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All Solutions of Type: Lab Methods

GR8677 #39

Lab Methods $\Rightarrow$ }Log-Log Graph

The key phrase is most accurately expressed. Choice (E) works because it applies decently well to the following two key points, one of them given, and the other one right next to the given: $(\omega=3 E 5,1E2)$ and $(1E6,1E1)$

Try a power-law relation: $Gain=K\omega^n\Rightarrow$ $1E2=K(3E5)^n$ and $1E1=K(1E6)^n$ . Divide the two equations to get $1E1=(3E-1)^n Rightarrow 10=(1/3)^n$ . The closest answer would be $n=-2$ , which is case (E).

(Can't be an exponential relation, since the Gain would decrease exponentially as $\omega$ increases. It's not decaying nearly as fast.)

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GR8677 #27	Lab Methods $\Rightarrow$ }Log Graphs Log graphs are good for exponential-related phenomenon. Thus (A), (C), and (E) are appropriate, thus eliminated. The stopping potential has a linear relation to the frequency, and thus choice (B) is eliminated. The remaining choice is (D). Click here to jump to the problem!

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GR8677 #28

Lab Methods $\Rightarrow$ }Oscilloscope

A superposition of two oscillations has the form $\sin \omega_1 t + \sin \omega_2 t = 2 \cos(\frac{\omega_1-\omega_2}{2}) \sin(\frac{\omega_1+\omega_2}{2})$ . This implies that the cosine term is the amplitude of the combined wave.

Similarly, one can see one the lower frequency as the contribution towards the bigger envelope-like wave and the higher-frequency as the zig-zag-gish motion along the envelope.

One oscillation must have a high frequency and the other has a relatively lower frequency. Only choices (D), (A), and (B) show this trait. The high frequency oscillation should have a smaller amplitude than the lower frequency oscillation. Only choices (A) and (D) show this trait. Finally, the amplitude of the lower frequency wave forms the envelope, and the amplitude from that is only about 1 cm; on a 2V scale, this is about 2V---which is closest to choice (D).

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GR8677 #45

Lab Methods $\Rightarrow$ }High-pass filter

Recall the impedance formulae for capacitors $X_c = \frac{1}{\omega C}$ and inductors $X_L=\omega L$ . The complex impedance is $Z=-iX_C + iX_L + R$ , and the ac-version of Ohm's Law becomes: $V=IZ$ .

For choice (E), one has $Z=R + iX_C \Rightarrow V_{in}=I\left( R + i(X_L - X_C) \right) \Rightarrow I=\frac{V_{in}\left(R+iX_C\right)}{R^2+X_C^2}$ , where in the last step, one multiplies top and bottom by the complex conjugate of the denominator impedance Z. The voltage across the resistor is the voltage from ground, thus $V_{out}=IR=\frac{V_{in}\left(R+iX_C\right)R}{R^2+X_C^2}$ .

For high frequencies, one has $\omega \rightarrow \infty \Rightarrow X_C \rightarrow 0 \Rightarrow V_{out} = V_{in} R^2/R^2$ .

For low frequencies, one has $\omega \rightarrow 0 \Rightarrow X_C>>1 \Rightarrow V_{out}=\frac{V_{in}R}{X_C}\rightarrow 0$ .

Circuit (E) meets the given conditions.

(Incidentally, choice (D) is a low-pass filter giving $V_{out}\rightarrow 0$ for high frequencies.)

(For more on this, check out Horowitz' The Art of Electronics.)

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GR8677 #72

Lab Methods $\Rightarrow$ }Negative Feedback

Negative feedback, according to Horowitz's The Art of Electronics, has to do with canceling out some of the input in the output. Although that might seem like redundantly adding noise to the system, it actually reduces the amplifier's gain, increases stability (by decreasing nonlinearity and distortion).

From that bit of info, two choices remain. Choice (A) and (B). Choose (A) because negating the feedback should not increase the amplitude.

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GR8677 #15	Lab Methods $\Rightarrow$ }Precision The most precise measurement might be the wrong value, as long as after meany measurements, each measurement is very close to the mean. Thus, the width of the height vs number of trials graph must be as thin as possible. The only choice that shows this characteristic is choice (A). Click here to jump to the problem!

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GR8677 #16

Lab Methods $\Rightarrow$ }Sample

The mean of the ten number is $\bar{x}=2$ . Thus, the standard deviation of the sample is $\sqrt{\bar{x}}\approx 1.44$ . (Search for Poisson Distribution on the site for another problem similar to this.)

