Physics 7E study notes, UCI summer 2003

by Nasser M. Abbasi

June 27, 2003

1 chapter 15 fluids
2 chapter 13 Simple Harmonic Motion
2.1 Undamped spring SHM motion
2.2 Damped spring SHM oscillations
2.3 Damped and external force spring SHM motion
2.4 Spring K arrangments
3 Chapter 16 Mechanical waves (Transverse)
4 Chapter 17 Mechanical waves (longtitudenal)
5 Chapter 17 Mechanical waves (doppler)
6 Chapter 18 Superposition and standing waves
7 Chapter 38 Diffraction and polarization
8 Chapter 37 interference (2 slits)
9 Misc

1 chapter 15 fluids

Bernulli

$1 2 1 2 p1 + 2ρv1 + ρgy1 = p2 + 2ρv2 + ρgy2$

continuity equation

$A1v1 = A2v2$

2 chapter 13 Simple Harmonic Motion

$x = A cos(ωt + ϕo)$

$2πf = ω$

$T = 1f$

we can find $ω$ from:

For spring given K and Mass attached to spring, $∘ -k ω = m$
For normal pendilum, $∘ -- ω = gL-$
For physical pendilum, $∘ ---- ω = mgd. I$ where d is distance from center of mass to pivot, and I is moment of inertia. For a rod, $I = 1∕3 M L2$

Parallel axis theorm: moment of interia= $I + M R2 cm$

2.1 Undamped spring SHM motion

- kx = ma

solution is $x = A cos(ωt + ϕo)$

$√ ------- v = - A ωsin(ωt + ϕo) = ± ω A2 - x2$ (use this form if given x and A and asked to find v)

$a = - A ω2cos(ωt + ϕo)$

Total energy in a spring = $1kA2 2$ where A is the amplititude.

KE+PE=total energy.

KE= $1 2 2mv$

PE= $12kx2$

2.2 Damped spring SHM oscillations

retadring force $R = - bv$

- kx- bv = ma

Solution is $x = Ae- b2m-tcos(ωt+ ϕo)$ where $ω = ∘ km---(b2m-)2-= ∘ ωo---(b2m-)2$

There are 3 cases of damped oscillations:

underdamped: $Rmax = bvmax < kA$
critical damped: $( ) km-= 2bm-$ i.e. $ω = 0$
over dampled: $( ) km-> 2bm-$

2.3 Damped and external force spring SHM motion

Assume we apply a force $Fextcos(ωt)$ , in addition to damping force $- bv$ ,hence the ODE now is

2 Fextcos(ωt )- kx - bdx-= m d-x- dt dt2

Solution is $x = Acos (ωt + ϕ)$ where $Fext A = ∘------m-2----2 (ω2-ω2o) +(bωm-)$ where $ωo = km-$

We see that as $ω - → ωo$ then $-Fexmt Fext A = (bωm) = bωo$ ,Now in the absense of damping force, this means $b = 0$ , then we see that $A - → ∞$

2.4 Spring K arrangments

For springs in series, $--1- -1 1- ktotal = k1 + k2$ , so $k1k2- ktotal = k1+k2$

For springs in parallel. $ktotal = k1 + k2$

3 Chapter 16 Mechanical waves (Transverse)

see printout

4 Chapter 17 Mechanical waves (longtitudenal)

see printout

5 Chapter 17 Mechanical waves (doppler)

see printout

6 Chapter 18 Superposition and standing waves

If the path difference $Δr = 0,λ,2λ,3λ,⋅⋅⋅$ then the result sum of the waves is constructive (more amplitude). If $Δr = 0.5λ,1.5λ,2.5λ,⋅⋅⋅$ then the result is destructive.

In a pipe, which is open on one end and closed on the other end, we can only have the fundamental harmoic, then the 3rd then the 5th, etc... i.e. we skip the second and the fourth, etc.... so we get $f ,3f ,5f ,⋅⋅⋅ 1 1 1$ . here, $f = v- 1 4L$

If closed (or open) on both ends, i.e. we have standing waves, then we get $f1,2f1,3f1,4f1,⋅⋅⋅$ . here, $f1 = 2vL$ wher $∘ -- v = Tμ-$ or $v =$ speed of sound depending on instrument.

given $y1 = A sin (kx- ωt )$ and $y2 = A sin(kx+ ωt)$ i.e. same waves but travelling towards each others (i.e. in opposit directions), their sum is $y = (2A sin kx)cosωt$

In a standing wave (both ends are closed) and when the waves are in phase, and have the same frequency and amplitude, then the locations of the nodes and antinodes are given by

nodes: $x = 0, 0.5λ, λ, 1.5λ,⋅⋅⋅$

antinodes: $x = 0.25λ, 0.75λ, 1.25λ, 1.75λ,⋅⋅⋅$

So, to find locations of nodes/antinodes, simply calculate $λ$

Also, we can write $1 2 3 4 L = 2λ1 = 2λ2 = 2λ3 = 2λ4 = ⋅⋅⋅$ we can use this to find L if given two different and successive $λ$

7 Chapter 38 Diffraction and polarization

Diffraction is the divergence of light from its initial line of travel (as it leaves a slit). To see diffraction use single slit. (to see interference of light, use 2 slits).

For diffraction, slit width is used. For interference, distance between slits is used.

The angle $θ$ at which dark bands are seen on the screen is found from $asinθ = m λ$ , where $m = ±1,±2, ⋅⋅⋅$

Also notice that the central maxiumum band is twice as wide as the second maxima.

To find intensity at different maxima and minima use this:

$(sin(β))2 I = Imax --β2- 2$ here, $β -2$ is the phase difference

where $β 2 = m π$ , $m = ±1,±2, ⋅⋅⋅$ for minima (we get I=0)

and $β 1 2 = (m + 2)π$ , $m = ±1,±2, ⋅⋅⋅$ for maxima (we get I=0)

For example, for $m = 1$ , maxima intensity is found

$( ) sin(π2) 2 ( 1)2 I = Imax π2 = Imax π2 = Imax$ $(4-) π2 =$ $0.40528 Imax$

for $m = 2$

$( ) sin(5π2 ) 2 I = Imax 5π2 = Imax$ $-2 1.6211 × 10$

for $m = 3$

$I = Imax (sin(77π2π2 ))2 = Imax$ $8.2711 × 10-3$ $\text{[math]}$

For angular resolution of 2 sources, use $θmin = λ a$ where $a$ is the width of the slit.

If the slit was circular, with diameter $D$ then $θ = 1.22λ- min D$

Diffraction granting is used to resolve/find wave length.

Resolving power $R$ of a diffracting granting is $average λ R = --Δ-λ---$

$R$ is also given by $R = N m$ , where $N$ is the number of lines/grids, and $m$ is the order number. $m = 0,1,2,...$

Polrization: $I = I cos2 θ max$

Polarization by relfection: When tan $θp = n$ , then reflected light is polarized.

8 Chapter 37 interference (2 slits)

One course, but 2 slits. Hence, we consider the light comming from the slits as 2 coherent sources (since from the same original source). (So, same frequency $ω$ and the same amplitude $E0.$

Now, as rays leave the slits, they travel different directions. And the path difference travelled causes a phase difference between the rays. Let this pahse difference be $ϕ$