Lecture 9

Simple Harmonic Motion

The Restoring Force

In order to stretch a spring that obeys Hooke's law, a stretching force F = kx must be applied. This creates a tension in the spring, so that the spring pulls backward on whatever it is attached to, which experiences a

Restoring force: F = -kx

A point mass m connected to a spring obeying Hooke's Law, experiences a force F = -kx, and it performs a motion called simple harmonic motion.

A graph of the displacement as a function of time has the shape of a sine or cosine function. The maximum value of the displacement is called the amplitude. It is fixed by the distance the mass is pulled down before it is let go.

The pattern of the displacement repeats itself after a characteristic time that is called the period T of the oscillation.

The frequency f is just one over the period

f = 1/T

The SI unit for frequency is the hertz.

E.g. a radio station has a frequency of 1060 kHz. The period of the oscillations of the radio waves is

T = 1/f = 1/(1060×10³ Hz) = 0.943 × 10^-6 s

or about 1 microsecond.

The angular frequency is

w = 2p/T = 2pf

(no separate SI unit)

The Reference Circle

Imagine a point mass performing uniform circular motion, with angular velocity w. Shine a light in the plane of the circle to project a shadow of the point mass on a flat screen. Then that shadow performs uniform circular motion.

Radius of the circle is A, angle turned through is q , then length of the shadow is A cos(q). But q = wt so

x = A cos(wt)

This is simple harmonic motion. The angular frequency of the simple harmonic oscillator has the same value as the angular velocity of the point on the reference circle.

The tangential velocity of the point performing uniform circular motion is

v_T = wr = wA

The velocity of the harmonic oscillator is the projection of this. Since the tangential velocity is at right angles to the radius, there is a sine factor. Also, the horizontal component of the velocity points backwards, so

v = -v_Tsin(wt) = -wA sin(wt)

When the displacement is at its maximum positive or its maximum negative value, the mass instantaneously comes to rest, and the velocity is zero. The mass has its maximum speed when it is moving through the center of its swing, i.e. when the displacement is zero.

Acceleration of the harmonic oscillator is a projection of the acceleration of uniform circular motion, which is the centripetal acceleration

a_C = v²/r = w²r = w²A

- always in the opposite direction to the displacement vector - cosine factor and a negative sign

for SHM: a = -w² A cos(wt) = -w²x

Newton's second law:

F = ma

-kx = m × (-w²x) = -mw²x

so

and, from this

and

If m is increased, the object responds more slowly. If the spring constant is increased, the mass goes through its oscillation more rapidly.

Energy of Harmonic Oscillator

The work done in stretching a spring from its relaxed length through an extension x is ½kx². And the elastic force is conservative.

The elastic potential energy of a spring with a force constant k that is stretched through an extension x is

PE(elastic)=½kx²

The mass that is performing simple harmonic motion also has kinetic energy, the usual ½ mv².

Total energy of the oscillator is

E(total) =PE + KE = ½kx² + ½mv²

with:

k = mw²

x = A cos(wt)

v = -wA sin(wt)

so

E(total) = ½mw² A(cos²(wt) + sin²(wt))

From Pythagoras' theorem:

for any angle.

Hence

E(total) = ½mw²A²

The total energy is constant, independent of time. As the mass oscillates on the spring there is a continually switching of the energy between potential and kinetic energies, with the net amount of energy staying constant.

The Simple Pendulum

Mass on the end of a string - simple pendulum - not exactly a simple harmonic oscillator. Good approximation to one if the amplitude of the oscillations is small.

When the pendulum is turned through an angle q, the mass has traveled through a circular arc of length s = Lq. Forces acting on the mass are the tension T in the string, and the weight mg of the mass. The tension is perpendicular to the arc, so it has zero component along the path.

The weight is vertically downwards. Its component along the path is -mg sin(q). Equation of motion is

ma = -mg sin(q)

- does not have a simple solution. Make an approximation.

sin(q) » s/L

then

ma = -mgs/L

a = -(g/L)s

This has the same form as the equation for simple harmonic motion of

for SHM a = -w²x

with

The period of a simple pendulum is not strictly independent of the amplitude.

Fluids

Use the word fluid to encompass both gases and liquids.

Mass density

The mass density is the mass m of a substance divided by its volume V:

r = m/V

The gram was chosen so that the density of water was 1.00 gm/cm³. In SI, the density of water is approximately 1000 kg/m³. (It depends slightly on temperature). Some densities are

Water (4^oC) 1000 kg/m³

Blood (37^oC) 1060

Wood (yellow pine) 550

Quartz 2660

Aluminum 2700

Iron 7860

Lead 11300

Gold 19300

Specific gravity = Density of substance/Density of water at 4^oC

E.g. specific gravities of wood and lead are 0.55 and 11.3 respectively (no units).

The Earth, Moon, and Sun are all, near enough, spheres. The volume of a sphere is

V = (4p/3)R³

Astronomical from the inside front cover of the textbook (and an extra line for Jupiter)

Mass (kg) Radius (m) Density (kg/m³)

Earth 5.98×10²⁴ 6.38×10⁶ 5500

Moon 7.35×10²² 1.74×10⁶ 3330

Sun 1.99×10³⁰ 6.96×10⁸ 1410

Jupiter 1.90×10²⁷ 7.15×10⁷ 1240

Earth is part rock and part iron, the Moon is almost all rock. Sun is hydrogen gas, highly compressed. Jupiter is similar.

