Simple Harmonic Motion

Sinusoidal Curve and Circular Motion

A mass is oscillating on a spring

Position inequal time intervals:

This type of motion, is called Simple Harmonic Motion. The system is trying to place the mass at its static equilibrium position

Repetitive Motion

Repetitive motion can be modelled as circular motion

A is the equivalent to x the displacement from equilibrium and is also the radius of the circle

What Can Algebra Tell Us?

a sF F����������������������������

Circle Spring

2

2

2

2

4

4

cma kx

rm kxT

rm krT

2

2

2

4

4

2

mT

k

mT

k

m

k

This tells us that the period only depends upon the mass and the spring constant, not on the distance the spring is pulled back. In English, the time for the mass to move back and forth is ALWAYS the same even if the amplitude decreases.

What Can Algebra Tell Us?

a sF F����������������������������

Circle Spring

2

2

22

cma kx

vm kxr

vm krr

krv

m

kv r

m

This tells us that the velocity depends upon the inverse of the mass, the spring constant, and the distance the spring is pulled back. In English, the velocity of the spring depends on how far it was pulled pack and the spring constant as well as the bigger the mass the slower the velocity of the spring.

We could has also come up with the equations by setting the Kinetic Energy equal to the Spring Potential Energy and solving for v

What Can Algebra Tell Us?

a sF F����������������������������

Circle Spring

c

c

c

ma kx

ma kr

kra

m

This tells us that the spring experiences no acceleration when r=0, this is, at the relaxed position. But we have to use logic to determine which way the acceleration is pointing.

Calculus Can Tell us More

sF F����������������������������

Circle From Force by Spring

2

2

2

2

ma kx

d xm kxdt

d x kx

dt m

Therefore x is a function of time, that when given a time value, t, I can tell you the position x, where the mass is.

But, the question is: What function’s second derivative is equal to this value?

sink

x t tm

The force by the spring is called a Restoring Force

Circular Motion

Yes, But what if I didn’t Know Calculus?

Vertical position versus time:

Period T

Period T

Sinusoidal motion

Time (s)

Displacement (cm)

Period T

Sine function: mathematically

x

y

2π

π/2 π 3π/2 2π 5π/2 3π 7π/2 4π 9π/2 5π

-1

1

y=sin(x)y=cos(x)

Sine function: employed for oscillations

x

y

π/2 π 3π/2 2π 5π/2 3π 7π/2 4π 9π/2 5π

-1

1y=sin(x)

Time t (s)

Displacement y (m)

T/2 T 2T

-A

Ay= A sin(ωt)

Sine function: employed for oscillations

1. Maximum displacement A

2. ωT = 2π

3. Initial condition y(t=0)2

T

Angular frequency in rad/s

Amplitude A is the maximum

distance from equilibrium

Starting from equilibrium:

y=A sin(ωt)

Starting from A:

y=A cos(ωt)

Starting from -A:

y=-A cos(ωt)

sin sin sin sin2 2

2

y A t A t A tmk

Ak

mt

T

k

m Therefore

Note:

Position, Velocity, Acceleration

sin s sin2

inT

x t Ak

mt A t A t

cos c cos2

osT

x t Ak

mt A t A t

cos cos c2 2

osv t A t A t Ak

T Ttk

m m

2

22

sin4 2

sin sina t A t A tk

T

k

mA

Tt

m

2

22

cos4 2

cos cosa t A t A tk

T

k

mA

Tt

m

sin i2

s n2s n i

k k

m mv t A t A A t

Tt

T

Example 1 - determine y(t)

2( ) 15 sin

4y t cm t

s

y(cm)

t(s)5 10 1530

Period?

T=4 s

Sine/cosine?

Sine

Amplitude?

15 cm

Where is the mass after 12 seconds?

2(12 ) 15 sin 12 15 sin 6 0

4y s cm s cm cm

s

Example 2 – graph y(t)

2( ) 3 cos

2y t cm t

s

Amplitude?

3cm

-3

3

y (cm)

y(t=0)?

-3cmPeriod?

2s

2 4 6 8 t(s)

When will the mass be at +3cm?

1s, 3s, 5s, …

When will the mass be at 0?

