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draft_page_2 [2014/04/05 19:12]
wikimanager
draft_page_2 [2014/05/18 22:33] (current)
tom_grass [Practice/work here if other pages are temporarily locked by other editors]
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 +====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=== ===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? 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?
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     * 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.     * 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.
 {{ :​figs:​lec01fg7.png?​nolink |}} {{ :​figs:​lec01fg7.png?​nolink |}}
 +
 +
 +
 +
 +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_2.1396725146.txt.gz · Last modified: 2014/04/05 19:12 by wikimanager