1. A One should not typically rely on figures for precise answers, but Figure 2 does reveal that the period of oscillation is at least approximately constant. The key to this question is to notice that the term inside the argument of the cosine is of the form (ωt), or a constant (for any given set of values k, b, and m) times t, and thus the frequency of the cosine is itself constant (ω/2π).
2. B Equilibrium occurs when the net force acting on the system is zero. Thus to find the equilibrium length of a spring with a hanging mass, we find the displacement ∆s for which the restoring force of the spring equals the weight of the mass. This give us k s mg s mg
k . Adding this to the equilibrium length of the unencumbered hanging spring L gives the correct answer.
3. B As always, slope is rise over run, which in this case yields y/t. Taking the units of this quantity we find [y]/[t] = m/s, which are of course the units of velocity.
4. A This is a slight variation on the classic two-by-two Process of Elimination question, because in this case the amplitude decays either more or less rapidly than in the experimental set up with a less viscous liquid, whereas the frequency can, among the choices, decrease, increase, or remain the same. The passage de-fines b, the damping coefficient, as “a measure of the frictional effects on the system due to the movement of the submerged plate through the fluid.” Because a more viscous liquid has provides greater fluid fric-tion, b should increase with increased viscosity. A larger b means a larger negative exponent in Equation 3, meaning a more rapid decrease in amplitude. It also means a smaller coefficient of t in the argument of the cosine function (because the term containing b is subtracted from k/m), which means a lower fre-quency of oscillation.
5. C The total mechanical energy of a harmonic oscillator, the sum of its kinetic and potential energies at any time t, is equal to its maximum potential energy at that time t, or (1/2)kAt2 (this is just another way of saying that the energy in an oscillating system like a wave goes as the square of the amplitude). For the case of a simple harmonic oscillator, that energy is constant because amplitude is constant, but for this damped harmonic oscillator the amplitude varies as the exponential term in Equation 3. The question asks at what time the energy will be one quarter its original value, which means we are looking for the time at which the amplitude will be half its original value (the square root of a quarter). Thus we have
Ae bt2m = 1
2A ln e bt2m = ln
( )
21 = ln(2) 2mbt = ln(2) t = 2m ln(2)b .
Note that choice A corresponds to the time at which the oscillator first crosses its equilibrium length (that is, when cosine is zero).
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6. D The rate of decay of the amplitude of oscillations is determined by the exponential term, which contains b and m but not k. Thus the stiffness of the spring has no effect on the amplitude. Choices B and C, chang-ing the area of the plate or drainchang-ing the liquid from the tank, would each alter the value of b.
7. C If the hanging mass is not interacting with the liquid in the container, we can approximate its motion as simple harmonic. Thus we use Equation 1 in the passage and simply plug in given values for k and m (the given value for A does not influence our answer because for simple harmonic motion, period is indepen-dent of amplitude). We have the following:
2 4 6 1 6 1.2 seconds
100 25 5
T .
OSCILLATIONS, WAVES, AND SOUND
Drill
MCAT COMPLETECHAPTER 43 PRACTICE PASSAGE
Early in the 20th century, several crucial experiments established the dual wave-particle nature of light and the ways that electromagnetic radiation interacts with matter. In 1923, Arthur H. Compton conducted an important experiment showing how photons in the X-ray and gamma-ray portion of the spectrum interact with free electrons. Compton’s experiment provided further verification of the photonic nature of electromagnetic phenomena.
Compton scattering occurs when incident photons strike electrons in elastic collisions, imparting momentum and kinetic energy to the electrons and thereby reducing the momentum and energy of the scattered photons. The reduced energy of the photon implies a reduced frequency according to the photon energy equation E = hf, where h is Planck’s constant h = 4.14 × 10−15 eV∙s). This change in frequency implies a corresponding
change in wavelength; this shift in wavelength of Compton scattered photons is given by the Compton Shift equation:
= h
mec(1 cos ) Equation 1
Here φ represents the scattering angle, or the angle between the photon’s original path and its path after the collision, and me is the rest mass of the electron.
To study Compton scattering in the lab, a student conducts the following experiment. Using a radioactive source (Americium-241, a commonly manufactured radioactive isotope used in household smoke detectors with a peak gamma emission of 59.5 keV), a metallic target shaped into a circular arc, and an X-ray detector that converts incident X-ray photons into electric current by means of the photoelectric effect, the student counts the number of scattering events at various angles over set durations. So long as the source, detector, and target all lie on a circle on a flat surface, the student can be relatively assured of receiving only photons scattered at particular angles due to the geometric rule that inscribed angles in a circle are equal to half the intercepted arc. The experimental set-up is shown below.
The student conducts experimental runs at four different scattering angles over four different durations. The following data record the scattered energy peaks in terms of scattering angle and photon counts (only the peak counts are included, not the entire energy spectrum for each scattering angle).
Scattering
2. If incident photons from a green light source undergo Compton scattering off of electrons at rest, which of the following could NOT be true of the scattered photons?
A) They appear blue.
B) They appear yellow.
C) They have a longer wavelength than the incident photons.
D) They have a lower energy than the incident photons.
Drill
4. During experimental runs, the student places a thin sheet of lead directly between the Amercium-241 source and the scintillation detector. Which of the following best explains the purpose of this measure?
A) It prevents electrons scattered in the air between the source and the detector from entering the detector and corrupting the data.
B) It ensures that photons scattered off the metallic target at angles different from the desired angle do not enter the detector.
C) It prevents high energy photons from cosmic radiation from entering the detector and corrupting the data.
D) It prevents photons emitted from the source from entering the detector directly without scattering off the target.
5. According to Table 1, for which scattering angle was the rate of detection events at the peak energy highest?
A) 60°
B) 75°
C) 90°
D) 120°
6. Suppose another trial of the experiment were conducted set up to record scatterings at an angle of 105°. If the experiment were allowed to run overnight (for a duration of 12 hours), which of the following would most likely correspond to the peak energy detected and the number of detections?
A) 51.8 ± 0.3 keV and 1320 detections B) 51.8 ± 0.3 keV and 1935 detections C) 54.3 ± 0.6 keV and 1320 detections D) 54.3 ± 0.6 keV and 1935 detections
7. Inverse Compton Scattering occurs when photons scatter off of relativistic electrons in extremely fast-moving gases around astrophysical phenomena like black holes and gamma ray bursts. Which of the following might you expect to observe as a result of such scattering?
A) An increase in the number of electrons detected from that direction in the sky
B) An increase in the velocity of the scattered photons C) An increase in the energy of the scattered photons D) An increase in the relative population of radio photons to
gamma ray photons
LIGHT, OPTICS, AND QUANTUM PHYSICS
MCAT COMPLETE