SIMPLE HARMONIC MOTION (1)danny/5_osc.pdf · Equilibrium 1 SIMPLE HARMONIC MOTION (1) ... Equilibrium 9 SIMPLE HARMONIC MOTION: EXAMPLE • A 747 is hit by turbulence. Upon landing,

5. Oscillations About Equilibrium

1

SIMPLE HARMONIC MOTION (1)

• Consider an air track with a cart of mass m attached to

a spring of force constant k

• When the spring is at equilibrium length (neither

stretched or compressed), the cart is at position x = 0

• If the cart is displaced from equilibrium by a distance x,

the spring exerts a restoring force given by F = -kx

• The spring exerts a restoring force whose magnitude is

proportional to the distance it is displaced from

equilibrium

• The force exerted by a spring is opposite in direction to

its displacement from equilibrium; this accounts for the

minus sign in F = -kx

• A restoring force is one that always points toward the

equilibrium position


2


• If the cart is released from rest at the location x = A, the

spring exerts a force on the cart to the left, causing the

cart to accelerate toward the equilibrium position

• When the cart reaches x = 0, the net force acting on it

is zero, but its speed is not zero so it continues to move

to the left

• As the cart compresses the spring, it experiences a

force to the right, causing it to decelerate and come to

rest at x = -A

• The spring continues to exert a force to the right; thus

the cart immediately moves to the right until it comes to

rest again at x = A, completing one oscillation in time T


3


• If a pen is attached to the

cart, it can trace its

motion on a sheet of

paper moving with

constant speed

• On this chart, we

obtain a record of

the cart’s motion as a function of time

• The motion of the cart looks like a sine or cosine

function

• The position of the cart as a function of time can be

represented by a sine or cosine function

• The position of the mass oscillates between x = +A and

x = -A, where A is the amplitude of the motion and is

the extreme displacement in each direction

• The amplitude is one half the total range of motion,

which repeats with a period T

• The position of the cart is the same at time t + T as it is

at time t

• Thus the mathematical description of position versus

time in simple harmonic motion is

• x = Acos(2πt/T) (ensure calculator set to radians)


4

SIMPLE HARMONIC MOTION:

EXAMPLE

• An air track cart attached to a spring completes one

oscillation every 2.4s. At t = 0 the cart is released from

rest at a distance of 0.1m from its equilibrium position.

What is the position of the cart at 0.3s, 0.6s, 2.7s and

3.0s?


5

CONNECTIONS BETWEEN

UNIFORM CIRCULAR MOTION AND

SIMPLE HARMONIC MOTION

• A peg is placed at the rim of

a turntable that rotates with

constant angular velocity

• Viewed from above, the

peg exhibits uniform

circular motion

• Viewed from the side

it appears to move

back and forth in a

straight line, as we

can see by shining a light to cast a shadow of the peg

onto a back screen

• The shadow of the peg moves with simple harmonic

motion

• Thus there is a connection between uniform circular

motion and simple harmonic motion


6


POSITION AS FUNCTION OF TIME

• Recall that angular position is simply θ = ωt

• Imagine drawing a radius vector of length A to the

position of the peg

• When we project the shadow of the peg onto the back

screen, the shadow is at a location x = Acosθ, which is

the x component of the radius vector

• Thus for an object undergoing a simple harmonic

motion, its position as a function of time is given by

x = Acosθ = Acos(ωt) = Acos(2πt/T)

• When dealing with simple harmonic motion, ω is called

the angular frequency

• ω = 2πf = 2π/T rad/s


7

VELOCITY IN SIMPLE HARMONIC

MOTION

• Recall that the velocity of an object in uniform circular motion of radius r is v = rω

• The velocity is tangential to the object’s circular path, as indicated above

• When the peg is at the angular position θ, the velocity vector makes an angle θ with the vertical

• Thus the x component of the velocity is –vsinθ

• Thus vx = -vsinθ = -rωsinθ

• Since we are only concerned with movement in the xdirection, that r = A and θ = ωt, the velocity in simple harmonic motion is v = -Aωsin(ωt) m/s

