tag:blogger.com,1999:blog-52444586741173631592024-03-06T01:20:04.422-08:00Daily GRETwo GRE questions a day. One Math. One Physics.Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.comBlogger35125tag:blogger.com,1999:blog-5244458674117363159.post-29840178447972522252011-09-05T21:08:00.000-07:002011-09-05T21:11:32.578-07:00Zai geen, sayonara, goodbye (for now at least).<br />
With the start of school looming so close, I fear that I won't have enough time to update this blog daily. Thus, I have decided to take a break on posting questions 'til winter vacation comes. I'll still be in the background gathering questions, but they won't be released to the public until December.<br />
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Thanks for all the comments, you guys really taught me a lot. I hope I was able to help all of you too (even if it was just the slightest bit).<br />
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- Paul<br />
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<span class="Apple-style-span" style="color: white;">PS: The title of this post is from a song my friend wrote. His alias is LittleColumbus. This text is in white so that I can properly attribute him while not turning this post into a bad advertisement.</span><br />
<span class="Apple-style-span" style="color: white;">PPS: It would be awesome if you guys spread the word about this blog. I'd love to get more people commenting so we can have more discussions.</span><br />
<span class="Apple-style-span" style="color: white;"><br /></span>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com5tag:blogger.com,1999:blog-5244458674117363159.post-39911670299978500662011-09-05T20:53:00.000-07:002011-09-05T20:53:51.727-07:00Physics GRE - # 34Eigenfunctions for a rigid dumbbell rotating about its centre have a $\phi$ dependence of the form $\psi(\phi)=Ae^{im\phi}$, where $m$ is a quantum number and $A$ is a constant. Which of the following values of $A$ will properly normalize the eigenfunction?<br />
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<ol>
<li>$\sqrt{2\pi}$</li>
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<li>$2\pi$</li>
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<li>$(2\pi)^2$</li>
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<li>$\dfrac{1}{\sqrt{2\pi}}$</li>
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<li>$\dfrac{1}{2\pi}$</li>
</ol>
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<b>Choice 4 is the answer.</b><br />
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For $\psi(\phi)$ to be normalized on $a$ to $b$, recall the normalization condition: \[\int_a^b{\left|\psi(\phi)\right|^2}d\phi=1.\] Since $\phi$ ranges from $0$ to $2\pi$, we have: \[\int_0^{2\pi}{\left|\psi(\phi)\right|^2}d\phi=\int_0^{2\pi}{\left|Ae^{im\phi}\right|^2}d\phi=\int_0^{2\pi}{|A|^2}d\phi=1\] where the second equality holds as $|e^{im\phi}|=1$ for any $m$ and any $\phi$.<br />
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Thus, we know that: \[\int_0^{2\pi}{|A|^2}d\phi=2\pi|A|^2=1\implies |A|=\dfrac{1}{\sqrt{2\pi}}.\]<br /></div>
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Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com2tag:blogger.com,1999:blog-5244458674117363159.post-36221823458095719872011-09-04T21:43:00.000-07:002011-09-04T21:43:54.957-07:00Physics GRE - #33A radioactive nucleus decays, with the activity shown in the plot below. The half-life of the nucleus is:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVJuOG2SyU1Rx7vuk3rYftQ1Y547P3UcXSsfjhAZJwHAZrKNMkWDLJUNY5UGMNs9RoUoRF538CswQE3zyBu04Lf2vSTfp_GN8y6KVDKBbMgFu4ffgykVfQJNhb4DVpZl4hE-_fIA0V-gFk/s1600/DecayPlot.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="395" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVJuOG2SyU1Rx7vuk3rYftQ1Y547P3UcXSsfjhAZJwHAZrKNMkWDLJUNY5UGMNs9RoUoRF538CswQE3zyBu04Lf2vSTfp_GN8y6KVDKBbMgFu4ffgykVfQJNhb4DVpZl4hE-_fIA0V-gFk/s400/DecayPlot.png" width="400" /></a></div>
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<ol>
<li>2 min</li>
<li>7 min</li>
<li>11 min</li>
<li>18 min</li>
<li>23 min</li>
</ol>
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Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div>
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<b>Choice 2 is the answer.</b><br />
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The initial activity is $6\cdot 10^3$ (yes, 6 since the line marked $10^3$ is really $1\cdot 10^3$ and you start counting up from there). At $3\cdot 10^3,$ the time is around 5 to 10 minutes. Thus the only possible answer is choice 2.<br /></div>
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Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-87413184803143341332011-09-03T19:57:00.000-07:002011-09-03T19:57:42.605-07:00Physics GRE - #32As shown below, a coaxial cable having radii $a$, $b$, and $c$ carries equal and opposite currents of magnitude $i$ on the inner and outer conductors. What is the magnitude of the magnetic induction at a point $P$ outside the cable at a distance $r$ from the axis?
