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PHYSICS TEXTBOOK NOTES.docx

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Department
Physics
Course
PHYS 1070
Professor
Melissa Williams
Semester
Winter

Description
PHYSICS TEXTBOOK NOTES STUDY GUIDE 1 Mathematics of travelling and standing waves Vibrations and Waves 1.2 Simple Harmonic Motion Many objects in nature are subject when displaced slightly from their equilibrium position, to a force that is proportional to the displacement from equilibrium. If displaced slightly from equilibrium and then released, they will undergo a back and forth oscillatory motion known as simple harmonic motion (SHM). When the mass is at the equilibrium position (x=0), the spring exerts no force on it and if the mass is released from rest at eq, it remains there. If it is displaced to the right or left (+/- x direction) by a force F applied, the spring is stretched or compressed and exerts a force F on the mass in the direction opposite to the direction of displacement (ie force is directed back toward x=0). This is the restoring force. F(x)= -kx. A stiff spring has a large k and a soft spring has a small one. F and x are N and m, k= N/m (Expressed by a linear graph) Elastic Potential Energy Potential energy is expressed in joules. Suppose a force is applied such that the mass is displaced from equilibrium at a constant velocity. With constant velocity, acceleration of the mass is zero so the net force must be zero (F=ma). Thus, the applied force must be equal to the force exerted by the spring but in the opposite direction. As the mass moves, work is done by the applied force. If the force is constant, W= Fapp x delta x. Work is area under a Fapp,x vs x graph. Area= 1/2kx^2. W= ½ kx^2 This work is positive for elongation of the spring (x>0) and for compression (x<0), and may be considered as elastic potential energy U stored in the spring. U=1/2kx^2. Elastic potential energy refers to energy that is stored in any object as a result of twisting, stretching or compressing. Angular Frequency, Period and Frequency of SHM If the mass is pulled aside from eq to x=A, in a truly frictionless system, the mass will oscillate indefinitely between the positions x=A and x=-A. x=Asin(wt) (in radians) w is a constant related to the force constant k and the mass m and A is the amplitude of the oscillation which the max distance that the mass moves away from eq. w=angular frequency of the oscillation and it is measured in rad/s. Set calculator to radians W is closely related to the period, T, of the oscillation, which is the time for one complete oscillation of cycle of the particle (from x=A to x=-A back to x=A). During a time interval of one period, the argument (wt) of the sine function in x=Asin(wt) must increase by 2pi radians in order that the value of x to return to the value at the beginning of the period. (deltaw=2pi). W=2pi/T The number of complete oscillations per second is the frequency. F=1/T (s^-1). W=2pif T=2pisqrroot(m/k) Example The vibration of the tympanic membrane in the ear is essentially shm. A tympanic membrane having a mass=2.4x10^-5kg is vibrating with frequency of 550Hz. What is the force constant associated with the membrane? What are the angular frequency and period? Solution F= 1/2pi x sqrroot(k/m) K=4pi^2f^2m =4pi^2(550)^2(2.4x10^-5) =2.9 x10^2 N/m w=2pif=2pi(550)=3.5x10^3 rad/s T=1/f=1/550= 1.8x10^-3 Position, Velocity and Acceleration in SHM X=Asin(2pit/T) X=Asin(2pift) t=0 can be chosen to be at any point in the cylclic motion of the particle. The graph will be sine but it can be shifted up to ½ period to the left or right with the exact shift depending on just where t=0 is chosen to be x=Asin(wt+phase angle). S is dimensionless called the phase angle and the exact value depends on the point in the cycle where t is chosen to be zero (depends on the amount of shift in the sine curve. If set t=0 and then S=0, sine is unshifted. Another choice is to let t=0 when the particle is at x=A. This gives A=Asin(S) where S=1 , pi/2. When shifted ¼ period left, it turns to cos. Objects that oscillate with the same period and