• Plotting x and v for SHM shows that when the displacement from equilibrium is a maximum, the velocity is zero

• When x = +A and x = -A, object is momentarily at rest

• The speed is a maximum when the displacement from equilibrium is zero, i.e. vmax = Aω (since largest value of sinθ =1)


8

ACCELERATION IN SIMPLE

HARMONIC MOTION

• The acceleration of an object in uniform circular motion

is given by acp = rω2

• Direction of acceleration: toward centre of circular path

• When the angular position of the peg is θ, the

acceleration vector is at an angle θ below the x-axis,

and its x component is –acpcosθ

• Setting r = A and θ = ωt the acceleration in simple

harmonic motion is a = -Aω2cos(ωt) m/s2

• Position and acceleration for SHM are plotted below

• Acceleration and position vary with time in the same

way, but with opposite signs

• When the position has a maximum positive value, the

acceleration has a maximum negative value

• After all, the restoring force is opposite to the position,

hence the acceleration a = F/m must also be opposite

to the position, thus a = -ω2x

• Maximum acceleration occurs when –cosθ = 1, thus the

maximum acceleration amax = Aω2


9


EXAMPLE

• A 747 is hit by turbulence. Upon landing, data from the

airliner’s black box indicates that the plane moved up

and down with an amplitude of 30.0m, and a maximum

acceleration of 1.8g. Treating the up and down motion

of the plane as simple harmonic, find the time required

for one complete oscillation. Also find the plane’s

maximum vertical speed.


10

THE PERIOD OF A MASS ON A

SPRING

• Net force exerted by a spring with constant k, acting on

a mass m at position x is F = -kx

• Since F = ma, it follows that ma = -kx

• Substituting: m[-Aω2cos(ωt)] = -k[Acos(ωt)]

• Which yields ω2 = k/m or ω = √(k/m)

• Noting that ω = 2π/T, it follows that the period of a

mass on a spring is T = 2π√(m/k) seconds

• The period increases as the mass increases, and

decreases with the spring’s force constant

• A larger mass has greater inertia, and thus takes longer

for the mass to move back and forth through an

oscillation

• A larger value of the force constant k indicates a stiffer

spring, which means a mass would complete an

oscillation is less time that it would on a softer spring

• Example: A 0.120kg mass attached to a spring

oscillates with an amplitude of 0.075m and a maximum

speed of 0.524 m/s. Find the force constant and the

period of motion.


11

A VERTICAL SPRING

• Up to now, we have considered springs that are

horizontal

• When a mass m is attached to a vertical spring, it

causes the spring to stretch

• The vertical spring is in equilibrium when it exerts an

upward force equal to the weight of the mass, i.e the

spring stretches by an amount y0

• Thus, for vertical springs: ky0 = mg

• Or y0 = mg/k

• A mass on a spring oscillates around the equilibrium

point y = y0, and also exhibits simple harmonic motion

• Example: A 0.26kg mass is

attached to a vertical spring.

When the mass is put into

motion, its period is 1.12s.

How much does the mass

stretch the spring when it

is at rest in its

equilibrium position?


12

THE SIMPLE PENDULUM

• A simple pendulum consists of a mass m suspended by

a light string or rod of length L

• The pendulum has a stable equilibrium when the mass

is directly below the suspension point, and oscillates

about this position if displaced from it

• Consider the potential energy of the system

• When the pendulum is at an angle θ with respect to the

vertical, the mass m is above its lowest point by a

vertical height L(1 – cosθ)

• If the potential energy is 0 at θ = 0, then the potential

energy for general θ is: U = mgL(1 – cosθ)


13

FORCES ACTING ON A PENDULUM

• The forces acting on the mass are due to gravity and

the tension in the string (both are vectors)