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgVDVQZZP9i54FA1QaLrhg6c5mvaziaQRUuMBdgIGIJJhDpuaUm_rgV24qNjXcQSqIdhCRY5AzF-LKkH0AW0ZVx-CMk59O5YWHVv0G3DWl-R-Mlc6itTZVSU1ztrwzGjzZRwmwq5n-hrVY7/s1600/AmperesLawCable.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="277" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgVDVQZZP9i54FA1QaLrhg6c5mvaziaQRUuMBdgIGIJJhDpuaUm_rgV24qNjXcQSqIdhCRY5AzF-LKkH0AW0ZVx-CMk59O5YWHVv0G3DWl-R-Mlc6itTZVSU1ztrwzGjzZRwmwq5n-hrVY7/s320/AmperesLawCable.png" width="320" /></a></div>
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<ol>
<li>$0$</li>
<li>$\dfrac{\mu_0 ir}{2\pi a^2}$</li>
<li>$\dfrac{\mu_0 i}{2\pi r}$</li>
<li>$\dfrac{\mu_0 i}{2\pi r}\dfrac{c^2-r^2}{c^2-b^2}$</li>
<li>$\dfrac{\mu_0 i}{2\pi r}\dfrac{r^2-b^2}{c^2-b^2}$</li>
</ol>
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Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div>
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<b>Choice 1 is the answer.</b></div>
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The currents cancel out the magnetic fields created by each other, resulting in a net field of zero. More formally, this is an application of <a href="http://en.wikipedia.org/wiki/Amp%C3%A8re's_circuital_law#Integral_form">Ampere's Law</a>.<br /></div>
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Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-86545157443379747052011-09-01T22:33:00.000-07:002011-09-01T22:34:55.441-07:00Physics GRE - #31Which of the following is nearly the mass of the Earth? The radius of the Earth is about $6.4\cdot10^6\,\text{m}$.
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<ol>
<li>$6\cdot10^{24}\,\text{kg}$</li>
<li>$6\cdot10^{27}\,\text{kg}$</li>
<li>$6\cdot10^{30}\,\text{kg}$</li>
<li>$6\cdot10^{33}\,\text{kg}$</li>
<li>$6\cdot10^{36}\,\text{kg}$</li>
</ol>
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<b>Choice 1 is the answer.</b><br />
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If you don't remember the mass of the earth off the bat, recall that at ground level, \[mg=\frac{GmM_E}{R_E^2}\implies M=\frac{gR_E^2}{G}\] where $g$, $G$, and $R_E$ are given constants (a table of constants is given during the exam). We can plug in the numbers and compute \[M_E\approx6\cdot10^{24}\,\text{kg}.\]</div>
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Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com1tag:blogger.com,1999:blog-5244458674117363159.post-67829939569314804072011-08-31T23:02:00.000-07:002011-08-31T23:02:47.658-07:00Physics GRE - #30The energy levels of the hydrogen atom are given in terms of the principal quantum number $n$ and a positive constant $A$ by the expression:<br />
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<ol><li>$A\left(n+\frac{1}{2}\right)$</li>
<li>$A(1-n^2)$</li>
<li>$A\left(\frac{1}{n^2}-\frac{1}{4}\right)$</li>
<li>$An^2$</li>
<li>$-\frac{A}{n^2}$</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 5 is the answer.</b><br />
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Recall that the energy levels of the hydrogen atom is given by \[E_n = -\frac{13.6\,\text{eV}}{n^2}.\] It's just one of those things that you've got to remember, as it takes time you don't have to derive it from first principals (I only ever remember that $E_n\propto \frac{1}{n^2}\,$ since the constant of 13.6 eV is usually unimportant).</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-12566930506061120222011-08-30T21:52:00.000-07:002011-08-31T17:29:20.935-07:00Physics GRE - #29Solid argon is held together by which of the following bonding mechanisms?<br />
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<ol><li>Ionic bond only.</li>
<li>Covalent bond only.</li>
<li>Partly covalent and partly ionic.</li>
<li>Metallic bond.</li>
<li>Van der Waals bond</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 5 is the answer.</b><br />
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Argon is not charged and stable as a monoatomic element (due to it's full orbitals). Thus, because it is stable alone, it does not ionically bond with itself since argon does not exist naturally as an ion. This eliminates choice 1.<br />
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Because covalent bonding occurs with electron sharing and the stability from filling unfilled orbitals, Argon does not covalently bond with itself since it already has full orbitals. This eliminates choice 2.<br />
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It cannot be partly ionic and partly covalent as we just stated above that it is to stable to do either. This eliminates choice 3.<br />
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Argon is not a metal and so it does not participate in metallic bonding (metallic bonding occurs in metals due to massive electron sharing among many metal atoms, but again, due to argon's completely filled orbitals, it is too stable to do this). This eliminates choice 4.<br />
