physics notes 2010-2

8/8/2019 physics notes 2010-2

1/30

PHYSICS

2010 1st semester

NATASHA KELBAS


2/30

UNIT 1: Vibrations, Waves and Sound

Retart Stand

String

Rubber stopper

What factors effect the period of the pendulum?

-rotation of the Earth

* -weight (mass) of the rubber stopper (M)

* -length of sting (L)

- air density/temperature

* -amplitude of the vibration (A)

-material used for string

-gravitational force (w/o it the pendulum won't move)

(NOTE: * means that these are the factors with the biggest impact/ easily measurable)

For an ideal pendulum:

T=2L/g


3/30

Technical terms:

Vibration: particles moving back and forth in time.

Period (T) time for 1 complete cycle.-measured in seconds (s) usually

-from A to B and then back to Ai.e. A-> B -> A

A B

Frequency (f): # of vibrations/ unit time- measured in Hz. (Hertz or 1/s)

Waves : vibrations that travel in space AND time-ex. water waves

Therefore():

time for n vibrations

T=---------------------------

# of vibrations

1. T= t/N

2.f=N/ t

3. (s)

4. f= 1/T (Hz)

constant

crestamplitude

x-axis

time(s)

displacemnet from

equilibrium (m)

Period (T)


4/30

trough

The example above is the ideal vibration

-the amp and the T remains fixed for all time.

Realistic vibrations:

-damped vibrations

What vibrations look like in reality.

A vibration that has been damped in a very

short amount of time.

Ex. Automobile shock absorber

- after the initial shock (a pothole) the car has

to be able to dampen the vibration as soon as

possible. so the shock absorber is there to

minimize the time it takes for the vibration to


5/30

be equal to 0.

The wavelength (-lambda)

The time is fixed

The period (T)

The position is fixed

UNIVERSAL WAVE EQUATION:

V=f derived from:

speed of wave= distance travelled/time

V= / T V= x 1/T V= f

Interpretation of V= f-it is assumed that every wave travels through a medium

V- is controlled by a medium (medium fixes the speed of the wave)

you can change for however once fis chosen then is automatically set. (and visa versa)Ex. wave moving through a coil.

Medium= coil

given the medium, you can easily adjust f(i.e. move your hand faster/slower is automatically

set)

Units:

V= m/s

= mf=Hz


6/30

Example: Radio

V - fixed

f- chose

Frequency

Waves in 1-D, 2-D, and 3-D1-D wave

mechanical wave moving through a coil

2-D wave

Water wave on the surface of water

3-D wave

sound waves from a speaker

Transverse waveparticles in medium move perpendicular to the direction in which the wave travels.

Tunning

FM

AM

Volume

MHz

KHz

Amplitude


7/30

Ex. wave in a coil

Longitudinal waveparticles move parallel to the medium

Ex.

ResonanceEvery object has a "natural" frequency of vibrations

Ex. period of pendulum

T=2L/g

Resonance: if we "force" a system to vibrate @ same frequency the largest amp. response will be

when the "forced" frequency is approximate to the natural frequency.

Example:

Boyfriend has to push his girlfriend on a swing. The best time for him to push so close to the natural

frequency of the girlfriend on the swing in order to get the largest amplitude response of the swing.

Push happnes

here to


8/30

REAL life application

-if you car is stuck in a rut during the winter. You could just press gas and get out eventually.

However if you tap the gas pedal to get the car rocking and then using the motion tap on the gas

each time its going up, so using the natural frequency to increase the amplitude. This saves yourengine, time and gas.

Think of a wine glass breaking due to high frequency waves.

-The waves match the natural frequency of the glass and there for increase the amplitude to the

point that the glass can no longer keep itself together.

Interferencewaves undergo a special phenomena called interference - waves can travel though each other and

form new waves

Ex 1. SUPER CREST: Total constructive interference.

two waves (crests) moving towards each other with the same

amplitude @ t=0s at with the same speed.

