Physics 1200 Final Examhumanic/p1200_lecture25.pdf · THIN RING =MR 2 τ EXT =Iατ=FlI=mr ......

Physics 1200 Final Exam

Date & Time Lecture time Final Exam date & time 3:00 pm Friday 4/27/2018 4:00 pm – 5:45 pm 4:10 pm Monday 4/30/2018 4:00 pm – 5:45 pm Place Rooms to be listed on Carmen

Physics 1200 Final Exam -- continued

•  105 minutes •  Comprehensive – covers all of the course material (more

or less) uniformly •  35 multiple-choice questions •  Equations and constants will be provided on the exam

of equations given on the Final Exam

Final Exam -- Physics 1200 Spring 2014 -- 10:20 am lecture section --Humanic

Tuesday, April 29, 2014, 8:00 am - 9:45 am

Name _________________________________Rec. Instructor_____________________ * Exam is closed book and closed notes. * Write your name and the name of your recitation instructor on every page of the exam. * This exam consists of 35 multiple-choice questions (5 points each, 175 points total) * You have 1 hr 45 minutes to complete the exam.

Fx =max Fy =may F =G m1m2

r2∑ G = 6.67×10−11 Nm2 kg2 g(Earth)∑ = 9.80 m s2

vx = v0 x + axt x = 12 v0 x + vx( ) t x = v0 xt + 1

2 axt2 vx

2 = v0 x2 + 2axx

vy = v0 y + ayt y = 12 v0 y + vy( ) t y = v0 yt + 1

2 ayt2 vy

2 = v0 y2 + 2ayy W = F cosθ( )s

fk = µkFN h2 = ho2 + ha

2 sinθ = hoh

cosθ = hah

tanθ = hoha

F = −kx

!v = Δ!rΔt

!a = Δ

!vΔt

F =mg+ma aC =v2

rFC =

mv2

rfsMAX = µsFN

PEgravity =mgh PEspring = 12 kx

2 KE = 12mv

2 WNC = Ef −E0 E = KE +PE P =Wt

!Pf =

!P0

!P =!p1 +!p2

!p =m!v

!J =

!F∑( )Δt = !p f −

!p0

ω =ω0 +αt θ = 1

2 (ω0 +ω)t ω 2 =ω02 + 2αθ θ =ω0t + 1

2αt2 xCM =

m1x1 +m2x2

m1 +m2

vCM =m1v1 +m2v2

m1 +m2

m1v01 +m2v02 =m1vf 1 +m2vf 2 v01 − v02 = − vf 1 − vf 2( ) 12 m1v

201 + 1

2 m2v202 = 1

2 m1v2f 1 + 1

2 m2v2f 2

s = rθ ω =ΔθΔt

α =ΔωΔt

vT = rω aT = rα aC = rω2 1 rev = 2π rad = 360o ITHIN RING =MR

2

τ EXT = Iα τ = Fl I = mr2( )∑∑ IDISK = 1

2MR2L = I ω

Lf =

L0 τ EXT∑( )Δt = ΔL

ISOLID SPHERE = 25MR

2 WR = τθ KER = 12 Iω

2 WR = ΔKER E = 12 Iω

2 + 12Mv

2 +Mgh Ef = E0

of equations given on the Final Exam Name ______________________________Rec. Instructor_____________________ Humanic Physics 1200 Final Exam April 29, 2014 8:00am-9:45am

ρ =mV

P = FA

P2 = P1 + ρgh FB =Wdisplaced fluid = ρVg

A1v1 = A2v2 Q = Av = ΔVΔt

ρwater =1000 kg/m3 Patm =1.013×105 Pa

P1 + 12 ρv1

2 + ρgy1 = P2 + 12 ρv2

2 + ρgy2 x = Acosωt vx = −Aω sinωt

ax = −Aω2 cosωt f = 1

Tω = 2π f ωspring-mass =

km

ωpendulum =gL

vsoundair = 343 m/s v = f λ I1

I2

=r2

2

r12 I = P

Aβ = 10 dB( ) log I

I0

!

"#

$

%&

I0 =1.00×10−12 W/m2 fn = nv

2L!

"#

$

%&, n =1, 2,3,.... v = F

m Lfn = n

v4L!

"#

$

%&, n =1,3, 5,.....

sinθbright =mλd

, m = 0,1, 2,3,.... sinθdark = m+ 12( ) λd

, m = 0,1, 2,3,.... fbeat = f1 − f2

sinθ1

v1

=sinθ2

v2

θ(radians) ≈ λD

sinθdarksingle-slit =m λ

W, m =1, 2,3,.... sinθdark

aperture =1.22 λD

source moving towards stationary observer: fo = fs1

1− vs v!

"#

$

%& , away: fo = fs

11+ vs v!

"#

$

%&

observer moving towards stationary source: fo = fs 1+ vov

!

"#

$

%& , away: fo = fs 1− vo

v!