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Van der Waals bonding occurs in all chemicals and compounds as they are mostly caused by instantaneously inducted dipoles due to movement of electrons in the atom. Argon is no exception to this rule. Hence choice 5 is the answer. Note that although Van der Waals bonding is always present, it is by no means the strongest bonding mechanism. If this question was applied to other elements, Van der Waals may not be the answer since it could be negligible compared to other bonding mechanisms.</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-37911096681767808252011-08-29T22:07:00.000-07:002011-08-29T22:08:48.289-07:00Physics GRE - #28lalalala test element $^A_ZX$ decays by natural radioactivity in two stages to $^{A-4}_{Z-1}Y$. The two stages would most likely be which of the following?<br />
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<table><tbody>
<tr> <td></td> <td><b>First stage</b></td> <td><b>Second stage</b></td> </tr>
<tr> <td>1.</td> <td>$\beta^-$ emission with an antineutrino</td> <td>$\alpha$ emission</td> </tr>
<tr> <td>2.</td> <td>$\beta^-$ emission</td> <td>$\alpha$ emission with a neutrino</td> </tr>
<tr> <td>3.</td> <td>$\beta^-$ emission</td> <td>$\gamma$ emission</td> </tr>
<tr> <td>4.</td> <td>Emission of a deuteron</td> <td>Emission of two neutrons</td> </tr>
<tr> <td>5.</td> <td>$\alpha$ emission</td> <td>$\gamma$ emission</td> </tr>
</tbody></table><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 1 is the answer.</b><br />
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For choice 1, we have \[^A_ZX\rightarrow ^A_{Z+1}X+e^-+\bar{v}_e\] for the <a href="http://en.wikipedia.org/wiki/Beta_decay#.CE.B2.E2.88.92_decay">$\beta^-$ emission</a>, and then \[^A_{Z+1}X \rightarrow ^{A-4}_{Z-1}Y+^4_2\text{He}\] for the <a href="http://en.wikipedia.org/wiki/Alpha_decay">$\alpha$ emission</a>. Everything works out, charge/lepton/baryon numbers are conserved as they should be.<br />
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For choice 2, lepton number is not conserved as the $\beta^-$ emission does not occur with an antineutrino.<br />
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For choice 3, the atomic number remains unchanged under both emissions.<br />
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For choice 4, the emissions specified are extremely rare. In other words, this type of emission is impossible for the 'natural radioactivity' specified in the question.<br />
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For choice 5, the atomic numbers do not match up (there are fewer protons than needed with this decay path).</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-11348439248989562712011-08-28T23:33:00.000-07:002011-08-28T23:33:02.574-07:00Physics GRE - #27Suppose that there is a very small shaft in the Earth such that a point mass can be placed at a radius of $R/2$ where $R$ is the radius of the Earth. If $F(r)$ is the gravitational force of the Earth on a point mass at a distance $r$, what is: \[\frac{F(R)}{F(2R)}?\]<br />
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<ol><li>32</li>
<li>8</li>
<li>4</li>
<li>2</li>
<li>1</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 3 is the answer.</b><br />
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A very simple question today.<br />
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Recall that the gravitational force between two objects is given by \[F(r)=\frac{GMm}{r^2}=\frac{k}{r^2}\] where $k$ is a constant.<br />
Hence we have: \[\frac{F(R)}{F(2R)}=\left.\frac{k}{R^2}\right/\frac{k}{(2R)^2}=4.\]<br />
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On the same exam, they also asked for \[\frac{F(R)}{F\left(\frac{R}{2}\right)}.\] Try it if you want some extra practice.</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com6tag:blogger.com,1999:blog-5244458674117363159.post-59827549449201568352011-08-27T22:24:00.001-07:002011-08-27T22:25:02.604-07:00Physics GRE - #26An engine absorbs heat at a temperature of 727 degrees Celsius and exhausts heat at a temperature of 527 degrees Celcius. If the engine operates at maximum possible efficiency, for 2000 J of heat the amount of work the engine performs is most nearly:<br />
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<ol><li>400 J</li>
<li>1450 J</li>
<li>1600 J</li>
<li>2000 J</li>
<li>2760 J</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 1 is the answer.</b><br />
<br />
Recall that the <a href="http://en.wikipedia.org/wiki/Carnot_cycle#The_Carnot_cycle">Carnot Efficiency</a> of a heat engine is: \[\eta=1-\frac{T_C}{T_H}\] where $T_C$ and $T_H$ are in Kelvins. For this problem, the efficiency is: \[\eta\approx1-\frac{527+273}{727+273}=1-\frac{800}{1000}=0.2.\] Thus the maximum amount of work we can get is \[W=\eta\cdot2000\,J=400\,J.\]</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-22872024438001300652011-08-26T18:20:00.001-07:002011-08-26T18:21:52.678-07:00Physics GRE - #25A 2-kilogram box hangs by a massless rope from a ceiling. A force slowly pulls the box horizontally until the horizontal force is 10 N. The box is then in equilibrium as shown below.<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_wZr4zwI6lwzFEVPE_p5Ca_0A3MtPvunkzTkDMF7TD-tdbpKWUXg070DA9A6yLx_nczAyKalwGecL2tiX5_H33Evk5JcaBHaQkXSEh20SBpY-JovXDsY6AJmIrBsIhju-1I2U6QdJDcik/s1600/BoxInEqm.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_wZr4zwI6lwzFEVPE_p5Ca_0A3MtPvunkzTkDMF7TD-tdbpKWUXg070DA9A6yLx_nczAyKalwGecL2tiX5_H33Evk5JcaBHaQkXSEh20SBpY-JovXDsY6AJmIrBsIhju-1I2U6QdJDcik/s1600/BoxInEqm.png" /></a></div><br />