• The tension acts in the radial direction, and keeps the

mass moving along its circular path

• The net tangential force acting on m is just the

tangential component of its weight: F = mgsinθ

• The direction of the net tangential force is always

toward the equilibrium point, thus F is a restoring force

• For small angles sinθ ≈ θ, so the arc length s = Lθ

• Equivalently θ = s/L

• If the mass is

displaced from

equilibrium by a small

arc length s, the force

is F = mgsinθ ≈ mgθ

= (mg/L)s

• Comparing to a mass

on a spring, F = kx

• Letting x = s and

k = mg/L

• Period of a pendulum

is the same as that of

a mass on a spring

• T = 2π√(m/k) = 2π√(m/(mg/L)) = 2π√(L/g) seconds


14

PENDULUM EXAMPLE

• A pendulum is constructed from a string 0.627m long

attached to a mass of 0.25kg. When set in motion, the

pendulum completes one oscillation every 1.59s. If the

pendulum is held at rest and the string is cut, how long

will it take for the mass to fall through a distance of

1.0m?


15

DAMPED OSCILLATIONS (1)

• Up to this point, oscillating systems in which no

mechanical energy is gained or lost (due to friction, air

resistance) have been considered

• In most physical systems, there are energy losses

• As the mechanical energy of a system decreases, its

amplitude of oscillations decrease as well

• This type of motion is called a damped oscillation

• An oscillating mass may lose its mechanical energy to

a force such as air resistance that is proportional to the

speed of the mass and opposite in direction, which can

be defined as

• The damping constant is b and is a measure of the

strength of the damping force (kg/s)

• If the damping constant is small, the system will

continue to oscillate, but with a continuously decreasing

amplitude – called underdamped motion

• In underdamped motion, the amplitude decreases

exponentially with time: A = A0e-bt/2m where A0 is the

initial amplitude, m is the mass, time t

• As the damping is increased, a point is reached where

the system no longer oscillates, but simply relaxes back

to the equilibrium position – critically damped

• The system is overdamped if the damping is increased

beyond this point, and the system still returns to

equilibrium without oscillating, but takes longer

vFrr

b−=


16

DAMPED OSCILLATIONS (2)

Shock absorbers are

designed to be critically

damped, so that when a

car hits a bump, it returns

to equilibrium quickly

without oscillations


17

DRIVEN OSCILLATIONS

• Applying an external force to a

system furnishes it with energy

– positive work is done

• If you hold the end of a

string from which a small

weight is suspended, and

the weight is set in motion,

holding your hand still will

cause it to stop oscillating

• If you move your hand back and forth, the weight will

oscillate indefinitely

• This motion is called a driven oscillation

• Moving your hand at the wrong frequency (too high or

too small) will cause the weight to oscillate erratically

• Oscillating your hand at an intermediate frequency will

result in large amplitude oscillations for the weight

• This frequency is called the natural frequency, f0, of

the system

• Natural frequency of a pendulum: f0 = 1/T = (1/2)√(g/L)

• For a mass on a spring: f0 = 1/T = (1/2)√(k/m)

• Generally driving any system near its natural frequency

results in large oscillations


18

RESONANCE

• Consider a plot of amplitude,

A, versus driving frequency

f for a mass on a spring

• There are large amplitudes

for frequencies near f0• This type of large response

is known as resonance

• The curves are referred

to as resonance curves

• Systems with small damping

have a high narrow peak on their resonance curve

• Tuning a radio changes the resonance frequency of an

electrical circuit (made up of capacitors, inductors and

resistors) in the tuner

• When its resonance frequency matches the frequency

being broadcast by a station, that station is picked up

• Bridges affected by high winds may swing with large

amplitudes, especially if the frequency of the wind

matches the natural frequency of the bridge as a

system (recall the Millennium Bridge fiasco)

Documents

SIMPLE HARMONIC MOTION (1)danny/5_osc.pdf · Equilibrium 1 SIMPLE HARMONIC MOTION (1) ... Equilibrium 9 SIMPLE HARMONIC MOTION: EXAMPLE • A 747 is hit by turbulence. Upon landing,