# Physics 203 at Portland State 2014

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• $v_{13}=.21c$
• $v_{23}=.34c$
• $v_{21}=?$
1. $v_{21}=\frac{v_{23} - v_{13}}{1-\frac{(v_{22})(v_{13})}{c^2}}$
2. $=\frac{.34c -.21c}{1-\frac{(.34c)(.21c)}{c^2}}$
3. $=\frac{.13c}{.9286} =.14c$
• Equations:
• $d\sin\theta=m\lambda$
• $y=L\tan\theta$
• To solve for $y$, the distance between bright fringes, find $\theta$ when $m=1$.
1. $\sin\theta=\frac{7.5\times 10^{-7}}{0.001}$
2. $\theta=\sin^{-1}\left(\frac{7.5\times 10^{-7}}{0.001}\right)$
3. $\theta=7.5\times 10^{-4}\,$rad
4. $y=L\tan\theta$
5. $y=15\,{\text m}\cdot\tan\left( 7.5\times 10^{-4}\right)\approx 0.011\,{\text m}=1.1\,{\text{cm}}$

## ...or for working on your edit if the main page is locked by another user

1. Focal length
• $\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$
• $f = \left[\frac{1}{5\,{\text m}} + \frac{1}{0.08\,{\text m}}\right]^{-1}$ = 7.8 cm
2. Image Size
• $h_i = h_o\left(\frac{-d_i}{d_o}\right)$ $=1.7\,{\text m}\left(\frac{-0.08\,{\text m}}{-5\,{\text m}}\right)$ = 27.2 mm
3. A positive number means an upright orientation

#### Yeah, welcome to the "sandbox"

 Heading 1 Heading 2 Row 1 Col 2 Row 1 Col 3 no colspan this time Row 2 Col 2 Row 2 Col 3

$\frac{n!}{k!(n-k)!} = \binom{n}{k}$ ## Announcements

past announcements have moved to sidebar and can be found here.

### Almost ready for Apr 1!

• Everyone who edits this will automatically get 5 points extra credit — Nicholas Kuzma 2014/04/01 00:01
• Good luck!

I need to figure out how to insert a video, instead of the hyperlink? Tom. How to embed a YouTube video in Wiki Page? Copy YouTube URL Edit your page. Click on Widgets icon in the editor toolbar Click on YouTube Video in widget dialog box Paste the YouTube URL in text box. Click on Insert button and Save Page.

Figured it out.

## Lectures and Notes

Use WolframAlpha for calculations

## Ch.13: Oscillations (Apr 1)

### Constants Ch.13

• Standard gravity $g=9.80665\,\frac{\text m}{ {\text s}^2}\approx 9.8\,\frac{\text m}{ {\text s}^2}$ ### Math Ch.13

Addition of angels Doubling of angels Squaring of trick functions
$\sin(\alpha+\beta)=\sin\alpha+\sin\beta$ $\sin(2\alpha)=2\sin\alpha$ $\sin^2\alpha =\sin(\alpha^2)$

### Examples Ch.13

#### Problem 3

Without glass, one can see at x meters

#### Problem 13.13 A mouse's heart rate increases from 588 Batman wingflaps (BMW) to 612 BMW in two minutes. What is the change in the heart period?

Steps:

1. Convert units to SI:
• $f_1=588\,{\text{BMW}}$ $=\frac{588}{60}\,{\text{Hs}}=9.8\,{\text{Hs}}$
• $f_2=612\,{\text{BMW}}$ $=\frac{612}{60}\,{\text{Hs}}=10.2\,{\text{Hs}}$
2. Convert frequencies to period:
• $T_1=\frac{1}{f_1}=\frac{1}{9.8\,{\text{Hs}}}$ $\approx 0.102\,{\text s}=102\,$ms
• $T_2=\frac{1}{f_1}=\frac{1}{10.2\,{\text{Hs}}}$ $\approx 0.098\,{\text s}=98\,$ms
3. Find the change:
• $\Delta T=T_2-T_1=-4\,$ms
4. Make sense of the answer:
• The mouse heart started to beat faster, its period decreased by 4 ms (by about 4%). #### Problem 13.13.13

A “pocket balance” (shown to the right) extends by 1/4 of an inch when loaded with a 3 kg mass. Find the frequency with which this mass will oscillate around the equilibrium. Hint: Find the spring constant first.

Steps:

1. Convert units to SI:
• $\Delta l=\frac{1}{4}\,{\text{inch}} = \frac{0.0254\,{\text m}}{4}$ $=0.00635\,{\text m}=6.35\,$mm
2. Find the spring constant from the balance of gravity and spring forces on the mass:
• $mg-k\Delta l=0\;\;\Rightarrow$
• $k=\frac{mg}{\Delta l}=\frac{3\,{\text{kg}}\,\cdot\,9.8\,\frac{\text m}{ {\text s}^2}}{0.00635\,{\text m}}$ $=4630\,\frac{\text N}{\text m}$
3. Find the frequency:
• $f =\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ $\approx\frac{1}{6.283}\sqrt{\frac{4630\,\frac{\text N}{\text m}}{3\,{\text{kg}}}}$ $=6.25\,$Hz
4. Make sense of the answer:
• This is still slow enough to see by eye. Anyone has such a scale at home?

#### Problem 1.3

It takes 5 lb of force to charge a small spring-loaded toy gun. If fired straight up, it shoots a 10-g projectile 40 feet high. Find the frequency of oscillation if the projectile gets stuck to the spring.

Steps: 1. Convert all units to SI:
• 10 g = 0.01 kg
• 40 ft = $40\,{\text{ft}}\times 12\,\frac{\text{inch}}{\text{ft}}\times 0.0254\,\frac{\text m}{\text{inch}}$ $=12.19\,$m
• 5 lb of force = $5\,{\text{lb}}\cdot 0.454\,\frac{\text{kg}}{\text{lb}}\cdot 9.8\,\frac{\text m}{ {\text s}^2}$ $=22.2\,$N
2. Use the charging force to relate the spring compression $\Delta l$ to the spring constant $k$:
• $k\cdot|\Delta l|=F=22.2\,$N
3. Instead of getting bogged down in kinematics, use the conservation of energy:
• Pot. Energy of loaded spring (before the shot) =
= Kin. Energy of projectile (right after the shot) =
= Pot. energy of projectile (at max. height)
• $\frac{1}{2}k\,(\Delta l)^2=mgh$
4. We have two equations with two unknowns (k and $\Delta l$). Let's square of the force equation and divide it by the equation above:
• $\frac{k^2(\Delta l)^2}{\frac{1}{2}k\,(\Delta l)^2}=\frac{F^2}{mgh}$ $\;\Rightarrow$
• $2k=\frac{F^2}{mgh}\;\Rightarrow$
• $k=\frac{F^2}{2mgh}$ $=\frac{(22.2\,{\text N})^2}{2\,\cdot\,0.01\,{\text{kg}}\,\cdot\,9.8\,\frac{\text m}{ {\text s}^2}\,\cdot\,12.19\,{\text m}}$ $=206\,\frac{\text N}{\text m}$
5. Using this spring constant and the mass, find the frequency:
• $f =\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ $\approx\frac{1}{6.283}\sqrt{\frac{206\,\frac{\text N}{\text m}}{0.01\,{\text{kg}}}}$ $=22.8\,$Hz
6. Make sense of the answer:
• This is at the lower range of what human ear can detect: “Wrrrrrr!”
• The compression $|\Delta l|$ comes out to be 0.108 m, or about 4 inches. Quite reasonable for a toy gun.

#### 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 