Physics 4
Near-sighted
able to focus clearly on close objects, but not on distant objects. The image is formed in front of the retina.
Doppler Effect
∆f/fs=v/c ∆λ/λs=v/c Doppler shift perceived by the observer is dependent upon the relative velocity between the source and the observer. The greater the relative velocity the greater the shift in frequency or wavelength.
The four Lens/Mirro Rules-single lens systems only!
1) Object distances are always + 2) Image distances or focal point distances are + if they are on the same side as the observer and - if they are on the opposite side. 3) The observer and object are on the same side for a mirror and on opposite sides for a lens (you have to be behind your glasses to see through them to view the object on the other side) 4) PRI/NVU: Positive, Real, Inverted and Negative, Virtual, Upright always stay together.
Intensity in DB
=10*log(I/I₀), where I is the intensity of the sound wave in W/m² and I₀ is the threshold of human hearing (I₀=1 x 10⁻¹² W/M²)
Far-sighted
Able to focus clearly on distant objects, but not on close objects. The image is formed behind the retina.
Velocity of sound waves in a solid
Although densities of solids are typically thousands of times higher than the densities of gases, the elastic moduli are even increased by an even larger factor. This property causes solids to spring back extremely quickly following deformation. Sound (compression) waves in solids are therefore typically much faster than sound waves in gases.
Diverging Lenses
Concave, negative=always produces a negative, virtual, upright image.
Superposition of Waves
Constructive-Regions where the amplitudes of superimposed waves add to each other, increasing amplitude Destructive-Regions where the amplitudes of superimposed waves subtract from each other, decreasing amplitude
Diffraction
Diffraction is the tendency of light to spread out as it goes around a corner or through a slit. Without diffraction the characteristic interference patterns would not be formed.
Energy of a Photon
E=hf
string or pipe open or closed at both ends
L=nλ/2 f₁=v/2L, f₂=2(v/2L) λ₁=2L, λ₂=2L/2 all harmonics 1,2,3...
String or pipe open at one end and closed at other
L=nλ/4 one node and one antinode f=v/4L fn=n(v/4L) Harmonics 1,3,5
Red to violet
Longer wavelength and lower frequency (less energy) to shorter wavelength and higher frequency (more energy)
Which of the following would be correct units for the proportionality constant, C?
Looking at the units of the quantities in the equation, we have [m] = [units of C] [1/s] [1/s] /[kg/m^3]. Solving for the units of C, we get [units of C] = [m] [s^2] [kg/m^3] = [kg s^2/m^2], choice D.
Magnification-Two lens system
M=m₁m₂
Which of the following best describes the propagation of a sound wave at the molecular level?
MCAT absolutely LOVES questions like these. This is why we always tell you to know the why and how as much as the what. Answer A is false, but is how most people think of compressional waves based on the drawings they see. We erroneously think that the compressions are actually pushing bands of matter outward from the source; which is not the case. Answer B is false because the molecules do not vibrate up and down perpendicular to the direction of propagation, that is what happens in a transverse wave. Answer D would be true of individual bonds within a molecule, but not of air as a whole. Answer C is the correct answer and describes what actually happens. The molecules only get displaced for a short distance near their original origin; they then quickly run into other molecules, causing them to bounce back to their original positions and the other molecules to continue forward.
Electromagnetic Waves
No medium required, capable of propagating in a vacuum; transfer energy and momentum ( visible light, microwaves, radio waves) Transverse only
Beat Frequency
Occurs when two waves with close to the same frequency interfere. fbeat=|f₁-f₂|
Power-Two lens system
P=p₁+p₂
A jet with a transmitter mounted to its side is flying away from a stationary receiver. Which of the following will increase the ratio of the change in wavelength to the original wavelength?
The "ratio of the change in wavelength to the wavelength" means Δλ/λ. Answer B is false because according to Δλ/λ = v/c, using a faster wave (variable c in the formula) will decrease the ratio of change in wavelength to original wavelength. By contrast, using a slower wave will increase that ratio, making A the correct answer. Answer C is false because decreasing the speed of the jet will decrease the relative velocity v and thus the ratio in question. Answer D is false because increasing the wavelength of the original wave would decrease the ratio.