The angle that the rope makes with the vertical is closest to:<br />
<br />
<ol><li>$\arctan0.5$</li>
<li>$\arcsin0.5$</li>
<li>$\arctan2.0$</li>
<li>$\arcsin2.0$</li>
<li>$45^\circ$</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 1 is the answer.</b><br />
<br />
Since the block is in equilibrium, we must have $\vec{F}+\vec{F_g}=\vec{T}$ as shown below.<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhp8f7vpmrh3RPkwiSQaZUH-pepcUj6hCwG4ONukwvrWI0j6TItVC7hdB1ibmpOlgPThZiozz-3Q75Toiv8U5Y2AIYpobGPdYfDb9YSktntGuqHwuoOmGsd62bodR6YgWnt3iLVu5ABmXPD/s1600/Force+Diagram.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="241" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhp8f7vpmrh3RPkwiSQaZUH-pepcUj6hCwG4ONukwvrWI0j6TItVC7hdB1ibmpOlgPThZiozz-3Q75Toiv8U5Y2AIYpobGPdYfDb9YSktntGuqHwuoOmGsd62bodR6YgWnt3iLVu5ABmXPD/s320/Force+Diagram.png" width="320" /></a></div><br />
Hence $\theta=\arctan\frac{|\vec{F}|}{|\vec{F_g}|}=\arctan\frac{10}{2g}\approx\arctan\frac{10}{20}=\arctan0.5$.</div></div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-26869726288806114392011-08-25T22:52:00.000-07:002011-08-25T22:53:12.244-07:00Physics GRE - #24Assume $A$, $T$, $\lambda$ are positive constants. The equation \[y=A\sin\left[2\pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\right]\] represents a wave whose:<br />
<br />
<ol><li>amplitude is $2A$</li>
<li>velocity is in the negative x-direction</li>
<li>period is $\frac{T}{\lambda}$</li>
<li>speed is $\frac{x}{t}$</li>
<li>speed is $\frac{\lambda}{T}$</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 5 is the answer.</b><br />
<br />
Choice 1 is incorrect because $\sin\theta\leq 1$ for any $\theta$ so $y\leq A$.<br />
<br />
To see that choice 2 is incorrect, suppose that we are following the minimum wave as time passes (by changing $x$ with respect to time to a specific $x(t)$) in such a way that the wave seems to stand still (i.e. we are travelling at the same velocity as the wave). For the wave to 'stand still', we need $x(t)$ to satisfy \[2\pi\left(\frac{t}{T}-\frac{x(t)}{\lambda}\right) = const. = C\] Satisfying this relationship ensures that the wave always looks the same to us as: \[y=A\sin\left[2\pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\right] = A\sin C = constant.\] Since we are following one of the minimums of the wave, we can choose $C = 0$ to get the minimum at $\sin 0$. Thus we get: \[2\pi\left(\frac{t}{T}-\frac{x(t)}{\lambda}\right) = 0 \implies x(t) = \frac{\lambda}{T}t.\] To get the sign of the velocity, we take the derivative of $x(t)$: \[\frac{dx}{dt} = \frac{\lambda}{T}>0\] since $\lambda,T > 0$. This also tells us choice 5 is correct since the speed is $\frac{\lambda}{T}.$<br />
<br />
Choice 3 is wrong as the units do not work out.<br />
<br />
Choice 4 is wrong because we just calculated the speed to be $\frac{\lambda}{T}.$</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-37980921544199429232011-08-24T03:45:00.001-07:002011-08-24T03:45:29.025-07:00Physics GRE - #23A particle is initially at rest at the top of a curved frictionless track.<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEisODhIROHdgeBWfRzRDheeQfk-e5AXlodzQp0q4zHcRsov9G140IrhUmD0-_AcCMD3WGcVw-yTQJ_N4k52p6j9Nh3zxbkClf9DLi-PMrkU2dNUMelZr-J_BHT-QuE5cq2UaF6YrO9Nhifz/s1600/FrictionlessParabolicTrack.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="152" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEisODhIROHdgeBWfRzRDheeQfk-e5AXlodzQp0q4zHcRsov9G140IrhUmD0-_AcCMD3WGcVw-yTQJ_N4k52p6j9Nh3zxbkClf9DLi-PMrkU2dNUMelZr-J_BHT-QuE5cq2UaF6YrO9Nhifz/s200/FrictionlessParabolicTrack.png" width="200" /></a></div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;">The x and y coordinates of the track are related in dimensionless units by $y=\frac{x^2}{4}$, where the positive y-axis is in the vertical downward direction. As the particle slides down the track, what is its tangential acceleration?</div><div class="separator" style="clear: both; text-align: left;"></div><ol><li>$0$</li>
<li>$g$</li>
<li>$\frac{gx}{2}$</li>
<li>$\frac{gx}{\sqrt{x^2+4}}$</li>
<li>$\frac{gx^2}{\sqrt{x^2+16}}$</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 4 is the solution.</b><br />
<br />
First we note that choice 2 cannot be the answer as there is a normal force from the track.<br />
<br />
Now as $x\rightarrow\infty$, we should expect that the acceleration goes to $g$ as the tangent becomes nearly vertical. This makes choice 4 is the only possible answer.<br />