pass through x=0 at the same time in the same direction oscillate in phase (phase angles must be identical). Amplitudes can be different If one object is at its maximum positive position when the other is at its negative max, their phase angles differ by pi radians or 180 degrees and they are said to oscillate pi radians out of phase. Range from 0 to +/- radians. As an object oscillates back and forth, its velocity changes with time t. Vmax= Aw at equilibrium position The velocity is zero at those times when the object is farthest from eq. that is x=+/-A. When x is negative, acceleration is positive and vice versa. amax=Aw^2 Example The wind is causing a leaf to vibrate, undergoing shm with a period of 0.40s and an amplitude of 1.2 cm. At t=0 the leaf is passing through its equilibrium position and travelling in the +x-direction. What is the equation relating x and t? Sketch a graph At t=0.72 seconds, how far is the leaf from the eq. Solution Since x and t=0 and the velocity is positive, then x is related to t by a sine function with no phase shift. x=Asin(wt) Since period T is given, use w=2pi/T x=Asin(2pit/T) x=(0.012)sin(5.0pit) see 1-11 for graph x=(0.012)sin((5.0)t) =(0.012)sin(5 x 0.72) =-0.011 m Example A rubber raft is bobbing up and down beside a dock (SHM). The simple harmonic motion is described by x=0.30sin(5.0t +0.40). What is the amplitude, period and frequency? Solution x=Asin(wt+S) x=0.30sin(5.0t +0.40) w=2pif so f=w/2pi=5.0/2pi=0.80Hz T=1/f=1/0.80=1.3 seconds Damped and Forced Harmonic Motion and Resonance Friction causes the amplitude to decrease with time. This damped harmonic motion. If the friction is large, the particle will not oscillate at all but will simply return slowly to eq (looks like ski slope). It is possible to attempt to drive a system into oscillation at any frequency by applying a driving force that varies periodically with time (forced harmonic motion) If the system is driven at its free oscillation frequency, oscillations of very large amplitude can easily be built up. This is called resonance. And the free oscillation frequency is resonant frequency. Think of a child’s sing. It has a natural oscillation and it is hard to push it back and forth at any other frequency. However, gentle pushes applied at the natural oscillating frequency can easily cause the swing to swing back and forth with a ver large amplitude. 1.3 Traveling Waves waves on the water. In travelling waves, the disturbance moves forward across the water although the particles move approx. up and down at one location. The water does not move forward though the wave disturbance does. Traveling wave-a disturbance which moves through space carrying energy without the bulk forward movement of matter. Sound is another example. In water, the particles move up and down perpendicular to the direction (horizontal) that the wave travels (transverse waves). Sound waves, the particles oscillate back and forth in the same direction of the wave (longitudinal waves). Light is transverse. Equation of a Traveling Wave Y is the size of the wave disturbance (displacement). In a water wave, y represent vertical displacement. On these graphs there can be numerous sine waves for different times (time is a phase shift, graph is a snapshot at a time). The wave is sine and repeats itself after a length (wavelength). The wavelength is the distance between the two successive crests or troughs. As the wave moves along, the disturbance y at any particular point at a fixed value of position x oscillates in shm about its eq. Frequency is the number of wavelengths that pass per second. T=wavelength=1/f Y=Asin ((2pi/T)t – (2pi/wavelength)x)) for a wave traveling in the POSITIVE DIRECTION K is the wave number k=2pi/wavelength (m^-1). Y=Asin(wt +/- kx) Wave Speed The speed of a wave can be expressed in terms of wavelength and period. (v=wavelength/T=wavelength x frequency). The speed depends on the properties of the medium through which it moves. Once a wave is launched, its frequency remains constant but as it passes through mediums, the speed and thus the wavelength change. Example A wave moves along a