Real
There is actual light at the image (an image formed on your retina)
Transverse waves
Transverse waves displace the medium perpendicular to their direction of travel (electromagnetic waves, a wave on a string);
Wave speed
V=λƒ
Lens/Mirror Calculations
f=1/2R (for mirrors only) 1/f=1/di+1/do M=-di/do=hi/ho
Concave mirrors
follow same rules as converging lenses f=(+)
3 Cardinal Wave Rules
1) Wave speed (velocity) is determined by the medium and sometimes (for a dispersive medium) wavelength and frequency. 2) Frequency never changes when a wave moves from medium to medium 3) Wavelength does change when a wave moves from medium to medium
Intensity
Many waves such as sound travel outward from their origin in all directions simultaneously-creating a wave front in the shape of a growing sphere. Because intensity is measure per square meter, its magnitude decreases according to the area of the growing sphere (ie the m² term increases, decreasing the intensity). A=4πr²
Velocity of a wave on a string
The elastic property that provides the restoring force in a string is its tension. The inertial property is mass per unit length, the linear mass density µ: v=√(T/µ). Increased tension always increases velocity. A thicker string (increased mass per length) always decreases velocity, provided the tension stays the same.
For air, the index of refraction is assumed to be approximately 1.00. A ray of light traveling in air approaches a flat solid surface (of unknown material) at an angle of 30 degrees to the normal. As the ray passes through the medium, the angle of the ray with respect to the normal is measured and found to be 24 degrees. Given this information, which of the following is NOT known and/or CANNOT be calculated?
The first angle described is the angle of incidence. The second angle given is the angle of refraction. The angle of reflection is ALWAYS equal to the angle of incidence. Finally, the index of refraction can be calculated by plugging the given info into Snell's Law. This makes D the best answer.
Snell's Law
n=c/v n₁sinθ₁=n₂sinθ₂ If n₂>n₁, the light ray will bend towards the normal, while if n₂<n₁, the light ray will bend away from the normal. Velocity will decrease in higher index of refraction. Velocity will increase in lower index of refraction.
Virtual
no actual light emanating from or reaching the image (image formed behind a plane mirror)
x=λL/d
where x is the distance between fringes, λ is the wavelength of light used, d is the distance between the two slits, and L is the distance between the double slit and the final screen (only when true when x is much smaller than L).
Oscillators
with matching ends, the wavelength of the second harmonic equals the length of the sting or pipe λ=L
EM spectrum
High to low frequency Gamma rays, x-rays, uv, visible, infrared, microwaves, radio waves
Sound Resonance
All objects have one or more natural frequencies at which they will vibrate when disturbed. Some objects produce a random array of different vibrational frequencies. Other objects (musical instruments) vibrate at non-random natural frequencies which are integer multiples of a number. These orderly frequencies are called harmonics. When one object is vibrating near another object it can cause the neighboring object to begin vibrating at that same frequency. If the exact frequency at which the second object is caused to vibrate happens to be one of its natural frequencies (harmonics), the two object are said to be in resonance and via constructive interference can produce a much louder sound. If the first object causes a vibration in the second object that is NOT a match to one of its natural frequencies, resonance does not occur.
Which of the following statements is/are true of single lens system?
All of these statements are true. REMEMBER THEM! Objects are always positive as long as it is a SINGLE lens system. As far as everything else is concerned, if it is on the same side as the observer, it is positive and if it is on the other side it is negative. Just make sure that you correctly decide which side is the observer. Statement IV reiterates the fact that NVU and PRI are never violated for single lens systems. All object distances are positive All images and focal points on the same side of the lens or mirror as the observer are positive. All images and focal points on the opposite side of the lens or mirror as the observer are negative. Negative, virtual images are always upright. NVU and PRI never violated for single lens systems.
A wave of light proceeds from object x and strikes the front side of a convex mirror. An image is formed:
Always remember the PRI/NVU rule. In this case, we can draw a convex mirror and predict that the light will be reflected outward or DIVERGED back toward the source. This will not create a focal point, so there must be a pseudo focal point that exists at the point behind the mirror where the two lines would cross if we traced them back. Because it is BEHIND the mirror it is NEGATIVE and therefore it must follow NVU and also be virtual and upright; making C the correct answer.
Which of the following changes must increase the velocity of the wave described?
