Simple Harmonic Motion

Periodic Motion:

any motion which repeats itself at regular, equal intervals of time.

Link to SHM Applet

Simple Harmonic Motion

It is actually anything but simple!

Simple harmonic motion is a special case of periodic motion

Characteristics of simple harmonic motion:1. the displacement is a sinusoidal function of time, it ranges from

zero to a maximum displacement (amplitude),2. Velocity is maximum when displacement is zero,3. Acceleration is always directed toward the equilibrium point, and is

proportional to the displacement but in the opposite direction,4. The period does not depend on the amplitude.

Equilibrium position

Positive amplitude

Negative amplitude

A

O

B

1. It is periodic oscillatory motion about a central equilibrium point,

Characteristics of simple harmonic motion:1. It is periodic oscillatory motion about a central equilibrium point, 2. The displacement is a sinusoidal function of time, it ranges from

zero to a maximum displacement (amplitude),3. Velocity is maximum when displacement is zero,4. Acceleration is always directed toward the equilibrium point, and is

proportional to the displacement but in the opposite direction,5. The period does not depend on the amplitude.

Dis

pla

cem

en

t

Time

A A

B B

O

Equilibrium position

Positive amplitude

Negative amplitude

A

O

B

Characteristics of simple harmonic motion:1. It is periodic oscillatory motion about a central equilibrium point, 2. The displacement is a sinusoidal function of time, it ranges from

zero to a maximum displacement (amplitude),3. Velocity is maximum when displacement is zero,(equilibrium

point)4. Acceleration is always directed toward the equilibrium point, and

is proportional to the displacement but in the opposite direction,5. The period does not depend on the amplitude.

Dis

pla

cem

en

t

Time

A A

B B

O

Velo

city

Time

A

O

B

O remember velocity is the gradient of displacement

Equilibrium position

Positive amplitude

Negative amplitude

A

O

B

Characteristics of simple harmonic motion:1. It is periodic oscillatory motion about a central equilibrium point, 2. The displacement is a sinusoidal function of time, it ranges from

zero to a maximum displacement (amplitude),3. Velocity is maximum when displacement is zero,4. Acceleration is always directed toward the equilibrium point,

and is proportional to the displacement but in the opposite direction,

5. The period does not depend on the amplitude.

Dis

pla

cem

en

t

Time

A A

B B

O

Velo

city

Time

A

O

B

OA

ccele

rati

on

Time

A

O

Bremember acceleration is the gradient of velocity

Equilibrium position

Positive amplitude

Negative amplitude

A

O

B

Characteristics of simple harmonic motion:1. It is periodic oscillatory motion about a central equilibrium point, 2. The displacement is a sinusoidal function of time, it ranges from

zero to a maximum displacement (amplitude),3. Velocity is maximum when displacement is zero,4. Acceleration is always directed toward the equilibrium point, and

is proportional to the displacement but in the opposite direction,5. The period does not depend on the amplitude.

Dis

pla

cem

en

t

Equilibrium position

Positive amplitude

Negative amplitude

A

O

B

Velo

city

Time

A

O

B

OA

ccele

rati

on

Time

A

O

B

Simple Harmonic Motion• Equilibrium: the position at which no net force acts on

the particle.• Displacement: The distance of the particle from its

equilibrium position. Usually denoted as x(t) with x=0 as the equilibrium position.

• Amplitude: the maximum value of the displacement with out regard to sign. Denoted as xmax or A.

What is the equation of this graph?

