Review for Midterm 3• Announcements:• I'm having a problem on #8 part A for tomorrow night's

homework. It doesn’t seem like my numbers work out geometrically. My "d" value is 2 meters, and my total string length is 3.7 meters, meaning that each side of the triangle has to be 0.85m. So when I tried to form a right triangle to figure out the tensions, I get a hypotenuse (0.85m) that is smaller than one of my sides (d/2 = 1m. Am I crazy or do these numbers not make sense? Thanks!

• This might seem weird, but I think problem #8 on the current web assign is wrong, at least with the numbers they give me.  It has a triangle with sides 2, 0.8 and 0.8, which isn't possible.  I tried the same problem in the book and got the right answer there, but I do not know how to solve it on the web assign if the numbers do not add up.

• I think something is wrong with the last webassign or im understading it wrong. Just the trig on it. The distance between the 2 pulleys is 2 meters for mine and the string lenght is less then 4 meters. There would not be enought string for it to reach the weight.

Conditions for static equilibrium

(For extended objects)

1. The net force acting on the particle must be zero.

2. The net torque about any axis acting on the particle must be zero.

3. The angular and linear speeds must be zero.

0 F

0

Young’s modulus:

iLL

AFY

/

/

strain tensile

stress tensile

Shear modulus:

hx

AFS

/

/

strainshear

stressshear

Bulk modulus:

ii VV

P

VV

AFB

//

/

strain volume

stress volume

Universal Gravitation

rr

mmGF ˆ

221

12

G… constant G = 6.673·10-11 N·m2/kg2

m1, m2 …masses of particles 1 and 2

r… distance separating these particles

… unit vector in r directionr̂

Gravitational potential energy

r

mmGrU 21)(

• Notice the – sign

• U = 0 at infinity

• U will get smaller (more negative) as r gets smaller.

• “Falling down” means loosing gravit. potential energy.

• Use only when far away from earth; otherwise use approximation U = mgh.

Kepler’s laws:

I. Planets move in elliptical paths around the sun. The sun is in one of the focal points (foci) of the ellipse

II. The radius vector drawn from the sun to a planet sweeps out equal areas in equal time intervals (Law of equal areas).

Area S-A-B equals area S-D-C

III. The square of the orbital period, T, of any planet is proportional to the cube of the semimajor axis of the elliptical orbit, a.

3

2

1

2

2

1

a

a

T

T

If density of object is less than density of fluid: Object rises (accelerates up)

If density of object is greater than density of fluid: Object sinks. (accelerates down).

Buoyant forces and Archimedes's Principle

Case 1:

Totally submerged objects.

gVFBF oofgtotal )(

Archimedes’ principle can also be applied to balloons floating in air (air can be considered a liquid)

Bernoulli’s equation

constant2

1 2 gyvP

22

2212

11 2

1

2

1gyvPgyvP

Conservation of energy

Properties of simple harmonic motion

Displacement: )cos()( tAtx

)sin()( tAtv

)cos()( 2 tAta

2

T2

1

Tf

Tf

22

Period T: Frequency: Angular frequency:

Units: 1/s = 1 Hz

Velocity:

Acceleration:

)(sin2

1

2

1 2222 tAmmvK

)(cos2

1

2

1 222 tkAkxU

constant2

1

2

1 2max

2 mvkAUKE

Energy of harmonic oscillator

Kinetic energy:

Potential energy:

Total energy:

The physical pendulum

mgd

IT

22

The pendulum

g

LT

22

Traveling waves

)(2

sin vtxAy

tkxAy sin

- Crest: “Highest point” of a wave

- Wavelength : Distance from one crest to the next crest.

- Wavelength : Distance between two identical points on a wave.

- Period T: Time between the arrival of two adjacent waves.

- Frequency f: 1/T, number of crest that pass a given point per unit time

2

number, veangular wa kT

2 frequency,angular

shift phase is

Reading quiz question

Can we just actually do a problem in class really quickly and find Phi? I still dont exactly get it but if you just go through it really quickly with actual numbers instead of just explaining the general idea again maybe I could get it. Thanks

Problem 1, from last night’s HW

B

v

B=Bulk modulus of medium, T=Tension, Y=Young’s modulus, =density of material

Y

v

In gas and liquids: In solids:

f

Tv : waveofVelocity

T

v

On string:

m/s 343 v:20 @273

1)m/s 331( :airin sound of Speed

S

CC

Tv C

TC … air temperature in degrees Celsius

Both ends open: One end closed:

Standing waves (both ends closed)

)sin(1 tkxAy )sin(2 tkxAy

tkxAyR cos)sin(2

+

=

)sin(1 tkxAy )Φsin(2 tkxAy

)2

sin(2

cos2

tkxAyR

+

=

Phase-shifted waves (traveling same direction)

... 3, 2, 1,n 2

L

vn

vf

nn

... 5, 3, 1,n 4

L

vn

vf

nn

Reading quiz question

In the following question, how do you find the the three smallest x values corresponding to the antinodes:

Two sinusoidal waves combining in a medium are described by the following wave functions, where x is in centimeters and t is in seconds.

y1 = (5.0 cm) sin (x + 0.20t)y2 = (5.0 cm) sin (x - 0.20t)

I don't understand the question conceptually

Moving detector Moving source

fv

vvf D

' f

vv

vf

S

'

+ detector→toward source

- detector→away from source

+ source→away from detector

- source→towards detector

fvv

vvf

S

D

'

Both detector and source

ss v

v

tv

tv

sin

The ratio vS/v is called the Mach number.

+ detector→toward source

- detector→away from source

+ source→away from detector

- source→towards detector

Doppler: slower than sound

Mach: faster than sound