Answer A will always lower wave speed because the speed of a wave in a gas is proportional to the square root of the temperature of the gas. Answer B will decrease wave speed because for a wave on a string v = sqrt(T/(mass per length). Answer C will also decrease speed because for compressional waves v = sqrt(B/density). Answer D is correct because for any mechanical wave, speed will increase as the square root of the elastic modulus. The greater the modulus, the faster the string will spring back to its original shape, sending the wave on down the string more quickly.
A non-standing transverse wave propagates within a perfectly elastic medium. All of the following statements about the wave are false, EXCEPT
Answer choice B is a true statement. Each discrete point along a wave can be represented on paper as a sine wave. Answer A is false because only a standing wave (and the question stem told us this is a non-standing wave) has no energy transfer. Answers C and D are both false because nodes and antinodes only exist in standing waves. Nodes are the points of zero amplitude that never move and the antinodes are the points of maximum amplitude that oscillate up and down continuously.
Optical Power
P=1/f Two Lens System: (binoculars, telescopes) The image formed by the first lens becomes the object for the second lens.
Pitch
Pitch is closely related to frequency, but not identical. Higher pitch sounds have higher frequencies and lower pitch sounds have lower frequencies.
An object X is located on the front side of a converging lens beyond the focal length, f. Which of the following describes the image made by the lens?
Recall that the observer goes "BEHIND" a lens and "IN FRONT" of a mirror. So this object is on the opposite side as the observer. Light rays will come toward the lens, converge on the same side as the observer, so the image must be positive and thus must be Positive, Real and Inverted (PRI).
Which of the following best account for the manner in which light passes through a prism?
Reflection is a wave bouncing off of a medium, refraction is a wave bending as it passes through a denser medium, deflection is a real term but has no specific relevance to waves. Diffraction is the bending/spreading of light as it pass by an object, through a pinhole, etc. This makes B the correct answer.
Mechanical Waves
Require a medium to propagate; transfer energy only Transverse (strings on a musical instrument(; Transverse mechanical waves require a fairly stiff medium to propagate and therefore cannot propagate in liquids or gases. Longitudinal (sound waves)
Production of Sound
Sound is always created by a vibrating medium. These vibrations propagate through liquids or solids, and generate pressure waves that propagate through gases such as air. As a mechanical wave, sound cannot propagate in a vacuum.
Velocity of sound waves in a gas
The elastic property is called the bulk modulus, B. The inertial property is normal density, p: v=√(B/p). The bulk modulus turns out to be both directly proportional to both density and temperature, however, so for gases the velocity ends up having this temperature dependence: v∼√(T).
For a pipe closed at both ends, the distance between the first and fourth node in a standing wave can be calculated using which of the following?
The distance between any two adjacent nodes will be one-half of a wavelength. The distance from node 1 to node 3 will be one wavelength. The distance from node 1 to node 4 will be 3/2 of a wavelength. This ratio only exists in Answer C. It is common for the MCAT to take things one step further by substituting in for a variable. Here, lambda has been replaced by v/f, which is entirely valid and is the same as 3/2λ.
If two waveforms share the same phase, they must also share all of the following, Except:
The easiest way to think of phase is as horizontal displacement of the sine wave from the y-axis on a graph. Two waves cannot have different wavelengths and share the same displacement cycle after cycle. Neither can they match if they have different frequency or period. It also helps to visualize that only waves EXACTLY in phase can add via totally constructive interference. Two waves with different amplitudes could still add entirely, but differences in frequency, period or wavelength would make them offset and unable to add via totally constructive interference.
Attenuation
The gradual loss of intensity as a wave passes through a medium. In non dispersive mediums this is due to scattering (ie reflection) of some waves and absorption of wave energy.
A child is playing with a string attached to a w all and creates a standing wave. If the string is 0/1 meters long and the speed of waves on this string is known to be 100 m/s, what is the frequency of the frist harmonic?
The key here is to recognize that the wavelength of the first harmonic is TWICE the length of the string. (It's the second harmonic where the wavelength equals the length of the string.) Most students will plug the length of the string into the wave speed equation and solve, getting answer D. For the first harmonic, with only one node, the wavelength will be twice the length of the string, or 0.2 meters. Plug this into the wave speed equation and solve to get 500Hz.
If the wavelength of the light emanating from the light source in Young's experiment was decreased, how would this affect the number of light and dark bands on the screen?
