Simple Harmonic Motion Intro

7/28/2019 Simple Harmonic Motion Intro

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Simple Harmonic Motion

• Simple harmonic motion (SHM) refers to acertain kind of oscillatory, or wave-like motionthat describes the behavior of many physicalphenomena:

– a pendulum – a bob attached to a spring

– low amplitude waves in air (sound), water, the ground

– the electromagnetic field of laser light

– vibration of a plucked guitar string – the electric current of most AC power supplies

– …



SHM Position, Velocity, and Acceleration




Periodic Motion: any motion of system

which repeats itself at regular, equal

intervals of time.







• Equilibrium: the position at which no net force acts onthe particle.

• Displacement: The distance of the particle from itsequilibrium position. Usually denoted as x(t) with x=0 asthe equilibrium position.

• Amplitude: the maximum value of the displacement without regard to sign. Denoted as xmax or A.



The period and frequency of a

wave• the period T of a wave is the amount of time it takes to

go through 1 cycle

• the frequency f is the number of cycles per second

– the unit of a cycle-per-second is commonly referred to as

a hertz (Hz), after Heinrich Hertz (1847-1894), whodiscovered radio waves.

• frequency and period are related as follows:

• Since a cycle is 2p radians, the relationship betweenfrequency and angular frequency is:

T

f 1

f p 2

T

t



• http://www.physics.uoguelph.ca/tutorials/s

hm/Q.shm.html

http://www.physics.uoguelph.ca/tutorials/shm/Q.shm.html






Its position x as a function of time t is:

where A is the amplitude of motion : the distance from thecentre of motion to either extreme

T is the period of motion: the time for one complete cycle

of the motion.

Here is a ball moving back and forth with simple

harmonic motion (SHM):



Restoring Force

Simple harmonic motion is the motion executed by a particle of mass m subject to

a force that is proportional to the displacement of the particle but opposite insign.

How does the restoring force act with respect to the displacement from the

equilibrium position?

F is proportional to -x



Springs and SHM

• Attach an object of mass m to the end of a spring, pull it out to adistance A, and let it go from rest. The object will then undergo simpleharmonic motion:

• What is the angular frequency in this case?

– Use Newton’s 2nd law, together with Hooke’s law, and the

above description of the acceleration to find:

)sin()( t At v

)cos()( t At x

)cos()( 2t At a

m

k



Springs and Simple Harmonic Motion



Equations of Motion

Conservation of Energy allows a calculation of the velocityof the object at any position in its motion…



Conservation of Energy For A Spring in Horizontal MotionE = Kinetic + Elastic Potential

E = ½ mv2 + ½ kx2 = Constant

• At maximum displacement, velocity is zero and all energy iselastic potential,

so total energy is equal to

½ kA2



• To find the velocity @ any displacement… doconservation of Energy… @ some point at max

displacement• ½ mv2 + ½ kx2 = 1/2kA2

• Solving for v

• This is wonderful but what about time??!

v=dx/dt and separate variables!

22 x Am

k

dt

dx

ct m

k A

dt

dx sin

2 2k

v A x m



2 2k v A x

m





SHM Solution...

• Drawing of A cos( t )

• A = amplitude of oscillation

p p 2p

T = 2 p/

A

A2p



SHM Solution...

• Drawing of A cos( t + )

p p 2p2p



SHM Solution...

• Drawing of A cos( t - p/2 )

A

= p/2

p p 2p

= A sin( t)!

2p



1989M3. A 2-kilogram block is dropped from a height of 0.45 m above an

uncompressed spring, as shown above. The spring has an elastic

constant of 200 N per meter and negligible mass. The block strikes the

end of the spring and sticks to it.

a. Determine the speed of the block at the instant it hits the end of thespring. 3 m/s

b. Determine the period of the simple harmonic motion that ensues. 0.63s

c. Determine the distance that the spring is compressed at the instant the

speed of the block is maximum. 0.098 m

d. Determine the maximum compression of the spring. 0.41 me. Determine the amplitude of the simple harmonic motion. 0.31 m

1989.M3 3.0 m/s, 0.63 s,

0.098 m, 0.41 m, 0.31 m



SHM So Far

• The most general solution is x = A cos( t + )

where A = amplitude

= angular frequency

= phase

• For a mass on a spring

– The frequency does not depend on the amplitude!!! – We will see that this is true of all simple harmonic

motion!

m

k



Pendulums

• When we were discussing the energy in asimple harmonic system, we talked about the

‘springiness’ of the system as storing the

potential energy

• But when we talk about a regular pendulum

there is nothing ‘springy’ – so where is the

potential energy stored?



