a. A mass moving up and down at the end of a spring or elastic string.

b. The depth of water at a point at sea varying with the tide.

c. Small oscillations of a pendulum.O is the

centre of motion

A and A_{1}are theextremepositions

P is the position at time t

The distance OA = OA_{1}= a is called theamplitudeof the motion

x = acos(nt + α) is a mathematical model for the motion.

Note: We could also use asin(nt + α)

This means in a simple harmonic motion that the acceleration is directed towards O and is proportional to the distance from O, thus it can be produced by a force with this property.

**Terminology**

**1.**The

**angle**(nt + α) is always measured in radians and is called the

**phase**of the motion.

(Remember that π radians = 180

^{o)}

When t = 0 the phase is α,called the **initial phase.**

If particle starts at A then α = 0

Note: Think what α would be if the particle starts at O or A_{1}

**2. **The **period**, **T**, is the time to complete one oscillation.

i.e. The** **phase increases by 2π

Therefore, n(t + T) + α = nt + α + 2π

Therefore,
T = 2π / n |

**3. **The **frequency, f, **is the number of oscillations in one second.

Therefore,
f = 1/T = n/ 2π |

Also note that v^{2} = n^{2}x^{2} = a^{2}n^{2} (sin^{2}(nt + α) + cos^{2}(nt + α))

Therefore, v^{2} + n^{2}x^{2} = a^{2}n^{2}

Therefore,
v^{2} = n^{2}(a^{2} − x^{2}) |

v = 0 when x = ± a

i.e. At A_{1} and A as we would expect

Also **v _{MAX} = ±an** at the centre O

**Examples**

**Example 1** A particle doing simple harmonic motion has a period of 6 seconds and an amplitude of 6 cm. Find its acceleration and velocity when it is 4 cm to the left of the centre.

T = 6 = 2π/n

Therefore, n = π/3

Use v

^{2}= n^{2}(a^{2}− x^{2}) with a = 6 and x = -4Therefore, v

^{2}= π^{2}/9(36 − 16) = 20π^{2}/9Therefore,

v = 4.68 cm/sAcceleration = -n

^{2}x= -π

^{2}/9.-4 (-4 because LEFT of O)

Acceleration = 4π^{2}/9 cm/s^{2}

**Example 2**

A particle doing simple harmonic motion is initially 4 cm to the right of centre and travelling away from the centre with a velocity of 3 cm/s

The period of the motion is 8 seconds.(i) Find a and α

(ii) Find the maximum speed

(iii) Find how long it takes the particle to first reach the centre.

(i)T = 2π/n = 8 Therefore, n =π/4Use v

^{2}= n^{2}(a^{2}− x^{2}) to findav = 3 when x = 4

Therefore, 9 = π

^{2}/16(a^{2}− 16)

Therefore, a^{2}= 9x16/π^{2}+ 16

Therefore, a = 5.53 cmTherefore, x = 5.53 cos(π/4.t + α)

When t = 0 then 4 = 5.53cosα

Therefore, cosα = 0.7232

Therefore, α = ±0.762 (cosine is positive in first/fourth quadrants)

But v = -5.53π/4.sin(π/4.t + α)But v = +3 when t = 0 (Moving away from the centre)

Therefore, sinαmustbe negative

Therefore, α =-0.762

(ii)v_{MAX}= an = 5.53π/4 =4.34 m/s(

iii)x = 5.53cos(π/4.t − 0.762)If x = 0 cos(π/4.t − 0.762) = 0

Therefore, π/4.t = π/2 + 0.762Therefore,

t = 2.97 seconds