Transcript Physics SAE - broward.k12.fl.us

Physics Subject
Area Test
WAVES
LIGHT & OPTICS
Vibrations and Waves
Simple Harmonic Motion
A restoring force is one that moves a system back to an
equilibrium position.
Example: mass on frictionless
table, attached to spring.
Example: gravity
acting on a mass
hanging from a string.
Example: gravity
acting on a mass
hanging from a spring.
Hooke’s Law
𝐹𝑒𝑙𝑎𝑠𝑡𝑖𝑐 = −𝑘𝑥
When the restoring force is linearly proportional to the
amount of the displacement from equilibrium, the force is
said to be a Hooke’s Law force.
When oscillations are small, the motion is called simple
harmonic motion (shm) and can be described by a simple
sine curve.
Wavelength, l, is the distance between
two
consecutive peaks.
Amplitude is the height of the wave
above or below the equilibrium point.
The wave period, P, this the time it take one wave to pass
the observer.
Frequency, f, is the number of waves passing a particular
point in one second.
In symbolic form
or
T
T 1T
Tff  Tf T
f f ff T
Waves transfer energy, not matter, from one place to another
A Vibrating source transfers a disturbance
Speed depends on type of vibrating source and medium
through which it travels
Wave speed = f x l
The same type of wave moves at the same speed
regardless of f or l
For any wave, f is inversely proportional to l
VIBRATION OF A PENDULUM
What does the period (T) depend upon?
Length of the pendulum (l )
Acceleration due to gravity (g ).
Period does not depend upon the bob mass
or the amplitude of the swing.
𝑙
𝑇 = 2𝜋
𝑔
Vibration of a
pendulum. The to-andfro vibratory motion is
also called oscillatory
motion (or oscillation).
Transverse waves vibrate across from direction of travel
Longitudinal waves vibrate along the direction of travel (as in a
spring)
Molecules in the air vibrate about some average position
creating the compressions and rarefactions. We call the
frequency of sound the pitch.
*
When two wave pass each other their
superposition
causes reinforcement or cancellation.
*Speed of sound (in air, 0⁰C, 1 atm) = 331 m/s
The standing sound
wave in the column
of air in a tube closed
at one end must
have a displacement
node at the closed
end and antinode at
the open end
Only odd multiples
are possible
λ1 = 4L, λ2 = 4/3L, λ3 = 4/5L, λ4 = 7/4L, …
Eigenfrequencies: (f = v/ λ)
f1 = v/4L, f2 = 3v/4L, f3 = 5v/4L, …
eigenfrequencies
*
http://www.astro.sunysb.edu/mzingale/software/astro/doppler.avi
A receiver will detect a higher frequency when the source is
approaching, and a lower frequency when the source is moving
away from the receiver.
Doppler shift, moving receiver
f’/ f = v’/v
f’ = f (1 ± VR/v)
* Example:
Suppose that a stationary siren emits a tone of frequency 440
Hz as the train moves away from it at 30.0m/s. What is the
frequency received on the train?
f’ = f (1 - VR/v)
f’ = 440Hz (1 – 30 m/s/331 m/s)=
400Hz
A motorboat speeding at 6.0 m/s is moving in the same
direction as a group of water waves of frequency 0.62 Hz and
speed 2.5 m/s (relative to the water). What is the frequency
with which the wave crests pound on the motorboat?
f’ = f (1 - VR/v)
f’ = 0.62Hz (1 – 6.0m/s/2.5 m/s)= 0.87Hz
*Lenses work because light slows down in media
other than a vacuum.
*The speed of light is given by:
c
vlight 
n
*n is the index of refraction
*
Substance
vacuum
air
water
glass
n
1
1.0003
1.3
1.5
Index of refraction is unitless
*When light encounters
a boundary between
two media, some of
the light is reflected
and some is
transmitted into the
new medium.
*If the light strikes the
boundary at an angle,
the transmitted light
is refracted.
*Ray diagram: rays
point perpendicular
to the wavefront.
• Wave diagram:
shows the crests of
the traveling waves
1. All angles are measured
from the normal. The
normal is the line
perpendicular to the
surface at the point of
reflection.
2. The reflected angle
is equal to the
incident angle.
 r  i
3. Snell’s Law for
refraction
ni sin  i  n f sin  f
*A ray of light strikes the
surface of a beaker of
hydrogen peroxide (n = 1.414)
making a 30º angle with the
surface normal.
*What angle does the
reflected ray make with the
normal?
*What angle does the
transmitted ray make with
the normal?
*a) The angle of the reflected ray is the same
as the incident ray, 30º
*b)
nair sin  i  n peroxide sin  f
1 sin 30  1.414 sin  f
sin  f  0.354
 f  20.7
* The critical angle of incidence results in a
transmitted ray that is parallel to the boundary
surface.
*If the angle of incidence is greater than
the critical angle, all the light is
reflected and none is transmitted.
 critical  sin
1
n2
n1
*Converging lens: focuses
parallel rays to a point a
distance F from the lens
*Diverging lens: causes
parallel rays to diverge,
as if emanating from a
point source a distance F
behind the lens
1 1 1
 