"#

$

%&

M =vobjectvsound

sinθ = vsoundvobject

=1M

Chapters 11, 12, 24

Refraction and Interference of

Waves

The Principle of Linear Superposition

When the pulses merge, the Slinky assumes a shape that is the sum of the shapes of the individual pulses.

The Principle of Linear Superposition

THE PRINCIPLE OF LINEAR SUPERPOSITION When two or more waves are present simultaneously at the same place, the resultant disturbance is the sum of the disturbances from the individual waves.

Constructive and Destructive Interference of Sound Waves

When two waves always meet condensation-to-condensation and rarefaction-to-rarefaction, they are said to be exactly in phase and to exhibit constructive interference.

λ =1 m


When two waves always meet condensation-to-rarefaction, they are said to be exactly out of phase and to exhibit destructive interference.

λ =1 m


Noise-cancelling headset

Outside noises are picked up by a microphone and electronics produces an out-of-phase version of the noise which cancels out the original noise.


If the wave patterns do not shift relative to one another as time passes, the sources are said to be coherent.

For two wave sources vibrating in phase, a difference in path lengths that is zero or an integer number (1, 2, 3, . . ) of wavelengths leads to constructive interference; a difference in path lengths that is a half-integer number (½ , 1 ½, 2 ½, . .) of wavelengths leads to destructive interference.


Example: What Does a Listener Hear? Two in-phase loudspeakers, A and B, are separated by 3.20 m. A listener is stationed at C, which is 2.40 m in front of speaker B. Both speakers are playing identical 214 Hz tones, and the speed of sound is 343 m/s. Does the listener hear a loud sound, or no sound?


Calculate the path length difference.

( ) ( ) m 1.60m 40.2m 40.2m 20.3 22 =−+

Calculate the wavelength.

m 60.1Hz 214

sm343===

fv

λ

Because the path length difference is equal to an integer (1) number of wavelengths, there is constructive interference, which means there is a loud sound.


Conceptual Example: Out-Of-Phase Speakers To make a speaker operate, two wires must be connected between the speaker and the amplifier. To ensure that the diaphragms of the two speakers vibrate in phase, it is necessary to make these connections in exactly the same way. If the wires for one speaker are not connected just as they are for the other, the diaphragms will vibrate out of phase. Suppose in the figures (next slide), the connections are made so that the speaker diaphragms vibrate out of phase, everything else remaining the same. In each case, what kind of interference would result in the overlap point?

Young’s Double Slit Experiment

In Young’s experiment, two slits acts as coherent sources of waves, e.g. light waves. Light waves from these slits interfere constructively and destructively on the screen.


The waves coming from the slits interfere constructively or destructively, depending on the difference in distances between the slits and the screen.


θsind=Δℓ

Bright fringes of a double-slit

Dark fringes of a double-slit

Δl =mλ ⇒ sinθ =m λd

m = 0,1, 2,3,....

Δl = m+ 12( )λ ⇒ sinθ = m+ 1

2( ) λd

m = 0,1, 2,3,…

order of the fringe


Example: Young’s Double-Slit Experiment Red light (664 nm) is used in Young’s experiment with slits separated by 0.000120 m. The screen is located a distance 2.75 m from the slits. Find the distance on the screen between the central bright fringe and the third-order bright fringe.


!951.0m101.20m106643sinsin 4

911 =⎟⎟

⎠

⎞⎜⎜⎝

⎛

×

×=⎟⎠

⎞⎜⎝

⎛=−

−−−

dm λθ

( ) ( ) m 0456.0951.0tanm 75.2tan === !θLy

Beats

Two overlapping waves with slightly different frequencies gives rise to the phenomena of beats.

Beats

The beat frequency is the difference between the two sound frequencies.

fbeat = f1 − f2

In this case, fbeat = 440− 438 = 2 Hz

The Derivation of Snell’s Law

sinθ1 =λ1h=v1fh

sinθ2 =λ2h=v2fh

sinθ1v1

=1fh=sinθ2v2

⇒sinθ1v1

=sinθ2v2

λ =vf

⇒ λ1 =v1f

and λ2 =v2f

f is same in each medium but v and λ are different

Snell’s Law

Wave transmitted from one medium to another

In this figure, v1 > v2

Example: A sound wave in air is incident on a pool of water at an angle of 10o with respect to the normal to the surface. Find the angle of the wave with respect to the normal to the surface after it is transmitted into the water.

sinθ1v1

=sinθ2v2

⇒sinθairvair

=sinθwatervwater

sinθwater =vwatervair

sinθair =1482343

sin10o = 0.750

∴θwater = sin−1 0.750 = 49o

Physics 1200 Final Examhumanic/p1200_lecture25.pdf · THIN RING =MR 2 τ EXT =Iατ=FlI=mr ......

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Transcript of Physics 1200 Final Examhumanic/p1200_lecture25.pdf · THIN RING =MR 2 τ EXT =Iατ=FlI=mr ......