The light and dark bands are created by interference between the two waves coming from the pinholes in the second board. If the wavelength of the light was very, very large, say bigger than the screen, then you would see no more than one band of interference, either completely dark or completely light. If we take this the other direction, to a very, very small wavelength, you would see tons of little light and dark bands. The key is that when two waves interfere, they do so in a repeating pattern whose distance between bands is proportional to the wavelength. This makes B the only logical answer.
All of the following statements correctly describe the optic characteristics of the human eye, Except:
The shape of the lens in the human eye is actually determined by a somewhat complex interaction between the ciliary muscles and the suspensory ligaments. The normal, default state of the lens, were it not attached to anything, would be highly spherical. The suspensory ligaments permanently flatten the lens, reducing its power (because it's less spherical) and increasing the focal length. The ciliary muscles work against these ligaments, reducing their effect when contracted (allowing the lens to revert to its more spherical shape) and thus decreasing the focal length and increasing the power (recall that power = 1/focal length). For the MCAT, all I would remember is that contracting = close and relaxing = farther away. Also be sure not to mix up the power stuff. A closer object requires a stronger, more powerful lens and very powerful lenses have very short focal lengths. The minimal focal length of the eye does increase with age as the lens become less flexible and the ciliary muscles become weaker; and the eye ALWAYS forms PRI images.
A guitar creates a sound when one of the strings is give a displacement x, either upward or downward, by the plucking motion of the player's finger. The string bounces back and a standing wave with two nodes is created. The relationship between x and the Amplitude, A, of the standing wave is:
This is a conceptual way of asking a very straightforward question: What is amplitude? This is a great example of why you must have solid conceptual definitions for ever term in the lessons. You can't just have a general feeling for what amplitude is, you must understand EXACTLY what it is and is not. Amplitude is the distance the wave rises above OR below the middle of the waveform. It is NOT the total distance from the bottom of a valley to the top of a crest (it would be half that). This makes A the correct answer.
Standing Waves
This is a special case of simultaneous constructive and destructive interference between two waves with identical frequencies moving through the same medium in opposite directions. At points of maximum destructive interference the waves cancel entirely to create a node. At points of maximum destructive interference the waves add completely to form an antinode. The result is a waveform in which the areas between the nodes appear to oscillate up and down between crests and troughs. A standing wave exhibits no net transport of energy and does not itself propagate. In other words, there is no translational movement of nodes or antinodes in either direction. By contrast, when a standard traveling wave propagates, the minima and maxima do move translationally in the direction of propagation.
A child is viewing an ant through a magnifying glass which approximates a single converging lens. the child holds the lens 0.5 meters away from his eye. Which of the following is true?
This is an excellent problem to test whether you are actually conceptualizing the lens and mirror principles we've covered. The magnifying glass (when used as a magnifier) is positioned such that the object is inside the focal distance. That causes a virtual, upright image to be formed on the far side of the glass. Think about it or try it out—where does it look like the enlarged image of the ant is? It appears to be behind the magnifying glass. Many students will choose answer A, thinking that the magnifying glass causes an image to be formed in the child's eye. That's not how magnifiers work! There IS an image formed on the retina—a real, inverted image—but that image is formed by the lens inside the eye, not by the magnifying glass.
A ray of light travels through air, bounces off of a stationary ideal reflector, and travels back toward its origin. Which of the following is the same before and after reflection of the light ray?
This is another important question. Because the reflector is "ideal" there should be no loss of energy or intensity to the wave as it bounces off the reflector. Had the question labeled it as a non-ideal reflector, you would assume some loss in amplitude and intensity. These facts make III and IV true statements. The rest can be evaluated given one of our cardinal rules: velocity is determined by the medium (unless dispersion is mentioned). In this case the wave is traveling in air both before and after the reflection, it must have the same velocity, making V a true statement. Finally, the other cardinal rule says frequency never changes. If it doesn't change and neither does velocity (in this case), then wavelength could not have changed (according to v = λf). D is thus the correct answer. Note that the reflector was stationary. The ONE AND ONLY time frequency can be thought of as changing is if the source or reflector were moving. Then it would be a Doppler problem.
A man stands 1000 meters in front of a sheer rock wall. Hed directs a wave with a wavelength of 0.5 meters at the wall and measures the time, t, required for the wave to return to a detector. If t=1 second, what is the frequency of the wave?