<br />
We can also solve this question elegantly without resorting to limits and boundary conditions (however it will take a good half page). Can you see how it can be done?</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-47141938383605685382011-08-23T22:07:00.000-07:002011-08-23T22:07:35.936-07:00Physics GRE - #22Two wedges, each of mass $m$, are placed next to each other on a flat floor. A cube of mass $M$ is balanced on the wedges as shown below.<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEil9PirRxITvvXe935lZHwv_vOvY5p-HIbd1gGIrtUDZrmnkotMBhUX_AZyggQQL0y4p7RdY6tWCQnstmN95H46wSlzPv9BHSWtX93AdR3-jlXBmW3yqgJWfe2x1yImvWyhuumciTQGOjvm/s1600/WedgeBlockWedge.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="167" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEil9PirRxITvvXe935lZHwv_vOvY5p-HIbd1gGIrtUDZrmnkotMBhUX_AZyggQQL0y4p7RdY6tWCQnstmN95H46wSlzPv9BHSWtX93AdR3-jlXBmW3yqgJWfe2x1yImvWyhuumciTQGOjvm/s320/WedgeBlockWedge.png" width="320" /></a></div><br />
Assume no friction between the cube and the wedges, but a coefficient of static friction $\mu < 1$ between the wedges and the floor. What is the largest $M$ that can be balanced as shown without motion of the wedges? <br />
<br />
<ol><li>$\frac{m}{\sqrt{2}}$</li>
<li>$\mu\frac{m}{\sqrt{2}}$</li>
<li>$\mu\frac{m}{1-\mu}$</li>
<li>$2\mu\frac{m}{1-\mu}$</li>
<li>All $M$ will balance.</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><br />
<div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 4 is the answer.</b><br />
<br />
We know immediately it's either choice 3 or 4 since any $M$ will balance if $\mu\rightarrow 1$, but it looks like we're going to have to do the work to figure out which one it is.<br />
<br />
Due to the symmetry of the problem, we may assume that one wedge holds up a mass of $\frac{M}{2}$. By breaking down the force of the block on the wedge into components, we can see that the normal force on the wedge is \[N = \left(m+\frac{M}{2}\right)g.\] So the friction force on the wedge is \[F_{fr} = \mu\left(m+\frac{M}{2}\right)g.\] Since the friction force has to equal the mass of half the block (recall that we are only considering one wedge holding up a block of mass $M/2$), we know that \[W_{1/2 - block} = \frac{M}{2}g = \mu\left(m+\frac{M}{2}\right)g.\] We can then solve this equation to obtain \[M=2\mu\frac{m}{1-\mu}.\]<br />
What if the wedges weren't angled at 45 degrees? Can you find a general expression for $M$ at any angle?</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-49437745443235222702011-08-22T01:40:00.001-07:002011-08-22T01:40:30.307-07:00Physics GRE - #21<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiaoBk682Qc6ltby7xH_iyyHKl64J3kmEO8dAnz91D8XCuQDvafevNAl1DWtp5bNRSKsaCd3mF3MGAKAFnzESHhyYo3TvbmTHozV5aAOBFq0IhniR1dLwIq4OYHcfaTbD870Syyk1dcUWnD/s1600/RingDiagram.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="172" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiaoBk682Qc6ltby7xH_iyyHKl64J3kmEO8dAnz91D8XCuQDvafevNAl1DWtp5bNRSKsaCd3mF3MGAKAFnzESHhyYo3TvbmTHozV5aAOBFq0IhniR1dLwIq4OYHcfaTbD870Syyk1dcUWnD/s320/RingDiagram.png" width="320" /></a></div><br />
The electric potential at point P (for the ring above), which is located on the axis of symmetry a distance $x$ from the centre of the ring, is given by:<br />
<br />
<ul><li>$\dfrac{Q}{4\pi\epsilon_0x}$</li>
<li>$\dfrac{Q}{4\pi\epsilon_0\sqrt{R^2+x^2}}$</li>
<li>$\dfrac{Qx}{4\pi\epsilon_0(R^2+x^2)}$</li>
<li>$\dfrac{Qx}{4\pi\epsilon_0(R^2+x^2)^{3/2}}$</li>
<li>$\dfrac{QR}{4\pi\epsilon_0(R^2+x^2)}$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">Recall that \[V=\int{\vec{E}\cdot d\vec{l}}=\int{\frac{dq}{4\pi\epsilon_0|\vec{r}|}}.\]<br />
Since we are essentially summing up the charge over the entire ring (i.e. we are integrating around the ring with respect to $q$), we know that $int{dq}=Q$, where $Q$ is the charge of the entire ring. Therefore, \[\int{\frac{dq}{4\pi\epsilon_0|\vec{r}|}}=\frac{Q}{4\pi\epsilon_0|\vec{r}|}=\frac{Q}{4\pi\epsilon_0\sqrt{R^2+x^2}}\] where $|\vec{r}|=\sqrt{R^2+x^2}$ as indicated in the diagram below (except they used $z$ for $x$).<br />
<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3qHaIbBLctc4EK-QtTdyz2Rnvfo895SxjNDe2gJNJz2Jb0whsrynCs2mMK2MP4OSsbC8wXMy1jrpdqF5b_yswNG3cy3yd02nuyJWK0QD7w1lZpwwhtmXZikUcVZwBh7HEvd0CnOCHD6-L/s1600/PotentialOfARing.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="282" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3qHaIbBLctc4EK-QtTdyz2Rnvfo895SxjNDe2gJNJz2Jb0whsrynCs2mMK2MP4OSsbC8wXMy1jrpdqF5b_yswNG3cy3yd02nuyJWK0QD7w1lZpwwhtmXZikUcVZwBh7HEvd0CnOCHD6-L/s320/PotentialOfARing.png" width="320" /></a></div></div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-71600181829060104602011-08-21T00:34:00.000-07:002011-08-21T00:34:00.826-07:00Physics GRE - #20According to the Standard Model of elementary particles, which of the following is <b>not</b> a composite object?<br />
<br />
<ul><li>Muon</li>
<li>Pi-meson</li>
<li>Neutron</li>
<li>Deuteron</li>