@ t=1s the two waves collide with each other creating a supercrest where the amplitude of the waves (x) is doubled to

become 2x

After the two waves @ t=2s they travel in their initial

directions with the initial speed and amplitude

Ex 2. Total destructive interference

Two waves a crest and a trough of equal amplitude (-x and x)


9/30

and speed(v) are travelling towards each other @ t=0s

When the two waves crest+trough meet the amplitudes cancel out to create a straight line (-x+x=0)

@ t=1s

@ t=2s the two waves continue with their initial speed, direction and amplitude.

Square wavesMany waves are sinusoidal in appearance

However due to the fact that it is so hard to figure out the size of the super crests, super troughs

and total destructive interference.

we use square waves (sharp angles)

Ex. 1

Wave 1

Wave 2

1st kink. B+G=0, B+G= 10+0=02nd kink. B+G=10+0=10, B+G=10+15=253rd kink. B+G=10+15=15, B+G= 0+15=154th kink. B+G=0+15=15, B+G=0+0=0

Step 1: place dots to the left and

right of the kink points for EACH

wave

Step 2: add the dots

Step 3: connect the new dots to form

the new wave.

15u

15u

10u

10u


10/30

more examples:

a)

b)

c)

final wave

wave 1

wave 2

Standing wave-basing physics of interference

-send a 2nd wave trains @ the same frequency in the opposite direction

^ a series of crests and troughs

medium

-where the two wave trains meet:

crest+crest=super crest trough+trough=super trough crest+trough=total destructive interference

result: standing wave

1st mode/fundamental mode: 2 nodes.

=L/2

2nd mode: 3 nodes

= L


11/30

3rd mode: 4 nodes

=3L/2

nth mode: n+1 nodes

n = 2L/n

Universal wave eqn:

V=f

f= V /

=V / (2L / n)

=Vn / 2L

Standing wave

fn= n(V / 2L)

V and L are fixed

n- changes

as fgoes up^, n goes up^

2-D waveswave eqn is still valid V=f

Plane wave.

Y

X


12/30

the crest/trough forms a plane with the other water molecules at that amplitude.

Interference: circular waves spread out and interfere with one another to produce an interference

pattern.

Sound-a vibration of molecules that make up a medium (air, water, metal[rail

road tracks]...)

-ears are sensitive to these slight vibrations of molecules

The person's vocal chords contract and expand causing the air molecules in your throat to vibrate.

The wave produced is a longitudinal wave with the compressed air particles and the rare fractions

where the molecules are spaced further apart. The ear then picks up on these slight vibrations and

because the ear drum vibrates the brain gets a signal of the amplitude (volume) and the wavelength

(pitch).

-looks like transverse wave: use the same mechanics

-V depends on the temp

Vsound = 331m/s + 0.59xT - (temp in Celsius.)

as T goes up V goes up.

NOTE: T is period and temperature. Be careful some abbreviations have more then one meaning

Sound Phenomena: "Beats"

->f1

f2


13/30

A is in tune and B is not certain. then strike same note (if trilling sound is noted then adjust B)

f(beat) = | f1 - f2 |

NOTE: if f1=f2; f(beat) = 0.

Closed and open vibrating air columnsclosed air column

one end is open for air

closed end

-many wind instruments can be viewed as either open/closed air columns (i.e. organ pipes, flute,

trumpet, etc.)

4 possible cases:

open column fixed length fixed frequency

closed column fixed length fixed frequency

closed air column open air column

fixed f Ln=(2n-1) L= /2

4

Ln=(n)/2

L= /2

fixed L fn= (2n-1)V fn=(2n-1)f1

4L

f1=V/4L

fn= n-(V/2L)

f1=V/2L

fn=f1 - n

L=L1-L2

closed, fixed length

L1= 1/4

L2= 32/4

L3= 53/4

Ln=(2n-1) /4

closed, fixed frequency

Open air column

open


14/30

Ln=(2n-1) /4

if L is fixed then changes.