string in the x direction with speed of 8.0m/s, a frequency of 4.0 Hz and amplitude of 0.050m. a. wavelength, wave number, period, angular frequency and equation? b. Value of y for point at x=3/4 m at time 1/8 s Solution V= wavelength/f = (8.0)/(4.0)= 2.0m K= 2pi/wavelength = (2pi)/(2.0)= pi T=1/f =1/(4.0)= 0.25s W=2pif= 8pi rad/s Y=0.050sin(8pit-pix) Y=(0.050)sin((8pi)(1/8)-(pi)(3/4)) =0.035m A standing wave is produced when two waves of the same wavelength travel in opposite directions through the same medium. Very often, one wave is simply the reflection of the other wave from a surface. The two waves interfere with each other, and by the principle of wave superposition, the resultant wave displacement y is the sum of the displacements of the two interfering waves. Thus, when a crest from one wave is at the same position as the crest from the other, the waves are undergoing constructive interference making a supercrest. Where a crest interferes with a trough, the two waves undergo destructive interference and the resultant is a small displacement. If the crest and trough are the same size, completely cancelled. Equation for a Standing Wave An equation for a standing wave can be developed by adding the equations for two waves travelling in opposite directions. When a travelling wave is reflected from a surface at which the net vibration is zero (fixed end of a string), it undergoes a phase change of pi radians. Thus, the wave changes not only in direction, but in phase too. Y1= A sin (wt-kx) Y2= -A sin (wt+kx) Y=A(-2cos(2wt)sin(kx)) (2Acos(wt)) is amplitude that varies with time. Wavelength = 2pi/k. The maximum amplitude is 2A and minimum amplitude is zero. Nodes are the place of zero displacement (two waves remember) and antinodes are maximum. Distance from node to node is wavelength/2. A traveling wave in the +x direction could also be expressed as y1= -Asin(wt-kx). A graph for this has positive y values for the first ½ wavelength whereas y1=Asin(wt-kx) has negative. Also, for the negative direction, Y2=+Asin(wt+kx) has positive for the first half and y2=-Asin(wt+kx) has negative. Y=+/-(2Acos(t))sin(kx) Negative sign for the standing wave, for the incident wave, the sign is positive and positive infront of the A (or –ve and –ve). Problem The equation y=0.10sin(2pit-1.57x) represents a travelling wave. The wave reflects from a surface at which the net vibration must be zero. What is the equation of the reflected wave, resultant standing wave? Y2= -0.10 sin(2pit+ 1.57x) Standing wave y= -0.20cos(2pit)sin(1.57x) See 1-24 for sketching Standing Waves on a String A string fixed at both ends, such as a guitar string, are standing waves. Each end of the string is fixed and unable to vibrate so there must be a node there. A number of possible standing wave patterns have a node at each end. Simplest has no nodes between the two ends. 1 harmonic is the fundamental tone. Simplest has the largest amplitude and wavelength and is the one that is heard. The others are present. Standing waves on a string fixed at both ends occur only for waves of specific wavelengths. L=n(wavelength:n)/2 where n is the harmonic number Wavelength:n= 2l/n F:n=v/wavelength:n=nv/2l=v/2l F:n=nf1 Example The wavelength of high E is 0.627m. What are the three longest wavelengths of standing waves? The frequency of the fundamental tone is 329.6Hz, what is the wave speed? Solution Wavelength:n=2l/n where the three longest are n=1,2,3 Wavelength 1= 1.25 2=0.627 3=0.418 The fundamental is known to be 1.25m wavelength. V=wavelength x f =(1.25)(329.6)=412m/s In an organ pipe that is closed at one end, there is negligible sound vibration (a node) at the closed end and a max vibration (antinode) at the open end. ¼ wavelength is contained in the length of f1 tube. l=n(wavelength:n)/4 where n=1,3,5 STUDY GUIDE 2 Acoustic Resonance The music produced by wind instruments is the result of standing waves in the air inside the circular cylindrical tubes forming them. Displacement antinode=pressure node and vise verca. wavelength 1=2l, wavelength 2=l , wavelength n= 2l/n In a closed tube, a node must be placed at the closed