Transcript review for test

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.

F 0
2. The net torque about any axis
acting on the particle must be
zero.

  0
3. The angular and linear speeds
must be zero.
Young’s modulus:
Shear modulus:
tensile stress
F/A
Y

tensile strain L / Li
shear stress F / A
S

shear strain x / h
Bulk modulus:
volume stress
F/A
P
B


volume strain V / Vi V / Vi
Gravitational potential energy
Universal Gravitation

m1  m2
ˆ
F12  G 

r
2
r
m1  m2
U (r )  G 
r
G… constant G = 6.673·10-11 N·m2/kg2
m1, m2 …masses of particles 1 and 2
r… distance separating these particles
r̂… unit vector in r direction
•
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).
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.
2
 T1   a1 
    
 T 2   a2 
Area S-A-B equals area S-D-C
3
Buoyant forces and
Archimedes's Principle
Case 1:
Totally submerged objects.
Ftotal  B  Fg  (  f   o )Vo  g
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).
Archimedes’ principle can also be applied to balloons floating in
air (air can be considered a liquid)
Bernoulli’s
equation
Conservation of energy
1 2
P  v  gy  constant
2
1
1
2
2
P1  v1  gy1  P2  v2  gy2
2
2
Properties of simple harmonic motion
Displacement:
Period T:
T
2

x(t )  A cos(t   )
Angular frequency:
Frequency:
2
  2f 
T
1 
f  
T 2
Units: 1/s = 1 Hz
Velocity:
v(t )  A sin( t   )
Acceleration:
a(t )   A cos(t   )
2
Energy of harmonic oscillator
Kinetic energy:
1 2 1
K  mv  m 2 A2 sin 2 (t   )
2
2
Potential energy:
U
Total energy:
E  K U 
1 2 1 2
kx  kA cos 2 (t   )
2
2
1 2 1
kA  mvmax 2  constant
2
2
The pendulum
2
L
T
 2

g
The physical
pendulum
2
I
T
 2

mgd
Traveling waves
y  A sin kx  t  
 2

y  A sin  ( x  vt   )


- Crest: “Highest point” of a wave
 is phase shift
- 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
2
angular wa ve number, k 
angular frequency,  

T
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
Velocity of wave : v 
In gas and liquids:
v
B

T
 f 
In solids:
On string:
v

T

v
Y

B=Bulk modulus of medium, T=Tension, Y=Young’s modulus, =density of material
TC
Speed of sound in air : v  (331 m/s )  1 
273C
@ 20C : vS  343 m/s
TC … air temperature in degrees Celsius
Both ends open:
fn 
v
n
n
v
2L
One end closed:
n  1, 2, 3, ...
Standing waves (both ends closed)
y1  A  sin( kx  t )
+ y2  A  sin( kx  t )
= yR  2 A  sin( kx) cost 
fn 
v
n
n
v
4L
n  1, 3, 5, ...
Phase-shifted waves (traveling same direction)
y1  A  sin( kx  t )
+ y2  A  sin( kx  t  Φ)



sin(
kx


t

)

2
2
= y R  2 A cos
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
Doppler: slower than sound
Moving detector
 v  vD 
f ' 
f
 v 
Moving source
 v
f '  
 v  vS
Both detector and source

 f

 v  vD 
 f
f '  
 v  vS 
+ detector→toward source
+ source→away from detector
+ detector→toward source
- detector→away from source
- source→towards detector
- detector→away from source
+ source→away from detector
Mach: faster than sound
- source→towards detector
v t
v
sin  

vs  t vs
The ratio vS/v is called
the Mach number.