The Simple Pendulum

• As we have already seen,the potential energy in asimple pendulum is stored inraising the bob up againstthe gravitational force

• The pendulum bob is clearlyoscillating as it moves backand forth – but is it exhibitingSHM?



• We can see that therestoring force is:

F = -mg sinθ If we assume that the

angle θ is small, then

sinθ ~ θ and our equation

becomes

F ~ -mgθ = - mgs/L

= - (mg/L)s



F ~ -mgθ = - mgs/L

= - (mg/L)s dividing both sides by m,

a = -(g/L)s.

Equate to a = -ω 2 x , and

ω = (g /L ) 1/2

The Physical Pendulum




• Now suppose that themass is not allconcentrated in thebob?

• In this case theequations are exactlythe same, but therestoring force actsthrough the center of mass of the body (C in

the diagram) which isa distance h from thepivot point



• Going back to our

definition of torque,we can see that

the restoring force

is producing a

torque around the

pivot point of:

sin g F L

•where L is the moment

arm of the applied

force



The Simple Pendulum

• This doesn’t appear too promising until we make thefollowing assumption –

• that θ is small…

• If θ is small we can use the approximation thatsin θ θ

• (as long as we remember to express θ in radians)

I F L g sin

If we substitute τ = Iα, we get:

The Ph sical Pend l m



• Making the substitution we then get:

I F L g

I

mgL


•So, we can reasonably say that the motion of apendulum is approximately SHM if the maximum

angular ampl i tude is smal l

which is the angular equivalent to

a = -ω 2 x

which we can then rearrange to get:




• Making the substitution we then get:

I F L g

I

mgL


•So, we can reasonably say that the motion of apendulum is approximately SHM if the maximum

angular ampl i tude is smal l

which is the angular equivalent to

a = -ω 2 x

which we can then rearrange to get:

Th Ph i l P d l




• So we go back toour previousequation for theperiod andreplace L with h to

get:

mgh

I

T p 2

Th Si l P d l



The Simple Pendulum

where I is the moment of inertia of the pendulummgL

I T p 2

The period of a pendulum is given by:

•If all of the mass of the pendulum is concentrated in the

bob, then I = mL

2

and we get:

g

LT p 2

A A l



An Angular

Simple Harmonic Oscillator

• The figure shows anangular version of asimple harmonic oscillator

• In this case the mass

rotates around it’s center point and twists thesuspending wire

• This is called a torsional

pendulum with torsionreferring to the twistingmotion



Torsional Oscillator • If the disk is rotated through an angle (in either

direction) of θ , the restoring torque is given by theequation:

kx F

withComparing

mreplacedhasIandk replacedhasWhere

2givesWhich

p I T

S l P bl



Sample ProblemA uniform bar with mass m lies symmetrically across two

rapidly rotating, fixed rollers, A and B, with distance L = 2.0cm between the bar’s center of mass and each roller. The

rollers slip against the bar with coefficient of friction µk = 0.40.

Suppose the bar is displaced horizontally by a distance x and

then released. What is the angular frequency of the resulting

horizontal motion of the bar?

ω = 14 rad/s



Sample Problem 1

• A 1 meter stick swings about a

pivot point at one end at adistance h from its center of

mass

• What is the period of oscillation?



A penguin dives from a uniform board that is hinged

at the left and attached to a spring at the right. The

board has length L = 2.0 m and mass m = 12 kg; the

spring constant k = 1300 N/m. When the penguindives, it leaves the board and spring oscillating with a

small amplitude.

Assume that the board is stiff enough not to bend,

and find the period T of the oscillations.

T = 0.35 s

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Simple Harmonic Motion Intro

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