L= (2n-1) n/4 ---------> n = 4L/(2n-1)

V=f

fn=V/ = V(2n-1)/4L

Equation: fn=V(2n-1)/4L

Open, fixed frequency

L1= /2

L2 =

L3 = 3/2

Ln=n/2

Open, fixed Length

Ln=nn/2

n= 2L/n

V=f

fn=V/n = n(v/2L)

Ex. 1

Bugle 20 C

-acts like an open air column.

a) determine fundamental frequency

b) determine 2 higher frequencies

open, L fixed

a) fn=nV/2L=340/2.65 Hz

= 64Hz

Ex. 2

3rd resonance L of a closed air column resonating to a tuning fork is 95cm (L3)

determine the 1st and 2nd resonance

b) f2= 2f1-128Hz

F3= 3f1= 192Hz


15/30

closed, fixed f.

Ln=(2n-1) /4

L3= 5/4

= 4L3/5

= 4 x 0.85m/5= 0.76m

Intensity of Sound Wavesquantity related to the Amplitude of the wave

- if we have a point source of sound

ex. a small speaker

in general, as distance from the source increases, the intensity of the sound decreases

inverse square law: Intensity proportional to 1/(distance from source)^2

I 1/r^2

as r goes down I increases

as r goes up I decreases

NOTE: unit for intensity is W/m^2 (Watts/ m^2)

in practice, however, intensity is called using dB (decibel) levels

10dB= 1B (bell)

dB = 10 x log(intensity/threshold intensity)

"threshold intensity" - 10^-12 W/m^2

dB scale is similar to the Richter scale for earthquakes.

ex. 7.0 earthquake compared to 8.0 eq. The 8.0 is 10x more intense then the 7.0 earthquake

same for dB 10 units, Intensity has increased by a factor of 10.

source intensity dB

threshold of hearing 10^-12 0

normal breathing 10^-11 10

vacuum cleaner 10^-4 80

pneumatic chisel 10^-1 110


16/30


17/30

d1= initial position= +40m

d2= final position= +110m

time interval= 2.0s

Vavg= (d2-d1)/t = (110m-40m)/2.0s = +35m/s (+means that the object is moving to the

right)

Ex.1

Vavg= (d2-d1)/t = (-20+60)/2 = +40m/2.0s = +20m/s

Ex. 2

Vavg. should be

-ve! car is moving to the left

Vavg. = (d2-d1)/t = (30m-60m)/3.0s = -10m/s

- in your car the speed-o-meter only measure the size of the velocity.

i.e. speed = |velocity| (magnitude)

a speed-o-meter measures the instantaneous speed which is |average velocity| over very small time

interval. (t =/= 0)

Vinst= (Vavg as t starts to approach 0)

Origin

(O)

Initial

position

40m

Final position

(110m)

-ve +ve

O

-60m i.p. -20m f.p

2.0s

ANS: Vav should be +ve! b/c car is moving

towards the right.

O 30m f.p. 60m i.p.

3.0s


18/30

V

slope of the secant line of d-t (graph-Vavg)

slope of tangent line of d-t (graph- Vinst.)

2nd Kinematic quantity- Acceleration

Aavg = (V2-V1)/(t2-t1) = V / t units are = m/s/s = m/s^2

d2

t1 t20 Time (s)

Pos

itio

n(m)

Straight

line

Vavg. Is the slope of thesecant line connect

points (t1,d1) and (t2,d2)

Slope of this

line is the

Vav .

The secant line

approaches the

tangent line

(touches the curve

t -> 0


19/30

similarly as t approaches 0 the Aavg becomes Ainst.

Difference b/w velocity and acceleration

if V1 is +ve and A


20/30

NOTE: starting point must be given.

A(m

/s^2)

3

2

1

1 2 3

V(m

/s)

3

2

1

1 2 3

d(m)

5

4

3

2

1

1 2 3

A(m

/s^2)

3

2

1

Area

under

the

graph

Slope


21/30

RULES: for going from v-t to d-t/ a-t to v-t

GENERAL RULE:

Five Useful Kinematic Equationsused to solve many problems

valid in the case where acceleration is constant

1. d = V1t + (1/2)At

2. d = V2t - (1/2) A (t)^2

3. d = t ((V1+V2)/2)

4. V2=V1+ A t

5. V2 ^2 = V1 ^2 + 2Ad

2-D worldso far we have looked at the 1-D world

i.e a line

1 2 3

Piece of a smiley parabola

Piece of a frowning parabola

Straight diagonal line

D V Aslope

area


22/30

we now expand our world into 2-D

i.e Cartesian plane

Vectors in the 2-D worldvector: an arrow

2 properties: the size (length of arrow) and the direction (where the arrow head is pointing)

where's waldo?