0.5s, 1.5s, 2.5s, 3.5 s …

Summary Harmonic oscillations are sinusoidal Motion is repeated with a period T Motion occurs between a positive and negative

maximum value, named Amplitude Position: y=A sin(ωt) or y=A cos(ωt) Velocity: v=Aωcos(ωt) or v=-Aωsin(ωt) Acceleration: a=-Aω2sin(ωt) or a=-Aω2cos(ωt)

Angular frequency:

2

T

2

mT

k

Energy of Harmonic Motion

According to the law of conservation of energy, when the mass of attached to a spring is released, the total energy of the system remains constant. This energy is the sum of the elastic potential energy and the kinetic energy of the spring

2 21 1

2 2TE kx mv

Energy Example

A 55 g box is attached to a horizontal spring (force constant 24 N/m). The spring is then compressed to a position A= 8.6 cm to the left of the equilibrium position. The box is released and undergoes SHM.

a) What is the speed of the box when it is at x=5.1 cm from the equilibrium position?

b) What is the maximum speed of the box?

Can you solve these questions two ways?

• Using energy?

• Using position and velocity formulas?

Energy Example

A 55 g box is attached to a horizontal spring (force constant 24 N/m). The spring is then compressed to a position A= 8.6 cm to the left of the equilibrium position. The box is released and undergoes SHM.

a) What is the speed of the box when it is at x=5.1 cm from the equilibrium position?

i fT TE E

i i f fp K p KE E E E

2 2 21 1 10

2 2 2i f fkx kx mv

2 2 2i f fkx kx mv 2 2 2f i fmv kx kx

2 2i f

f

kx kxv

m

2 224 0.086 24 0.051

0.055f

N Nm m

m mv

kg

1.4f

mv

s

a) What is the speed of the box when it is at x=5.1 cm from the equilibrium position?

Energy Example

Let the position function be:24

cos 0.086cos0.055

kx A t t

m

Let the velocity function be: sin 1.79648sin 20.8893k k

v A t tm m

Set the position function = to 5.1 and solve for the t the time.

240.051 0.086cos

0.055

0.051 24cos

0.086 0.055

t

t

0.044807t

1 0.051 24cos

0.086 0.055t

Plug this t value into the velocity function

1.79648sin 20.8893 0.044807

1.4465

v

m

s

Therefore the speed is 1.4 m/s

A 55 g box is attached to a horizontal spring (force constant 24 N/m). The spring is then compressed to a position A= 8.6 cm to the left of the equilibrium position. The box is released and undergoes SHM.

b) What is the maximum speed of the box?

Energy Example

The maximum speed will occur when it is at the equilibrium position (xf=0)

i fT TE E

i i f fp K p KE E E E

2 2 21 1 10

2 2 2i f fkx kx mv

2 2 2i f fkx kx mv 2 2 2f i fmv kx kx

2 2i f

f

kx kxv

m

2 224 0.086 24 0.

0.055f

N Nm m

m mv

kg

1.8f

mv

s

b) What is the maximum speed of the box?

Energy Example

The maximum speed will occur when it is at the equilibrium position (xf=0)

Let the position function be:24

cos 0.086cos0.055

kx A t t

m

Let the velocity function be: sin 1.79648sin 20.8893k k

v A t tm m

Set the position function = to 0 and solve for the t the time.

240 0.086cos

0.055

240 cos

0.055

t

t

0.075196t

24

2 0.055t

Plug this t value into the velocity function

1.79648sin 20.8893 0.075196

1.79648

v

m

s

Therefore the speed is 1.8 m/s

Conservation of Energy

When the spring is in Equilibrium with gravity, the forces are balanced (the downward force of gravity matches the upward for of the spring. So: g sF F mg kd

gF mg

sF kd

Where ‘d’ is the equilibrium extension of the spring and ‘g’ is the acceleration due to gravity. Therefore :

dmg

k

d

Conservation of Energy

We will now set the spring bouncing. At any time the total energy is:

Total K g SE E U U

To simplify things. I’m going to choose Ug = 0 at the extension equilibrium. I’m free to choose any location for this.

221 1

2 2TotalE mv mgx k x d

This gives us:

x=0+x

d

Ug=0

Conservation of Energy

Now, let’s expand out this equation:

221 1

2 2TotalE mv mgx k x d

2 2 21 1 1

2 2 2TotalE mv mgx kx kxd kd

mgd

k

2 2 21 1 1

2 2 2TotalE mv mgx kx kx kmg

kd

d

Conservation of Energy

Now, let’s expand out this equation:

2 2 21 1 1

2 2 2TotalE mv mg mgx kx kx d

2 2 21 1 1

2 2 2TotalE mv kx kd I’m a constant

Therefore, I’m a constant too!

Java Site for SHM

http://canu.ucalgary.ca/map/content/shm/springEnergy/simulate/page2.html