end since particles cannot oscillate there. wavelength 1=4l, wavelength 2= 4l/3, wavelength n= 4l/(2n-1) Example How long does a hollow cylinder open at both ends have to ve to make a pan pipe with fundamental frequency at 256Hz? At what frequency will its second harmonic be? Solution F1=v/2l 256Hz=340m/s/2l l=0.66m f2=2v/2l f2=2f1=512 Standing waves in tubes are examples of acoustic resonances. Beats Demonstrated by two or more waves of slightyly different frequencies. The slight diff produces constructive interference (loud sound) and destructive at regular intervals. Same direction waves. Y1+y2= 2Asin((w1+w2)t/2)cos((w1-w2)t/2) Fb=|f1-f2| The Ear and Hearing Intensity of sound is perceived as loudness and frequency as pitch. Distinguish low frequencies better than high. Example Sound has an intensity I=2.0 x 10^-5, what is the intensity level of this sound. Solution IL= 10log(I/Io) =10 log(2.0 x 10^-5/ 1.0 x 10^ -12) = 73 dB Example A sound has intensity, I1 which is twice the intensity of I2, what is the difference in their intensity levels? Solution IL1-IL2= 10log(I1/I2)= 10 log2 = 3.0 dB The human ear detects the frequencies present in sound and the intensity of sound through each structure. Read more in the textbook if needed Energy, Power and Intensity Sound waves result from some source causing a disturbance in the medium. It requires energy to move the medium and that energy comes from the source. The energy E is carried away from the source in the form of a wave. The rate at which the source delivers energy to the medium is called the power. P=E/t In watts Example A sound amplifier and speaker can produce power of 20W. How much energy will keep it there for 1 hour? Solution P=E/t E=Pt= 20x 60 x 60= 7.2 x 10^4 J The energy delivered to the medium by the source goes into the energy of molecules as they execute shm. When the potential energy is max, kinetic energy is zero and vice versa. Energy density= 1/2pA^2w^2 The energy density is carried by the wave at the speed of propagation of the wave. The power carried by the wave across an area is the intensity. I=2pi^2pvf^2A^2 Example For a sound of frequency 1000Hz, the intensity at the threshold is 1.0 x 10^-12. What is the amplitude of motion? Solution 1.0 x 10^-12= 2pi^2 x 1.2 x 340 x 1000^2 x A^2 A= 1.1 x 10^-11 Pspherical = I(4pir^2) Example An underater source of sound produces spherical waves. The intensity of the sound 5m from the source is 7.0x10^-3. What is the intensity 2m away? What is the total acoustic power? Solution I2r2^2=I1r1^2 I2 x (2)^2 = (7.0 x 10 ^-3)(5)^2 I2= 4.4 x 10^-2 Psource=I(4pir^2)= 4pi (5)^2 x 7.0 x 10^-3 = 2.2W Reflection and Refraction When a ray of light is incident on a transparent interface between two media, some of the light is reflected from the surface and some enters the second medium and has its direction changed (refracted). Laws of reflection: 1) the angle that the reflected ray makes with the normal is equal to the angle that the incident ray makes with the normal (angel ref= angle inc). 2) The incident ray, reflected ray and normal all lie in a common plane The refracted ray which enters the second medium has its direction changed so that, in general, angle refr does not equal angle inc. Refraction occurs because light travels more slowly through transparent materials than it does a vacuum. The ratio of speed of light in a vacuum to the speed in a medium is termed the refractive index (n). N=c/v. Frequency of light does not change in passing mediums. N= wavelength(vacuum)/ wavelength m (medium) ( all n are greater than one) Snells law: n1sin(theta)1= n2sin(theta)2 White light is made up of many colours (frequencies) and each colour is refracted by a different amount because the speed of light in the medium, and therefore refractive index, is slightly frequency dependent. This leads to the different image positions in a lens. When n1n2, the angle is further from the normal Example A light ray passes from air into water making an angle (in air) with the normal of 30 degrees. What is the angle the ray makes with the normal in water? What is the speed of light in water? A ray