This is one of those questions that is really pretty easy, but people seem to get mixed up on. Be confident and always double-check your reasoning, but never assume you have to be wrong just because it seems too easy; there are actually a good handful of easy questions on each MCAT. Because the wave has to travel to the wall and back, we must use 2000 meters as the distance traveled. 2000 meters in 1 second gives a wave speed of 2000m/s. Using the wave equation, v = fλ, we can now solve for f, giving f = v/λ. This is 2000/.5 or 4000Hz
Which of the following best explains why sound waves travel faster under water than through the air?
This question is testing your understanding of why sound waves (and other compressional waves) travel faster in solids/liquids than they do in gases. You should recall the relationship from your lessons of Vlongitudinal = sqrt(B/D), where B is the bulk modulus and D is the density. Note from this relationship that increased density still makes the wave travel slower, so it is not the density of the water that makes the sound travel faster. It is the bulk modulus, or lack of compressibility of the water that makes up for the effect of the density and overcomes it so that waves actually travel slightly faster in water than in air. Answers C and D are false because sound waves have a longer wavelength in water and a SHORTER wavelength in air.
A vehicle traveling away from an observer has a malfunctioning horn that sounds continuously. If the vehicle travels with a constant positive acceleration away from the observer, which of the following represents the magnitude of the change in frequency experienced by the observer over time?
To analyze this question, recall the equation for Doppler, Δf/f = v/c. This demonstrates that the change in frequency is directly proportional to the relative velocity. If the acceleration is constant, the velocity is changing at a constant rate, so anything directly proportional to that must be changing at a constant rate. Because no exponents are involved it must be linear, not exponential, making B the best answer.
In a vibrating guitar string the principle restoring force is provided by the:
When waves move through anything they ALWAYS displace the medium. The medium returns to its original shape/location due to some restoring force. In solids, this is the elastic force described by Hooke's Law. The displacement of the string is much like the stretching of a spring, which makes B a good possible answer. Answer A is false because gravitational potential energy would only cause a downward restoring force, and the actual restoring force must act both upward (when the wave displaces downward) as well as downward (when the wave displaces upward). Answer C is false because the linear mass density provides the inertia that resists the restoring force. Answer D is a nonsense term. If you know everything in the lessons, have the confidence to recognize such terms as nonsense. Don't assume its something you just don't know about or remember.
Which of the following is true of two waves most frequently used for Puls-Doppler Radar? (Passage) pg. 259
You'll need to know your electromagnetic spectrum to know that microwaves are higher frequency than radio waves. However, all electromagnetic waves travel at the speed of light, so Answers A and B must be false. Answer D is true in that radio waves are lower in frequency, but the equation in the passage tells us lower carrier frequency would DECREASE range. Answer C is correct and matches the relationships in the equation.
Young's Double Slit experiment
Young shone a monochromatic light through a screen with a single slit in it. The purpose of the slit was to create coherent wavefronts. Behind the first screen he placed a second screen with two narrow, parallel slits. These created the diffraction pattern. Finally, behind the second screen he placed a third screen. Light traveled through the first two screens and formed alternating pattern of light and dark bands on the third screen. For this experiment to work, the light traveling through each of the two slits in the middle screen must be coherent and have the same frequency and polarization.
Converging Lenses
convex, positive=usually produces a positive, real, inverted image, when the object is inside the focal point it produces a negative, virtual, upright image.
Intensity
defined as power per unit area. Waves have power because they transfer energy from one location to another within a specified time (energy flux). The intensity of any sound or mechanical wave is directly proportional to the amplitude squared and the frequency squared (i.e, I≈A²f²). For light waves, the intensity is also proportional to the amplitude squared, but not the frequency squared.
Longitudinal Waves
displace the medium parallel to their direction of travel (sound waves, p-wave earthquakes)
Convex mirros
follow same rules as diverging lenses f=(-)
Total Internal Reflection
light ray bends away from the normal as it moves from a medium of high refractive index to a medium of low refractive index. If this refracted light ray were to keep bending away from the normal as the incident angle increases, it would eventually reach a refracted angle that is 90⁰ away from the normal. This incident angle is called the critical angle.
Wave speed
is typically equal to the square root of an elastic property of the medium divided by an inertial property: v=√(elastic/inertial). The elastic property is often called a modulus. The inertial property is a type of density.
