# Physics 203 at Portland State 2014

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## Practice/work here if other pages are temporarily locked by other editors

### Equation Sheet Ch.28

28-1: Conditions for Bright Fringes (Constructive Interference) In a Two Slit Experiment:

• $d sin θ = m\lambda$. $d$ is the slit separation
• $m=0,±1,±2,±3.,..$
• $m=0$ occurs at $θ=0$, this is the central bright fringe.
• Positive values of $m$ are above the central bright fringe, negative values are below.
• Solving for $θ: θ = sin^{-1} (m\frac{\lambda}{d})$

28-2: Conditions for Dark Fringes (Destructive Interference) in a Two Slit Experiment:

• $d sin θ = (m-\frac{1}{2})\lambda.$ $m = 1,2,3...$ (above central bright fringe)
• $d sin θ = (m+\frac{1}{2})\lambda.$ $m = -1,-2,-3...$ (below central bright fringe)
• Solving for $θ: θ = sin^{-1} [(m ± \frac{1}{2})\frac{\lambda}{d}]$. $+ or -$ depending on location.

28-3 Linear Distance from Central Fringe:

• $y = L tan θ$. L is the distance to the screen.
• Solving for $θ$ of a bright fringe: $θ = tan^{-1} (m\frac{y}{L})$
• Solving for $\lambda: \lambda = \frac{d}{m}sinθ$
• Solving for $θ$ of a dark fringe: $θ = sin^{-1}[(m±\frac{1}{2})\lambda/d]$. $+ or -$ depending on location.

28-12 Conditions for Dark Fringes in Single-Slit Interference:

• $Wsin θ = m\lambda$. $m = ±1,±2,±3...$
• Solving for $\lambda: \lambda = \frac{Wsinθ}{m}$ \
• Solving for $θ: θ = sin^{-1}(\frac{m\lambda}{W})$

28-14 First Dark Fringe for the Diffraction Pattern of a Circular Opening:

• $sinθ = 1.22\frac{\lambda}{D}$

28-15 Rayleigh's Criterion:

• $θ_{min} = 1.22\frac{\lambda}{D}$
• Note: $\lambda$ is dependent on the diffraction of the material that the light is traveling through. If the diffraction is $n, \lambda$ becomes $\frac{\lambda}{n}$

28-16: Constructive Interference in a Diffraction Grating:

• $d sin θ = m\lambda$. $m = ±1,±2,±3...$
• Solving for $d: d = \frac{m\lambda}{sinθ}$

#### Problem 1.3.13.13

Two pendulums (or, pendula) are made of identical 1 kg masses suspended on two weightless strings, 40.0 and 40.5 cm in length. If these pendulums are deflected from vertical by 5 cm and released at the same time, how long will it take for them to get completely “out of step” with each other?

Steps:

1. Convert relevant quantities to SI units:
• $l_1$ = 40.0 cm = 0.400 m
• $l_2$ = 40.5 cm = 0.405 m
• masses and (small) deflections are irrelevant for finding periods and timing of oscillations
2. Find each period:
• $T_1=2\pi\sqrt{\frac{l_1}{g}}$ $\approx 6.283\sqrt{\frac{0.400\,{\text m}}{9.8\,\frac{\text m}{ {\text s}^2}}}=1.269\,$s
• $T_2=2\pi\sqrt{\frac{l_2}{g}}$ $\approx 6.283\sqrt{\frac{0.405\,{\text m}}{9.8\,\frac{\text m}{ {\text s}^2}}}=1.277\,$s
3. By the time $t$ the two pendulums are “out of step”, one would have completed an extra 1/2 periods compared to the other:
• $t = \left(n+\frac{1}{2}\right)T_1=nT_2$
4. Use this relationship to solve for $n$, then find $t$:
• first, open the parentheses: $t= nT_1+\frac{1}{2}T_1=nT_2$
• then, isolate $n$ by combining terms: $\frac{1}{2}T_1=nT_2-nT_1=n\,(T_2-T_1)$
• lastly, $n=\frac{T_1}{2(T_2-T_1)}$ $=\frac{1.269\,{\text s}}{2\,\cdot\,8\times 10^{-3}\,{\text s}}$ $=79$
• $t=nT_2$ $=79\cdot1.277\,{\text s}\approx 100\,$s
5. Make sense of the answer:
• This is much longer than the periods of oscillations, but short enough to observe the effect before the oscillations die down due to friction

### Homework Questions Ch.13

#### Problem 54 part B

A 0.550-kg block slides on a frictionless horizontal surface with a speed of 1.10m/s . The block encounters an unstretched spring and compresses it 21.0cm before coming to rest. For what length of time is the block in contact with the spring before it comes to rest? (Your actual numbers may vary)

• I looked at the homework solutions posted in D2L and for the solution, I don't understand why it states that “the period corresponds to one-fourth of a period.” Where does the 1/4-th come from? I know that the spring was compressed 0.25 m so are we assuming that the period is originally 1 meter and that's where the 1/4 comes from?
• If you are trying to map the motion described in the problem onto an oscillation, then, at the beggining, when the encounter happens, the potential energy of the spring is zero (unstretched spring), that means you start at an equilibrium point. When the block comes to rest, its velocity (and so its kinetic energy) is zero, that means you end up at the $x_\text{max}$ (maximum distance from equilibrium) position. Since a typical oscillation cycle goes like this: $0 \rightarrow x_\text{max}\!\rightarrow 0 \rightarrow -x_\text{max}\!\rightarrow 0...$ etc, the time between successive zeros (passing the equilibrium point) is half the period, and the time between a zero (equilibrium point) and $x_\text{max}$ is half that, or 1/4 period. If you plot a $\sin(x)$ function, it will be obvious. So, this is how you get 1/4 period.

Sound Propagation Problem:

Peak Blood Velocity in a fetal aorta is about 20cm/s. If we image it at 9MHz, find the Doppler shift. V=1500m/s f=(1+(Vobserver/V))/(1-(Vsource/V) = f(1+2(Vobserver/V) = f(1+2(.02m/s)/(343m/s)) = 9,010,000 Hz or about 10kHz

draft_page_2.txt · Last modified: 2014/05/18 22:33 by tom_grass