passes from glass to water making an angle, in the glass, of 20 degrees with the normal, what is the angle of refraction? Solution n1sin(theta)1= n2sin (theta)2 (1.00)sin(30)=(1.33)sin(x) x= 22 degrees Vwater= c/nwater-(2.998x10^8)/1.33= 0.3759 n1sinx1=n2sinx2 (1.50)sin(20)= (1.33)sinx x= 23 degrees The Human Eye as an Optical Instrument The eye must be spherical to rotate about an axis with a slight hemispherical buldge called the cornea where light enters the eye. The space between the cornea is the crystalline lens which is filled with transparent fluid- aqueous humour. The iris adjusts the amount of light entering the eye and lies in front of the lens. The ciliary muscle is attached to the lens and divides the eye. Posterior chamber is filled with virteous humour. The visual axis of the eye extends through the center of the refracting surfaces, through the geometrical center of the sphere and intersects the back surface in a region of specialized cells- the fovea. Extending for about 120 degrees in all directions about the fovea is the layer of photoreceptive cells, nerve cels and fibres-the retina. The nerve fibres converge into a nerve bundle which leaves the eye at about 15 degrees to the insdie of the fovea. This region of the retina contains no photoreceptors and acts as a blind spot. The cornea is a thin transparent film which is kept in its hemispherical shape by the slightly higher pressure of the aq humour in the anterior chamber. The cornea is transparent section of the white sclera, which encloses the rest of the eye. Radius of curvature is about 8.0mm. The lens is formed of regular layers of specialized cells. It is bounded by two spherical surfaces. In its relaxed state, the anterior surface has a radius of curvature 10mm, and posterior is 6 mm. The lens is flexible and constiruction at its edges by the ciliary muscle squeezes the edge of the lens and increases the curvature of the anterior surface, thus increasing the curve of the anterior and increasing refractive power (ability to focus at different distances). Aging adds more layers and inner layer compress and harden making the lens stiff. This is calld presbyopia and leads to longsightedeness. Center of the fovea, retina is 0.1mm thick, away from fovea, retina is 0.2mm. Surface of the retina facing the incoming light is covered with axons of output cells called ganglion, whose cell bodies are located just below the surface of the retina. These axons converge toward the blind spot of the eye where they form the optic nerve. The optic nerve exits from the eye at the blind spot and ends in a center in the midbrain, the lateral geniculate nucleus, where the nerve fibres make contact with other cells. Blind spot is also a region where blood vessels enter and leave the eye, spreading out and forming a network over the retina which shades the cells in the retina. Two types of photoreceptor cells lie outside the fovea, rods and cones. At the center of the fovea, only cone cells. Some cone cells in the fovea best respond to red light while the rest of the cone cells there respond best to green light. The cone cells of the fovea are responsible for human day vision. Cone cells that respond best to blue light and yellow light are responsible for night vision. Rod cells- the light sensitive part of the cell that races the epithelium is filled with membrane in the form of discs. In phospholipid bilayer. The disc membrane contains proteins involved in vision. Rods=vision, cones=colour Rods are responsible for vision at low light levels (scotopic vision). They do not mediate color vision, and have a low spatial acuity. Cones are active at higher light levels (photopic vision), are capable of color vision and are responsible for high spatial acuity. The central fovea is populated exclusively by cones. Rod cells dominate the light absorbing characteristics of the retina. Image Formation by Spherical Surfaces N2-n1/r=n1/p+n2/q P=n2-n1/r diopters (m^-1) Powers can be positive or negative. Positive powers converge the light rays and negative powers diverge them. Powers of surfaces add to give total power. Shortist distance at which a cornea alone can form an image is obtained when the objects distance p is equal to infinity, and therefor q=3.1 cm Sign Convention 1) the objects distance, p, is positive when the direction from the optical surface to the object is against the flow of light. 