If the student wants to obtain an uncertainty of 1 percent, then

$\sigma/\bar{x^{'}}=1/100=\sqrt{2 C}/(2C)=1/(\sqrt{2C})$ ,

where one assumes the average scales uniformly and C is the time to count. (Note: a good approximation of the uncertainty is given by the ratio of the standard deviation to the average, since that represents the deviation.)

Thus, one has $\sqrt{2 C}\approx 100$ . Thus the student should count C=5000 s.

A Sneak Peek of The Not-So-Boring Review of Undergrad Physics
(To be published in the to-be-posted library section of http://GREPhysics.NET in Feb 2006.)

The Poisson Distribution is intimately related to the raising and lowering operators of the (quantum mechanical) simple harmonic oscillator (SHO). When you hear the phrase ``simple harmonic oscillator," you should immediately recall the number operator $N=a^{\dagger}a$ , as well as the characteristic relations for the raising $a^{\dagger}$ and lowering $a$ and $a^{\dagger} |n\rangle = \sqrt{n+1} |n+1\rangle$ . And, don\'t forget the commutation relations that you should know by heart by now, $[a,a^{\dagger}]=1$ . (That\'s all part of the collective consciousness of being a physics major.)

Now, here\'s some quasi-quantum magic applied to the Poisson Distribution. I\'m going to show you how to arrive at the result for standard deviation, i.e., $\Delta n \equiv \sqrt{\bar{n}}$ from using the SHO operators.

Let\'s start with something easy to help jog your memory: The mean or average number in the distribution is just the expectation value of the Number operator, $\langle N \rangle = \langle n | N | n \rangle = \langle n | a^{\dagger}a|n\rangle \equiv \bar{n}$

Okay! So, on with the fun stuff: the standard deviation is given by the usual definition, $(\Delta n)^2 = \langle N^2 \rangle - \langle N \rangle^2$ .

The second term is already determined from the above expression for the mean, $\langle n \rangle^2 = \bar{n}^2$ .

The first term can be calculated from $\langle N^2 \rangle = \langle n | a^{\dagger}aa^{\dagger}a | n \rangle$ . Now, the commutation relation gives, $[a,a^{\dagger}]=1=aa^{\dagger}-a^{\dagger}a \Rightarrow aa^{\dagger}=a^{\dagger}a+1$ . Replacing the middle two of the four a's with that result, the expression becomes $\langle N^2 \rangle = \langle n | a^{\dagger} (a^{\dagger}a+1) a |n \rangle = \langle n | a^{\dagger}a^{\dagger}aa + a^{\dagger}a |n \rangle = \langle n | a^{\dagger}a^{\dagger}aa | n \rangle + \langle n | a^{\dagger}a | n \rangle = \langle N \rangle^2 + \langle N \rangle = \bar{n}^2 + \bar{n}$ .

Plugging the above results into the standard deviation, I present to you, this: $(\Delta n)^2 = \bar{n}^2 - ( \bar{n}^2+\bar{n} ) = \bar{n} \Rightarrow (\Delta n) = \sqrt{\bar{n}}$ .

It\'s no coincidence that the above works. The secret lies in the energy eigenfunction that you might not remember...

\subsection{The Poisson Distribution Function is just $P_x(n)=|\langle x | n\rangle|^2$ The Poisson Distribution for a parameter $m$ is given by $P_x(\lambda)\propto e^{-m}\frac{m^\lambd}{\lambda!}$ . But, wait, doesn\'t that look a wee bit too familiar? Indeed, the Poisson Distribution is merely the probability of obtaining $n$ photons at position $x$ : $P(n)=|\langle x | n \rangle|^2$ .\footnote{Note that since $E_n=\hbar\omega\left(n+\frac{1}{2}\right)$ , each time n increases, it is like we've created an extra photon, since the energy of a photon is $E=h\nu=\hbar\omega$ . Thus, $n$ represents the quanta.}

Why? So you ask. Well...

The energy eigenfunctions of the SHO is given by $|n\rangle = \frac{(a^{\dagger})^n}{(n!)^{1/2}} |0\rangle$ , where $\langle x |0\rangle \propto e^{-\bar{n}/2}$ .