VectorWaldo is @ 15m [E 20 N]

size direction

Vector: special notation

r = 15m (mag) [E 20 N] (dir)NOTE: when writing a vector notation by hand draw an arrow above the vector . when typing the

vectors should be bolded.

How to work w/ Vectors1. addition of vectors

w. given 2 vectors

A= 3.0m [E]

y

x

x

y

15m

20


23/30

B= 2.0m [E60* N]* is

find C****** C =/= 5.0m [E60*N]

graphically: 4.4m [E24*N]

2. Analytical (by calculation only)

Sine law:

(sinA)/a = (sinB)/b = (sinC)/c

Cosine law:

c^2 = a^2 + b^2 - 2(a)(b)cosC

mag: x= 2+ 3- 2(2)(3)cos 120*= 4.4m

dir: sin 120*/4.4m = sin / 2.0m

= 26*

HOWEVER note ambiguous case.

(the angle could be either in the I quadrant or II/ III or VI)

- don't use sine law only cosine law

cos C = (a+ b-c) / 2ab

3. Method of components

Ex. A = 3.0 [E]B = 2.0 [E 60* N]

Step 1: convert vector into bracket form

AB

60*


24/30

A = (3.0cos 0*, 3.0sin0*)B = (2.0cos60*, 2.0sin60*)Step 2: add the columns

A = (3.0cos 0*, 3.0sin0*) = (3, 0)B = (2.0cos60*, 2.0sin60*) = (1, 3)

= (4, 3)

Step 3: find the mag

#x + #y = 19 m = 4.4m

Step 4: find the direction

= |tan -1 (#y / #x)|

= 23*

A + B = 4.4m [E 23* N]quadrant I - [E _ N]

quadrant II - [W _ N]

quadrant III - [W _ S]

quadrant VI - [E _ S]

Why does this method work? (geometric proof)

Ex. 1

A = 2.0m [E 15* N] A = (2.0m cos 15*, 2.0m sin 15*)B = 3.0m [W 30* S] B = (3.0m cos 210*, 3.0m sin 210*)

A + B = (-0.67, -0.98)mag: (0.67)^2 + (-0.98)^2

= 1.2m

dir: |tan -1 (#y / #x)|

= 56*

because the #y and #x are both negative they are in quadrant VI [W_S]

1.2m [W 56* S]

Write the mag of each vector first and then the cos

and sin of the ccw (counter clockwise) angle from the E

N

E

S

WII I

III

x1

y2 Using the two vectors and making right angle triangles down to the

axis. Then you add the y and the x of the vectors (the brakets). Youget then the sides of a larger right angle triangle. Using Pythagorean

theorem you can find the magnitude of the blue vector. Now yo have

to find the dir. Knowing all three sides you can find the angle from


25/30

Subtracting vectorsyou can always make a -ve "look" like addition

Ex. 4-6 = 4 + (-6)

similarity for vectorsA - B = A + (-B)

Ex.

A = 3.0m [E40*N]B = 5.0m [S20*E]find A - B = A + (-B) = 3.0m [E40*N] + 5.0m [N20*W]A = (3.0mcos40*, 3.0msin40*)B = (5.0mcos110*, 5.0msin110*)= (1.64, 6.7)

mag = (1.64)^2 + (6.7)^2

= 7.00m

dir: |tan -1 (#y / #x)|

= 84.7*

A - B = 7.0m [E84.7*N]

Application of Vector TechnologyEx.

When the vector is -ve and you switch it to +ve.

You can change the direction to the opposite

quadrant.

Ex. the vector B is [E _ N] so when you change thedirection to be +ve its [W _ S]

E0

40m

45m

35*

20*

d


26/30

d- change in position

d2 - position2

d1 - position1

d = d2 - d1

= 45 [E20*S] - 40 [E35*N]

= 45 [E20*S] + 40 [W35*S]

(45cos340*, 45sin340*)

(40cos215*, 40sin215*)

(9.5, -38)

d = 39m [E76*S]Ex 2. car.