Hint

HIVO, HOVIS

A cos graph takes 2 radians to complete a cycle

Dis

pla

cem

en

t

Time

A A

B B

O

Relation between Linear SHM and Circular Motion

B

B A

A

O

Dis

pla

cem

en

t

Time

A A

B B

O

Velo

city

Time

A

O

B

O

Acc

ele

rati

on

Time

A

O

B

Relation between Linear SHM and Circular Motion

B

B A

A

O

Dis

pla

cem

ent

Time

A A

B B

O

At A, t = 0

the displacement is maximum positive x = xo

B

B

xo

O A

A

xo

T = periodic time

= the time to complete 1 cycle

Dis

pla

cem

ent

Time

A A

B B

O

At O, t =

the displacement is zero x = 0

B

B

xo

O A

A

T = periodic time

= the time to complete 1 cycle

Dis

pla

cem

ent

Time

A A

B B

O

At B, t =

the displacement is maximum negative x = -xo

B

B

xo

O A

A

-xo

T = periodic time

= the time to complete 1 cycle

Dis

pla

cem

ent

Time

A A

B B

O

At O, t =

the displacement is zero x = 0

B

B

xo

O A

A

T = periodic time

= the time to complete 1 cycle

Dis

pla

cem

ent

Time

A A

B B

O

At A, t = T

the displacement is maximum positive x = xo

B

B

xo

O A

A

xo

T = periodic time

= the time to complete 1 cycle

Dis

pla

cem

ent

Time

A A

B B

O

At A, t=0

the displacement is maximum positive x = xo

θ = 0°

B

B

xo

O A

A

xo

T = periodic time

= the time to complete 1 cycle

Dis

pla

cem

ent

Time

A A

B B

O

At O, t =

the displacement is zero x = 0

θ = 90°

B

B

xo

O A

A

Dis

pla

cem

ent

Time

A A

B B

O

At B, t =

the displacement is maximum negative x = -xo

θ = π 180°

B

B

xo

O A

A

-xo

T = periodic time

= the time to complete 1 cycle

Dis

pla

cem

ent

Time

A A

B B

O

At O, t =

the displacement is zero x = 0

θ = 270°

B

B

xo

O A

A

Dis

pla

cem

ent

Time

A A

B B

O

At A, t = T

the displacement is maximum positive x = xo

θ = 2π 360°

A

A

xo

O B

B

xo

T = periodic time

= the time to complete 1 cycle

M2 reminder

w =

Units

change∈angletime taken

=θt

rads per sec

Angular velocity =

Dis

pla

cem

ent

Time

A A

B B

O

At a random point M , the displacement is x

A

A

xo

O B

B

x

x

M

θ

since ω = θ = ω t

cos θ = cos (ω t) = Using trigonometry

x = xo cos (ω t)

and θ = θ°

T = periodic time

= the time to complete 1 cycle

xx o

Dis

pla

cem

en

t

Time

B

O

B A A

If timing begins when x = 0 at the centre of the SHM

Dis

pla

cem

ent

Time

A A

B B

O

If however timing begins when x = 0 at the centre of the SHM then the displacement equation is :

A

A

xo

O B

B

x

x

M

θ

x = xo sin (θ)

T = periodic time

= the time to complete 1 cycle

ω = θ = ω t

So x = xo sin (ω t)

Displacement Formulae

If x = x0 when t = 0 then use x = xo cos (ω t)

This is usually written as x = A cos (ω t)

where A is the amplitude of the motion

If x = 0 when t = 0 then use x = xo sin (ω t)

This is usually written as x = A sin (ω t)

where A is the amplitude of the motion

Memory aid Sin graphs start at the origin(centre)Cos graphs start at the top(end)

Dis

pla

cem

ent

Time

A A

B B

O

What is the displacement of a ball from the end undergoing SHM between points A and B starting at A where the distance between them is 1 m and the angle is rads

B

A

Amp

O B

A

x

x

M

( 𝜋7 )

x = A cos (ω t) when t = 0 A =

x = 0.5 cos = 0.45 m so it is 0.05m from the end

0.45

0.5 i.e the Amplitude

0.45

Periodic Time

ω = So t =

T =

So the time to complete one cycle is T =

To complete one cycle t = T and q = 2p

=

Tom is 2 pies over weight

What is the displacement of a ball from the centre undergoing SHM between points A and B starting at the A, after 2secs where the distance between them is 1 m and the periodic time is 6

Dis

pla

cem

ent

Time

A A

B

O

Amp

x = A cos (ω t) when t = 0 A =

x = 0.5 cos 2ω

To find ω use the fact that ω =

To complete one whole cycle then q = 2p and t = 6 So ω = = =

x = 0.5 cos 2ω = 0.5 cos 2 = -0.25 B O Ax

M

-0.25

2

0.5 i.e the Amplitude

Memory aid

Tom is 2 pies overweight T=

-0.25

What is the displacement from the of a ball going through SHM between points A and B starting at the centre, after 3secs where the distance between them is 0.8m and the periodic time is 8sx = A sin (ω t) when t = 0 A =

x = 0.4 sin 3ω

To find ω use the fact that ω =

To complete one whole cycle then q = 2p and t = 8 So ω = = =

x = 0.4 sin 3ω = 0.4 sin 3 = B O Ax

M

Dis

pla

cem

ent

A

B B

O

xo0.4

√23

0.4 i.e the Amplitude

Memory aid

Tom is 2 pies overweight T=

Calculus Memory Aid

s

v

a

Some Vehicles AccelerateThanks to Chloe Barnes

Starving Vultures Attack

Differentiate Integrate

Velocity Equation.