This result can be arrived at in the position basis as follows: $\langle x | 0 \rangle = \left(\frac{m\omeg}{\pi\hbar}\right)^{1/4} e^{\frac{-m\omega x^2}{2\hbar}}$ . The $x^2$ in the exponent can be re-expressed by the relation $x=\frac{\hbar}{2m\omeg}(a^\dagger+a)$ . Thus, $\langle x^2 \rangle = \frac{\hbar}{2m\omeg}\langle a^\dagger a + a a^\dagger \rangle = \frac{\hbar}{2m\omeg}(2\bar{n})$ , where the other terms like $\langle n |a^2| n\rangle=0$ and we\'ve used the result for $\langle n \rangle = \langle a a^\dagger \rangle=\bar{n}$ from above with the implicit assumption that its complex conjugate is the same, as the average photon number, $\bar{n}$ , is an observable. Thus, the exponent becomes $\langle x | 0\rangle=e^{-\frac{m\omega x^2}{2\hbar}} = e^{-\bar{n}/2}$ .

The probability in the $x$ -basis is thus, $P(n)=|\langle x | n \rangle|^2=\frac{\langle aa^\dagger \rangle}{n!}^{n} e^{-\bar{n}}=\frac{\bar{n}}{n!}^{n} e^{-\bar{n}}$ , where using the definition $\langle \bar{n} \rangle=\langle aa^\dagger \rangle$ , we\'ve recovered exactly the Poisson Distribution for a parameter $\bar{n}$ .

Making the following associations, $m\rightarrow \bar{n}$ and $\lambda \rightarrow n$ , you carve the first etchings in the Rosetta Stone between probability and photon statistics...

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GR8677 #17

Lab Methods $\Rightarrow$ }Oscilloscope

This problem can be solved by elimination. Since one is given two waves, one with twice the frequency of the other, one can approximate the superposed wave (which shows up on the oscilloscope) as $\sin(\omega t) + \sin(2\omega t)$ .

The summed wave no longer looks like a sine wave. Instead, it looks like a series of larger amplitude humps alternating with regions of smaller amplitudes.

However, since one is not supplied with a graphing calculator on the test, one can qualitatively eliminate the other choices based on the equation above. It is obviously not choices (D) and (E) since the superposition is still a one-to-one function. It isn't choice (C) or (B) since those are just sin waves (cosine waves are just off by a phase), and one knows that the superposed wave would look more complicated than that. Thus, one arrives at choice (A), which is a zoomed-in-view of the superposition above.

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GR8677 #18

Lab Methods $\Rightarrow$ }Coax Cable

Elimination time. The first-pass question to answer is why is it important that a coax cable be terminated at an end:
(A) Perhaps...

(B) Probably not. Terminating the cable at an end would not help heat dissipation and thus should not prevent overheating.

(C) Perhaps...

(D) Probably not, since termination should attenuate the signal rather than to prevent it.

(E) Probably not, since image currents should be canceled by the outer sheath.

Choices (A) and (C) remain. Now, use the second fact supplied by ETS. The cable should be terminated by its characteristic impedance. Characteristic impedance has to do with resonance. Thus, it should prevent reflection of the signal.

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GR8677 #26	Lab Methods $\Rightarrow$ }Log-Log graph Since initially, the counts per minute is $6E3$ , the half-count amount would be $3E3$ . This occurs between 5 and 10 minutes. Choice (B) seems a good interpolation. Click here to jump to the problem!

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GR8677 #48

Lab Methods $\Rightarrow$ }Uncertainty

The general equation for uncertainty is given by $\delta f/f=\sqrt{\sum_i (\frac{\delta x_i}{x_i})^2}$ , where $f=\Pi_i x_i$ and $\delta x_i$ is the generalized standard deviation of quantity $x_i$ .

So, for this case, one has $f=ma$ and thus its uncertainty is given by $\delta f/f = \sqrt{(\frac{\delta m}{m})^2+(\frac{\delta a}{a})^2}$ . (Basically, one has $x_1=m$ and $x_2=a$ .)

This is choice (C).

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GR8677 #86	Lab Methods $\Rightarrow$ }Oscilloscopes The discharge of the capacitor after it has been charged to $V_0$ is just $V(t) = V_0(1-e^{-\omega t})$ , where $\omega = 1/(RC)$ . One can find C by knowing $R$ and the sweep rate, which is related to t. (Solution due to David Latchman.) Click here to jump to the problem!

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