V1 = 2.0m/s [N]V2 = 3.5m/s [E]

t = 5.0s

numerator: V2 - V1= 3.5m/s [E] - 2.0m/s [N]

= 3.5m/s [E] - 2.0m/s [S]

(3.5cos0, 3.5sin0)

(2.0cos270, 2.0sin270)

(3.5, -2)

V = 4m/s [E30*S]Aavg = V/t

4m/s [E30*S]

5.0s

= 0.8 m/s^2 [E30*S]

the vector is pointing in the ES quadrant which means that the force acting upon the car travelling

Mag= (9.5)^2 + (-38)^2

= 39m

Dir= |tan -1 (#y / #x)|

= 76*

V1

V2

Aav = V/t

= (V2 -V1)/ t

Mag= (3.5)^2 + (-2)^2

= 4m/s

Dir= |tan -1 (#y / #x)|

= 30*

NOTE: the time will

only effect the

magnitude of the


27/30

N is pointing so the car can turn and travel directly E.

Acceleration due to Gravity - Free Fallprojective motion : motion in a vertical plane

projectile: any object thrown in the air. (includes all directions)free fall: objects in motion only under the influence of gravity

V = 0

A= 9.81m/s^2 [S]

d = V1t + 1/2 A (t)^2

d = 1/2A(t)^2

(t) = 2 d/A

= 0.505s

What goes up must come down-throw an object up its d increases while its v decreases. - the decrease is because of theacceleration downwards due to gravity.

V and A are opposite so the ball slows down

ball stops momentarily but due to acceleration (9.81m/s^2) [S] changes direction.

V and A are same so it speeds up.

1.25m

Ex. 1 If you drop a ball from 1.25m, how long will it take for the

ball to reach the ground?

V1 = 0 and A = 9.81m/s^2 [down] = +9.81m/s^2

b/c the ball is moving in 1 direction [S] the displacement is

9.81m/s^2 sinc V1= 0

NOTE: the time it takes for an object is directly

proportional to the root of the height

if there is no air resistance the time doesn't depend on

mass, volume or density

Vi A

A


28/30

Graphical representation: d-t

NOTE: Ball doesn't move in a parabolic arc.

Relative Velocity

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Vi A

Po

sition

(m)

Time (s)

N

S

Buttonville

Toronto Island Airport

Wind

[E]

Pilot ignored wind. Though the plane is

pointing [S] and all the instruments

show that he's flying due [S] due to the

N Buttonville

S TIA

This time the pilot saw that he wasn't

heading towards TIA so he pointed the

nose of his plane SW all the time.

However this is time consuming and time

= $

N Buttonville


29/30

Navigation: study of finding the direction of the heading (nose of plane in this case).

Relative Velocity eqn. : Vpg(plane wrt ground) = Vpa(plane wrt air) + Vag (air wrt ground)

term magnitude directionVpg ground speed (must be

calculated)

known!

i.e. your plane always has a

destination

Vpa air speed (of plane)

known!

heading (must be calculated)

Vag wind speed(Enviro Canada is

responsible for --------------

----->

wind direstion

these two numbers

Example

take off 2:00pm

|wind speed| = 20km/h

|air speed| = 80km/h

wind direction = [W]

a) what heading should the plane have?

b) Whats the estimated time of arrival in Niigata?

S TIA

The correct solution is to point the plane in the SW

quadrant so the nose of the plane cuts into the

wind which is blowing E. This way even though the

instruments on the plane don't read S you are

travelling in the correct direction.

Niigata [N]

Tokyo [S]

Wind [W]

Vag

Vpg

Vpa

20km/h

80km/h

opp

Sin = = 20/80Hyp

=14*

The heading is [N14*E] = [E76*N]


30/30

time of flight = ground distance / ground speed

= 300km / 80km/h^2 - 20km/h^2

= 3.9h

Arrival time: 5:54pm

SYMBOLS/ Abbreviations:

()- therefore

b/c- because

w/o- without

w/ - with

eqn - equation

wrt - with respect to

Documents

physics notes 2010-2