since v =

v = Velo

city

Time

A

O

B

O

x = A cos (ω t)

Characteristics of simple harmonic motion:Velocity is maximum when displacement is zero,

Dis

pla

cem

en

t

-ω A sin (ω t)

dxdt

since x = A cos (ω t)

Acceleration Equation.

a =

since a =

v = -ω A sin (ω t)

a =

Velo

city

Time

A

O

B

O

Acc

ele

rati

on

Time

A

O

B

Characteristics of simple harmonic motion:Acceleration is always directed toward the

equilibrium point, and is proportional to the displacement but in the opposite direction,

d vdt

-ω2 A cos (ω t)

-ω2 x

Equation for velocity where v depends on x rather than t

v = -ω Asin (ω t) v depends on t

Acceleration = rate of change of velocity =

dv

dt

Velocity = rate of change of displacement =

So acceleration =

dvv

dx

Chain Rule for acceleration in terms of displacement

=

dv dx

dx dt

Equation for velocity where v depends on x rather than t

v = -ω A sin (ω t) v depends on t

2 2.. dv

acceleration x x v xdx

∫ vdv=∫−ω2 xdx−ω2 x2

2+c

)

0 = c =

v2=ω2 Amp2−ω2 x2

Separate the variables

=

Equation for velocity where v depends on x rather than t

v2=ω2 Amp2−ω2 x2

v2=ω2(Amp¿¿2−x¿¿2)¿¿

The max velocity occurs at the centre where x = 0

Vmax = w Amp

x = A cos (ωt) if x = A when t = 0

v = -ω A sin (ωt) if x = A when t = 0

a = -ω2 x

T = Tom is 2 pies over weight

)

x = A sin (ωt) if x = 0 when t = 0

Vmax = ω A

Differentiate the displacement equation

Summary

Stop and look at Course notes

Horizontal elastic string

Consider a GENERAL point P where the extension is x .

Resolving 0 – T = m but

– = m so =

Comparing with the S.H.M eqn = –w2x w2 =

T+ve x

xl

xl

xl

x

m

l

m

l

T = for an elastic string

Ex.1 String l = 0.8m l = 20N m = 0.5kg

Bob pulled so that extension = 0.2m.

a) Find time for which string taught

b) Time for 1 oscillation

c) Velocity when ext. = 0.1m

T+ve x

x

l

Horizontal elastic string

Resolving 0 – T = mbut T = for an elastic string

– = 0 so =

Comparing with the S.H.M eqn = –w2x

T+ve x

x

l

xl

20 x

0.8

20 x50x

0.8 0.5

Hence w = 50

As the string goes slack at the natural length it then travels with constant velocity before the string goes taught again.

l = 0.8m l = 20N m = 0.5kg

w2 = 50

a) 2π2π

T= =50

This is the periodic time when the string is tight

b) Time for which initially taught is a T so t =2 50

As the initial extension is 0.2 m then

Amp = 0.2m

Velocity when string goes slack v = w Amp as x = 0

vslack = 500.2 = 1.414m/s

So the distance travelled when string is slack is 4 0.8

So time for which slack

So total time for 1 oscillation = + = 3.15secs

distance 4 0.8

speed 1.414

2

50

4 0.8

1.414

c) Vel. when ext = 0.1

v2 = w2(Amp2 – x2)

v2 = 50(0.22 – 0.12) = 1.5ms–1 v = 1.225ms–1

Two Connected Horizontal Elastic Strings

A and B are two fixed points on a smooth surface with AB = 2m. A string of length 1.6m and modulus 20N is stretched between A and B. A mass of 3kg is attached to a point way along the string from A and pulled sideways 9cm towards A. Show that the motion is S.H.M and find the max acceleration and max velocity.

AB

N1 N2E

0.8 0.2 0.8

T1 xT2

0.2

As x is increasing towards A make +vex is measured from the equilibrium position

1

x 20(0.2-x)T = =

0.8ll

T1 =

+ve direction T1 – T2 = 3a

2

x 20(0.2+x)T = =

0.8ll

- = 3

A B

N1 N2E

0.8 0.2 0.8

T1 xT2

0.2

T1 =

- = 3

-50x = 3

simplifying

- x = which is SHM

ω2=503

A B

N1 N2E

0.8 0.2 0.8

T1 xT2

0.2

T1 =

= - x ω2=503

250 0.091.5ms

3-- ´

=

Max velocity = wAmp = √ 503

0.09 = 0.37ms-1

A B