f di d0
*di =distance from lens to image
*d0 =distance from lens to object
*f = focal length of lens
*
1 1 1
 
f di d0
*f is positive for converging lens
*f is negative for diverging lens
*Negative di is an image on the same side of the
lens as the object
*Positive di
is an image on the opposite side of
the lens as the object
* If the light rays actually pass through the point
they appear to come from, the image is real.
* If the light rays are not actually coming from
this position, the image is virtual.
*
*
1.
Draw the lens, the object (arrow), and both focuses (F).
2. Draw a ray from the top of the object to the lens. This ray
will be parallel to the axis of the optical system until it
strikes the lens. Then is bends to pass through the focus. (Red
ray)
3. Draw a ray passing from top of the object through the
center of the lens. (Blue ray)
4. Draw a ray from the top of the object through the near focal
point. After the lens, this ray becomes parallel to the optical
axis. (Green ray)
1. Draw the lens, object, and focal length on the same
side as the object.
*
2. Draw a line from the top of the object to the lens, parallel
to the optical axis. After the lens, draw the line as if it had
come from the near focus. (Red ray)
3. Draw ray from the top of the object towards the far focus.
After the lens, this ray become parallel to the optical axis.
(Green ray)
4. Draw a ray from the top of the image through the center of
the lens. (Blue ray)
*The angle of reflection is equal to the angle of
incidence, measured with respect to the normal.
i   r
*The thin lens equation we used before can
also be applied to mirrors.
*In the case of mirrors, a positive di means the
image is on the same side of the mirror as the
object.
*This sign convention is opposite that used
for lenses.
1 1 1
 
f di do
*The focal point of a
plane mirror is at
infinity.
*The image formed by
a plane mirror is
virtual and upright.
*To trace rays
reflecting off a plane
mirror, just use law of
reflection.
*The image distance
and object distance
are the same for a
plane mirror.
1 1 1
 
f di do
f 
1
0

di  d o
*A spherical mirror is a
section of a sphere.
*If the outside surface
of the sphere is
reflecting, the mirror
is convex.
*If the inside surface is
reflecting, the mirror
is concave.
*
* 1. Draw ray from top of object to the curved mirror. After
reflecting off mirror, the ray passes through the focal point.
(Blue ray)
* 2. Draw a ray from the top of the object through the focal
point. The ray will be parallel to the optical axis after
reflecting. (Green)
The rays do not intersect, so we must extend them, resulting
in a virtual image.
1. Draw ray parallel to optical axis. After
reflection, this ray travels as if it is
coming from the far focus. (Blue ray)
2. Draw line from top of object to focus.
At the mirror, the ray becomes parallel to
the optical axis. (Green ray)