<li>Alpha particle</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">The answer is muon. A muon is an elementary particle (a lepton).<br />
<br />
Pi - mesons are a quark with an antiquark.<br />
A neutron is made of two down quarks and an up quark.<br />
A deuteron is a proton and a neutron (a proton is two up quarks and a down quark).<br />
An alpha particle is two protons and two neutrons.</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-25513972975034875872011-08-20T03:28:00.000-07:002011-08-20T03:28:30.509-07:00Physics GRE - #19As shown below, a ball of mass $m$, suspended on the end of a wire, is released from height $h$ and collides elastically, when it is at its lowest point, which a block of mass $2m$ at rest on a frictionless surface. After the collision, the ball rises to a final height equal to:<br />
<br />
<ul><li>$\dfrac{h}{9}$</li>
<li>$\dfrac{h}{8}$</li>
<li>$\dfrac{h}{3}$</li>
<li>$\dfrac{h}{2}$</li>
<li>$\dfrac{2h}{3}$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">Using conservation of energy, we know that: \[2mgh = mv_0^2 = mv_1^2+2mv_2^2\] where $v_0$ is the initial velocity and $v_1$, $v_2$ are the resulting velocities after the collision.<br />
<br />
Using conservation of momentum, we also know that \[mv_0 = mv_1+2mv_2.\]<br />
<br />
Solving these equations give \[v_1 = -\frac{v_0}{3}, v_2 = \frac{2v_0}{3}\] which make sense as the ball must bounce back after the collision. Now using conservation of energy again, we can determine the height that the ball will rise to: \[mgh_{new} = \frac{1}{2}mv_1^2 = \frac{1}{2}m\frac{v_0^2}{9}= \frac{mgh}{9}\implies h_{new}=\frac{h}{9}.\]<br />
<br />
There is a faster way to find $v_1$ using the concept of a centre of mass. Can you see it?<br />
</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com2tag:blogger.com,1999:blog-5244458674117363159.post-72145045829834607852011-08-18T19:35:00.001-07:002011-08-20T02:16:07.019-07:00Physics GRE - #18A string consists of two parts attached together.<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgScepDdWTwBIx74PEaN1ew5bEqYY4FGxtqppG3DzDlKdNxStk8z3jf1_OTQRDgPng3EwuUnbvvFyTp6cHqykOVgL26zm7fOiKjp2XqLgzlcy4lPVvXdTvAQ3rJQyupULrXksV2chg618Xa/s1600/WaveOnString.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgScepDdWTwBIx74PEaN1ew5bEqYY4FGxtqppG3DzDlKdNxStk8z3jf1_OTQRDgPng3EwuUnbvvFyTp6cHqykOVgL26zm7fOiKjp2XqLgzlcy4lPVvXdTvAQ3rJQyupULrXksV2chg618Xa/s1600/WaveOnString.png" /></a></div><br />
The right part of the string has mass $\mu_r$ per unit length and the left part of the string has mass $\mu_l$ per unit length. If a wave of unit amplitude travels along the left part of the string, what is the amplitude of the wave that is transmitted to the right part of the string?<br />
<br />
<ol><li>$1$</li>
<li>$\dfrac{2}{1+\sqrt{\mu_l / \mu_r}}$</li>
<li>$\dfrac{2\sqrt{\mu_l / \mu_r}}{1+\sqrt{\mu_l / \mu_r}}$</li>
<li>$\dfrac{\sqrt{\mu_l / \mu_r}-1}{\sqrt{\mu_l / \mu_r}+1}$</li>
<li>$0$</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">The correct answer is choice 3.<br />
<br />
To solve the problem we can look at some nice limiting cases.<br />
First, suppose $\mu_l =\mu_r$, then there should be no difference in amplitude as the wave travels (i.e. amplitude is 1 under this condition). <b>Choices 1, 2, and 3 satisfy this.</b><br />
<b><br />
</b><br />
Second, suppose $\mu_r\rightarrow\infty$. Then the wave should be completely reflected at the part where the two strings join (imagine a string attached to a wall). Thus, we should have amplitude equal to 0 under this condition. <b>Only choice 3 satisfy this.</b></div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com1tag:blogger.com,1999:blog-5244458674117363159.post-67960939978030792302011-08-17T02:16:00.000-07:002011-08-20T02:17:51.309-07:00Physics GRE - #17A block of mass $m$ sliding down an incline at constant speed is initially at height $h$ above the ground, as shown in the figure below.<br />
<br />
The coefficient of kinetic friction between the mass and the incline is $\mu$. If the mass continues to slide down the incline at a constant speed, how much energy is dissipated by friction by the time the mass reaches the bottom of the incline?<br />
<br />
<ul><li>$\dfrac{mgh}{\mu}$</li>
<li>$mgh$</li>
<li>$\dfrac{\mu mgh}{\sin\theta}$</li>
<li>$mgh\sin\theta$</li>
<li>$0$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">The answer is $mgh$.<br />
<br />
Recall conservation of energy. Since the velocity stays the same at the top and bottom, friction dissipated all of the gravitational potential energy. Thus the value of energy dissipated is $mgh$.<br />
In equation form: \[E_i = E_f \implies mgh+\frac{1}{2}mv_i^2=E_{fr}+\frac{1}{2}mv_f^2.\] But since $v_i=v_f$, \[mgh=E_{fr}.\]</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-25714019642313774292011-08-17T00:28:00.000-07:002011-08-18T04:15:51.991-07:00Physics GRE - #16Two circular hoops, X and Y, are hanging on nails in a wall. The mass of X is four times that of Y, and the diameter of X is also four times that of Y. If the period of small oscillations of X is $T$, the period of small oscillations of Y is:<br />
