xo2

At point, x = 0 :

U = 0

= maximum

= ½ m2xo2 + 0

= ½ m2xo2

constant

At point, x = xo :

K = ½ m2(xo2–xo

2) = 0

= minimum

-xo xo0 x

-xo xo0 x

At any displacement,x

E = K + U

½ m2xo2 = ½ m2(xo

2 – x2) + U

= ½ m2xo2 – ½ m2x2 + U

U = ½ m2x2

Kmax = ½ m2xo2 U = ½ m2x2Kmin = 0; ;

0

E

x-xo xo

Etotal=½ m2xo2

-xo xo0 x

K=½ m2(xo2 – x2)

U = ½ m2x2

-xo xo0 x

U = ½ m2x2

F = – dU dx

= – ½ (2) m2x

= – m2x

constant

Therefore: F x

0

F

x-xo xo

mxo

-mxo

-xo xo0 x

x = xo sin (t + )General solution

0 = xo sin ((0) + )At t = 0 , x = 0

0 = sin

= 0

x = xo sin t Therefore :The equation is

-xo xo0 x

x = xo sin t v = dx dt

= xo cos t

K = ½ mv2 = ½ m2xo2cos2 t

U = ½ m2x2 = ½ m2xo2 sin2 t

Etotal=½ m2xo2

T 2T

K

0

E

tU