Pressure

Pressure increases with depth in a fluid, because the weight of the fluid that is above provides a downward force. The pressure at any depth in a fluid is the weight of all the fluid above it that is contained in a column of unit cross-sectional area.

SI unit for pressure is the newton per square meter, called a pascal

1 pascal (Pa) = 1 N/m²

Then

1 atmosphere = 1.013 × 10⁵ Pa

Increase of Pressure with Depth in a Fluid

Imagine a vertical column of liquid, with a height h and an area A. Pressure at the top of the column is P₁. Downward force on the column corresponding to this pressure is P₁A. Pressure at the bottom of the column is P₂ and the upward force on the column at the bottom is P₂A. Other downward force is the weight of the fluid in the column. Volume of the column is hA, mass of the column is the density times the volume, or rhA. The weight of this mass of fluid is rghA. Hence

P₂A = P₁A + rghA

P₂ = P₁ + rgh

For example, if P₂ - P₁ is one atmosphere

rgh = 1.013 × 10⁵ N/m²

h = 1.013 × 10⁵/(9.8 × r)

If r = 1000, for fresh water, h = 10.3 m, about 34 feet. If is the density of mercury, 13600 kg/m³ h is 0.760 m, or 760 mm.

Pascal's Principle

The pressure is the same at all points in the fluid that are at the same depth. Used in hydraulic lift. (see fig. 11.16)

Pascal's Principle

Any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and the enclosing walls.

Archimedes' Principle and Buoyancy

Think about an object that is completely submerged in a fluid. It has a height h and a cross-sectional area A. The pressure at the top of the cylinder is P₁, the pressure at the bottom is P₂. The downward force on the upper surface is P₁A, the upward force on the lower surface is P₂A. The net upward force exerted by the fluid on the object is

P₂A - P₁A = (P₂ - P₁)A

but P₂ - P₁ = rgh

so the net buoyant force is rghA upwards.

Archimedes Principle

Any fluid exerts a buoyant force on an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces.

Problem 11-50 (**) One kilogram of glass (r = 2.60 × 10³ kg/m³) is shaped into a hollow spherical shell that just barely floats in water. What are the inner and outer radii of the shell? Do not assume the shell is thin.

The weight of the glass is 1 kg×9.8 m/s² = 9.8 N

Since it just floats, it displaces a weight of water of 9.8 N, which is a mass of water of 1 kg.

The volume of 1 kg of water is

V = m/(water) = (1 kg)/(1000 kg/m³) = 0.001 m³

The volume of a sphere is

V = (4p/3)R³

Radius of a sphere with a volume of 0.001 m³ is

(surprisingly small!)

The actual volume of 1 kg of glass is

V(glass) = m/(glass) = (1 kg)/(2.60×10³ kg/m³) = 0.000385 m³

so there must be a hollow space with volume 0.001 - 0.000385 = 0.000615 m³ at the center of the sphere. A sphere with this volume has a radius of

This gives the outer and inner radii.

Fluids in Motion

We distinguish between streamline, or steady, flow, and turbulent flow. In streamline flow, the fluid particles move along smooth trajectories. In turbulent flow, the particles follow unpredictable complicated paths, full of changes in direction and velocity. Turbulent flow is an example of chaos.

Equation of Continuity

Imagine an liquid moving along a tube with a cross-sectional area A, at a speed v. In a time t, each particle of the fluid moves a distance vt along the tube. In this time t, a volume Avt of the liquid goes by. If the density of the liquid at that point is r, then the mass of liquid passing in time t is rAvt.

If the cross-sectional area is changing from point to point, the mass of liquid entering the tube must equal the mass leaving the tube. Put subscript 1 on the values at the beginning of the tube and subscript 2 on the values at the end of the tube, then

r₁A₁v₁t = r₂A₂v₂t

r₁A₁v₁ = r₂A₂v₂

equation of continuity.

If the fluid is incompressible, as the tube narrows, the fluid must speed up.

Bernoulli's Equation

In a time t, the fluid at the left end travels a distance s₂ = v₂t, and a volume of fluid V = A₂v₂t enters the tube

The force exerted on this bit of the fluid from the fluid pushing it along is F = P₂A₂, and the work done by this force is W₂ = Fs₂ = P₂A₂s₂ = P₂V.

When the bit of fluid leaves the tube at the right end, it does work pushing the fluid ahead of it along the tube, and the work it does is W₁ = P₁V. The net work done on the fluid as it moves through the tube is W₂ - W₁.

As the bit of fluid enters the tube, it has kinetic energy ½mv₂² and it has potential energy mgy₂, where m = rV.

As it leaves the tube it has kinetic energy ½mv₁² and potential energy mgy₁.

For energy to be conserved

W₂ - W₁ = ½r Vv₁² + rVgy₁ - ½rVv₂² - rVgy₂

P₂V - P₁V = ½r Vv₁² + rVgy₁ - ½rVv₂² - rVgy₂

The factor of V cancels

P₂ + ½r v₂² + rgy₂ = P₁ + ½rv₁² + rgy₁

As the velocity of a fluid increases, its pressure falls.

- airplane's wing, wind can blow the roof off a building.