<ul><li>$T$</li>
<li>$\dfrac{T}{2}$</li>
<li>$\dfrac{T}{4}$</li>
<li>$\dfrac{T}{8}$</li>
<li>$\dfrac{T}{16}$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>The answer is</b> $\dfrac{T}{2}$.<br />
<br />
The key thing to note here is that the period of small oscillations of a circular hoop is the same as that of a normal pendulum with the mass where the centre of the hoop is.<br />
Thus, the period is given by the formula (we'll disregard the constant of $2\pi$ at the front, since it will not matter to our calculations):<br />
\[T=\sqrt{\frac{r}{g}}.\] For hoops X, we have \[T_x=\sqrt{\frac{r_x}{g}}.\] For hoop Y, we have \[T_y=\sqrt{\frac{r_y}{g}}.\] Since $r_y = \frac{1}{4}r_x$, we have: \[T_y=\sqrt{\frac{1}{4}\frac{r_x}{g}}=\frac{1}{2}\sqrt{\frac{r_x}{g}}=\frac{T}{2}.\]</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-40911729790821877712011-08-16T00:15:00.001-07:002011-08-16T00:17:16.212-07:00Physics GRE - #15Two positive charges of $q$ and $2q$ coulombs are located on the x-axis at $x=0.5a$ and $1.5a$, respectively, as shown below.<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg49rxGpr9v6RrknDyhEE1B7cLL76552ByKa9QfP6KtPsoqs8-J0UaFMUsnZERgRFLUlNi9frZe81lLJJ4RwznWmylttgo56WANx_Qp9sv2vipEsd-RMUuRKIbolR3XMEF5rr76EBtwU24A/s1600/ChargesOnALine.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg49rxGpr9v6RrknDyhEE1B7cLL76552ByKa9QfP6KtPsoqs8-J0UaFMUsnZERgRFLUlNi9frZe81lLJJ4RwznWmylttgo56WANx_Qp9sv2vipEsd-RMUuRKIbolR3XMEF5rr76EBtwU24A/s1600/ChargesOnALine.png" /></a></div><br />
There is an infinite, grounded conducting plane at $x=0$. What is the magnitude of the net force on charge q?<br />
<br />
<ul><li>$\dfrac{1}{4\pi\epsilon_0}\frac{q^2}{a^2}$</li>
<li>$\dfrac{1}{4\pi\epsilon_0}\frac{3q^2}{2a^2}$</li>
<li>$\dfrac{1}{4\pi\epsilon_0}\frac{2q^2}{a^2}$</li>
<li>$\dfrac{1}{4\pi\epsilon_0}\frac{3q^2}{a^2}$</li>
<li>$\dfrac{1}{4\pi\epsilon_0}\frac{7q^2}{2a^2}$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>The last choice is correct.</b><br />
<br />
Recall that Coulomb's Law states the force between two point charges to be: \[F= \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\] where $q_1$ and $q_2$ are the size of the two charges, $r$ is the distance between the two charges, and all other variables are constants.<br />
<br />
Since the conductor is grounded, <a href="http://en.wikipedia.org/wiki/Method_of_image_charges">image charges</a> of $-q$ and $-2q$ are induced at $0.5a$ and $-1.5a$ respectively. These image charges pulls charge $q$ left, while the real charge $2q$ pushes charge $q$ left. Therefore we have three contributions (counted from left to right) to the force on charge $q$: \[\begin{eqnarray*}<br />
\sum{F} & = & \frac{q}{4\pi\epsilon_0}\left(\frac{2q}{(-1.5a-0.5a)^2}+\frac{q}{(-0.5a-0.5a)^2}+\frac{2q}{(1.5a-0.5a)^2}\right) \\<br />
& = & \frac{1}{4\pi\epsilon_0}\frac{7q^2}{2a^2}.<br />
\end{eqnarray*}\]</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com5tag:blogger.com,1999:blog-5244458674117363159.post-62457594708775464372011-08-15T03:09:00.000-07:002011-08-15T03:09:05.810-07:00Physics GRE - #14The figure below shows a small mass connected to a string, which is attached to a vertical post.<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhxwJ4zS89dkYU42dJeaPfDwUtPA2TgxQEbfNjmgY2A2beQl4UwsQbdj5RAZXoldhQTlYhtgTsZxpcQU73C7T4RXNCM30Rlpmf3jy53BF8B5-7W-SxIla5-lQqoZWn_gDgt6v1flGKepDnq/s1600/MassOnString.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="154" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhxwJ4zS89dkYU42dJeaPfDwUtPA2TgxQEbfNjmgY2A2beQl4UwsQbdj5RAZXoldhQTlYhtgTsZxpcQU73C7T4RXNCM30Rlpmf3jy53BF8B5-7W-SxIla5-lQqoZWn_gDgt6v1flGKepDnq/s320/MassOnString.png" width="320" /></a></div><br />
If the mass is released when the string is horizontal as shown, the magnitude of the total acceleration of the mass as a function of the angle $\theta$ is:<br />
<br />
<ul><li>$g\sin\theta$</li>
<li>$2g\cos\theta$</li>
<li>$2g\sin\theta$</li>
<li>$g\sqrt{3\cos^2{\theta}+1}$</li>
<li>$g\sqrt{3\sin^2{\theta}+1}$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">We know that at $\theta=0$, the acceleration of the mass is exactly $g$. The only solution that satisfies this boundary condition is the last choice.</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-34777486884344513202011-08-14T01:08:00.001-07:002011-08-14T01:08:56.826-07:00Physics GRE - #13Two pendula are attached to a massless spring, as shown below. The arms of the pendula are both of length $l$, but the pendulum balls have unequal masses $m_1$ and $m_2$. The initial distance between the masses is the equilibrium length of the spring, which has spring constant $K$. What is the highest normal mode frequency of this system?<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj9ck5QXJT5jPWro9Py0AV3L8mAUUgMcwImHBbDoScLgrImOv8pR1malW-5QQO-uLXva95dC4HLi8Bl7dFe8edXOeN-8RwuYjRTvGqhHA3rp2TKCe2TY0x9lpEBowgb3ZJwUjIZNoUcPVr6/s1600/Normal+Mode+Springs.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj9ck5QXJT5jPWro9Py0AV3L8mAUUgMcwImHBbDoScLgrImOv8pR1malW-5QQO-uLXva95dC4HLi8Bl7dFe8edXOeN-8RwuYjRTvGqhHA3rp2TKCe2TY0x9lpEBowgb3ZJwUjIZNoUcPVr6/s1600/Normal+Mode+Springs.png" /></a></div><ol><li>$\sqrt{\frac{g}{l}}$</li>
