GR 8677927796770177 | # Login | Register

GR9677 #26
Problem
 GREPhysics.NET Official Solution Alternate Solutions
This problem is still being typed.
Electromagnetism$\Rightarrow$}Resonance Frequency

One wants to tune one's radio to the resonance frequency (a.k.a. the frequency at which impedance is matched). The resonance frequency of an LRC circuit is given by $\omega^2 = 1/LC$, where the quantities involved are angular frequency, inductance, and capacitance. Solving for C, one has $C=1/(L\omega^2) \approx 1/(2E-6 * 36 * 100E6) =1/(7.2E-11)\approx 0.1E-11 =1E-12$. This is choice (C). The hardest part of his problem, of course, is doing the math without a calculator. Easy.

Alternate Solutions
 There are no Alternate Solutions for this problem. Be the first to post one!
wittensdog
2009-10-30 13:47:27
I'm noticing that in a lot of cases, you can make some pretty liberal approximations and still be okay for the GRE. In this problem, if you say pi^2 = 10 , and that the broadcast frequency is 100 MHz and not 103.7 MHz, the math becomes amazingly faster. Now, with approximations like that, we're certainly adding some small errors, but in the case of the frequency it's in the third significant digit, and the difference between pi^2 and 10 in reality is only about 0.13. Now, the answer you get with this method is (1/8) * 10 ^ -11, which works out to be 1.25 * 10 ^ -12, or about a picoFarad. Without even knowing the decimal value of 1/8, you can see that we're on the order of 10 ^ -12. So we're probably looking at choice C.

The next closest choice is D, but that's a factor of 5 lower. Now, we may not know exactly how good our approximations are, but we're definitely not talking about factors of 5 in error here. And going the other way to choice B would be an error factor of 50!

So forget the extra significant digits, there's no point in getting your results to that kind of accuracy when the answer choices are so widely spaced out. The extreme case occurs when the answer choices are separated by like factors of 1000, then don't even bother with the factors at all, just look at powers of ten and shoot for the right ballpark.

Sometimes even when I've been trying to discriminate between two close choices, I've noticed that just paying attention to whether my approximation is shooting me too high or too low will let me get to the right answer.

So, despite my long rambling, my point is, I don't think the math in this problem is hard at all, so long as you toss out accuracy you don't need.
chemicalsoul
2009-10-17 01:31:46
w^2=(1/LC)-(R/2L)^2
Using this formula, I got ans approx equal to 0.5*10^-11.
 Prologue2009-10-20 12:44:18 $w$ is not equal to 103.7 MHz, $f$ is. And using $2 \pi f=w$ you see that you need to multiply 103.7 MHz by $2\pi$ to get omega.
 anum2010-11-08 09:07:23 that's the mistake i did.
 RusFortunat2015-10-20 14:07:36 not just you
Walter
2009-01-05 15:34:46
The resonant frequency of an LRC-circuit depends on R, in fact

$\omega_{d}^2 = \omega_{0}^2 - \frac{R^2}{4L^2}$

where $\omega_0$ is the resonant frequency of the undamped LC circuit, so before grinding out an answer you would need to check that $\omega_{0} \gg \frac{R^2}{4L^2}$ which you can do as for all the possible answers the magnitude of $L$ is such that the prior condition applies and thus $\frac{R^2}{4L^2}$ can be neglected.

It's a bit gung-ho given how mean some questions are to just ignore the resistance.
 wittensdog2009-10-16 19:37:45 This is indeed the natural oscillation frequency of an RLC circuit, but you still don't need to consider the resistance, for the following reason. The condition that, 1/w = sqrt(LC) actually comes from considering the impedance of an RLC circuit. For a given resistance, the largest possible power is dissipated across the circuit when the imaginary portion of the impedance, or the reactance, is zero. So if we specify that we are using a given resistance in the circuit, then the condition for maximum power absorption comes from setting the imaginary portion of the impedance to zero. It turns out that resistors have a purely real resistance, and capacitors and inductors have purely imaginary impedance. So we need to have the sum of the reactances of the capacitor and inductor be zero. Because these reactances are frequency dependent, this then becomes a condition that involves the frequency, inductance, and capacitance, but not the resistance.
 syreen2013-09-11 15:42:22 so, is there a QUICK way to tell whether or not R can be neglected?
 maxdp2013-09-24 22:04:50 Yup. Pretty sure as long as the circuit is hooked up to a power source (as a radio must be), then the power basically "pushes through" the resistor and gives the $\omega = 1/\sqrt{LC}$ that most people in this thread are using. And for the GREs, it seems unlikely that they'd expect you to remember the full formula for the frequency when a power source is not connected (called the damped frequency), which is what Walter is talking about.
gumpstuck
2008-09-22 21:39:47
Just to clarify and sum up previous comments:

$C=1/L\omega^2$
here $\omega=2\pi\cdot10^8$
$\omega^2\simeq5\times10^{17}$
so we have, $C=\frac{1}{(2\times10^{-6})(5\times10^{17})}=10^{-12}$ which is choice (C).
gatboja
2005-12-08 12:27:08
yosun, i believe you missed a squared sign for the 100E6.
StudyTime
2005-11-10 20:56:18
Mexicorn, you missed that you need to multiply the broadcast frequency by 2*pi to get the angular freqency.
Mexicorn
2005-11-10 17:20:49
woops, those x's should be multiplications and that -13 should be a positive at the in the last fraction I wrote
Mexicorn
2005-11-10 17:18:34
I'm not sure if i'm doing some math weird, but I can't understand how you got choice C for the problem. Using $C=\frac{1}{L\omega^{2}}$ I end up with $\frac{1}{(2x10^{-3})(1.03x10^{8})^{2}}$ which gives me $~1/2x10^{-13}$ or (D). Isn't the frequency supposed to be squared? Or am I missing something trivial?
 sabinscabin2008-10-04 14:51:09 you're squaring the frequency, not omega. it's not 1.03e8^2, but (2pi * 1.03e8)^2

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 30, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://grephysics.net/ans/9677/26", "fetch_time": 1544794341000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-51/segments/1544376825728.30/warc/CC-MAIN-20181214114739-20181214140239-00043.warc.gz", "warc_record_offset": 124410854, "warc_record_length": 8227, "token_count": 1879, "char_count": 6447, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2018-51", "snapshot_type": "longest", "language": "en", "language_score": 0.937942624092102, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00050-of-00064.parquet"}