<li>$\sqrt{\frac{K}{m_1+m_2}}$</li>
<li>$\sqrt{\frac{K}{m_1}+\frac{K}{m_2}}$</li>
<li>$\sqrt{\frac{g}{l}+\frac{K}{m_1}+\frac{K}{m_2}}$</li>
<li>$\sqrt{\frac{2g}{l}+\frac{K}{m_1}+\frac{K}{m_2}}$</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>The correct answer is choice 4.</b><br />
<br />
The easy way of solving this problem is to look at the limiting behaviour of the system in several ways. First we see that if $m_1\rightarrow\infty$ and $l\rightarrow\infty$, the resulting behaviour should be the same as a single mass, $m_2$, attached to a wall by a spring, thus we should have $\sqrt{\frac{K}{m_2}}$. This just <b>leaves options 3 and 4 as possibilities</b>. Then we note that if $K\rightarrow 0$, it would be as if there was no spring between the two pendula, and so we should get $\sqrt{\frac{g}{l}}$. This <b>leaves choice 4 as the right answer</b>. This solution sure beats deriving the normal frequencies from scratch (which I'm not even sure I remember how to do).<br />
<br />
<b><span class="Apple-style-span" style="font-size: large;">A note on looking at limiting behaviours</span></b><br />
Looking at limiting behaviour is a useful technique for doing a quick check of your answer or in multiple choice questions such as these. Sometimes looking at a system this way can also give you nice approximate answers. Common methods of looking at limiting behaviour usually involves making all masses/lengths/constants equal in a problem, or taking them to 0/∞.</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-23090629091228195202011-08-13T02:38:00.000-07:002011-08-13T02:38:54.887-07:00Physics GRE - #12A ball is dropped from a height h. As it bounces off the floor, its speed is 80% of what it was just before it hit the floor. The ball will then rise to a height of about:<br />
<br />
<ul><li>0.94 h</li>
<li>0.80 h</li>
<li>0.75 h</li>
<li>0.64 h</li>
<li>0.50 h</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">Conservation of energy tells us that $mgh = \frac{1}{2}mv^2$. Therefore, we know that $h\propto v^2$. Since $v_{new}=0.8v_{old}$, $h_{new} = (0.8)^2h_{old} = 0.64h$.<br />
<br />
It's a simple problem, but it may just mess you up if you aren't careful.</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-38299141935477640582011-08-12T01:29:00.001-07:002011-08-12T01:29:54.644-07:00Physics GRE - #11A pendulum of length $l$ is attached to the roof of an elevator near the surface of the Earth.<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgI6f_bhJUbHEOGDBmbnAdh2RgVxyZN_3z4EaVP-fXnAt8aV87tuWbiXIPrTbi7ELqmAKP8H8k77GIvebTb7DYYlSCl1i3Ws6AL546_T8dPQNabpB4tb9cMoG1wR9UQooyJad5yEM86D5wW/s1600/PendulumUnlabelled.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="181" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgI6f_bhJUbHEOGDBmbnAdh2RgVxyZN_3z4EaVP-fXnAt8aV87tuWbiXIPrTbi7ELqmAKP8H8k77GIvebTb7DYYlSCl1i3Ws6AL546_T8dPQNabpB4tb9cMoG1wR9UQooyJad5yEM86D5wW/s200/PendulumUnlabelled.png" width="200" /></a></div><br />
The elevator moves upward with acceleration $\frac{1}{2}g$. The angular frequency is:<br />
<br />
<ul><li>$\sqrt{\frac{3g}{2l}}$</li>
<li>$\sqrt{\frac{2g}{3l}}$</li>
<li>$\sqrt{\frac{g}{l}}$</li>
<li>$\sqrt{\frac{g}{2l}}$</li>
<li>$\sqrt{\frac{2g}{l}}$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Quick solution:</b><br />
The angular frequency of a pendulum is $\omega=\sqrt{\frac{g}{l}}$ when the gravitational field has value $g$. In this problem, the effective gravity is $g_{eff} = g + \frac{1}{2}g = \frac{3}{2}g$. \[\therefore \omega=\sqrt{\frac{g_{eff}}{l}}=\sqrt{\frac{3g}{2l}}.\]<br />
<br />
<b>Full solution: </b><br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEje86c4e0XHI3PNcfB4uJdWGZu8c8RAszX9TeAicjtQuvw14bEgpyIFAU18RAsjor4CIUHp2R-lptVtP0kyu1ENRRaTJRiE72ZWpTpy-4tg0eqaGuub298b1B8Qeb419LSTToPYqT3IxwyY/s1600/PendulumLabelled.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="186" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEje86c4e0XHI3PNcfB4uJdWGZu8c8RAszX9TeAicjtQuvw14bEgpyIFAU18RAsjor4CIUHp2R-lptVtP0kyu1ENRRaTJRiE72ZWpTpy-4tg0eqaGuub298b1B8Qeb419LSTToPYqT3IxwyY/s200/PendulumLabelled.png" width="200" /></a></div><div class="separator" style="clear: both; text-align: left;">By Newton's 2nd law for rotational motion, \[\sum{\tau}=I\alpha \implies -mg_{eff}l\sin\theta=I\ddot{\theta}.\]</div><div class="separator" style="clear: both; text-align: left;">Substituting $I = ml^2$ and $g_{eff} = \frac{3}{2}g$, we have</div><div class="separator" style="clear: both; text-align: left;">\[-\frac{3}{2}mgl\sin\theta=ml^2\ddot{\theta}\implies \ddot{\theta}=-\frac{3g}{2l}\sin\theta\approx-\frac{3g}{2l}\theta\] \[\therefore \omega^2 = \frac{3g}{2l} \implies \omega=\sqrt{\frac{3g}{2l}}.\]</div></div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0