2) The image distance, q, is positive when the direction from the optical surface to the image is with the flow of light 3) The radius of curvature, r, is positive when the direction from the optical surface to the center of curvature of the surface is with the flow of light. 4) The magnification, m, is positive when the image has the same orientation as the object. It is negative when the orientation of the object and image are opposite Look in text for more pictures Magnification For an object of finite height, y, the image height, y’ can also be determined. Y’/y=m=-n1q/n2p. Real image- results hen rays from the object actually converge at the image. It can be seen and caught on a screen. Real images have positive q. Virtual image- all light rays diverge in the image space but originate at the image. It can be seen but not caught on screen (mirrors), negative q. Inverted image and erect images Example Estimate the size of the image on the retina if the object is a man (2.0m tall) standing 5.0m in front of the eye. The image distance (cornea to retina) is about 2.5cm=0.025m. n2=1.34. Solution Y’=-n1q/n2p x y= - (1.00)(0.025)(2.00)/(1.34x5.0) - -0.0075m The Helmholtz Relaxed Eye Since the four refracting surgaces are not all close together, it is not sufficient to add all the powers. Instead, the position of the image formed by the cornea must be calculated. This image becomes the object for the anterior surface. Finally, the image formed by the anterior surface becomes the object for the posterior surface to form the final image on the retina. The relaxed eye is looking at infinite, so the first object distance is infinite. Therefore, the first two surfaces (the cornea) are dealt with by using or repeating the calculation. Nair/p1+naqh/q1= naqh-nair/r1 1/infinite + 1.34/q1= 1.34-1/0.0080= 31.5mm Next, find the image formed by the anterior lens surface. The image formed by the corneas is the object for this surface, therefore p2=q1- 0.0036=0.0220m which is also measure with the FOL (ie negative). Nlens/p3 + mvh/q3= nvh-nlens/r3 Q3=17.7mm. The final image is 17.7mm beyond the posterior surface of the lens or (17.7+3.6+3.6) 25mm behind the cornea. Reduced Eye Model In this model, the object distance= d+ 0.0017m and the image distance is (0.025-0.0017)=0.0233m. The reduced eye equation is P=1/(d+0.0017) + 1.32/0.0233 If d is infinite, P=57.5 1/m At d=5cm, more accommodation is required. Image Formation by Thin Lenses Where two refreacting surfaces are sufficiently close together that the total refracting power is just the simple sum of the two powers. P=P1+P2= (nL-n1)/r1-(nL-n2)/r2 n/p+n/q=(nL-n)(1/r1-1/r2)=P P=1/p+1/q=1/f (thin lens equation). The power and therefore focal length can be positive or negative. M=-q/p F is the focal length and the power equals 1/f and is a constant for a given lens. The focal length is the position where the image is formed of a very distant object. Example An object 1.5cm high is placed successively at 10cm, 5cm, 2cm from a lens in air with a focal length of 5 cm. Where is the image? Describe it. Solution 1/p+1/q=1/f 1/10+1/q=1/5 q=10cm far side of the lens from the object and the image is real M=-q/p=-10/10. Image height is =-1.0 x 1.5=-1.5 so it is inverted 1/5=1/5+1/q q= infinite (since it is greater than zero, the image is real) M=-q/p=-infinite (inverted and large) For 2, q= -3.3 so it is virtual and erect Example Find the magnification and image position if an object is placed 0.15m from the first lens of a two lens system. The focal lengths are 0.10m and 0.11m for the second. The two lenses are separated by 0.16m Solution For the first lens: 1/p1+1/q1=1/f1 1/0.15+1/q1= 1/0.1 q=0.30 The position of the image I1 at q1 yields the object distance for p2 for the second lens. Since q is downstream, p2 is negative. Thus P2=-(0.30-0.16)= -0.14m 1/p2+ 1/q2= 1/f2 1/-0.14 + 1/q2= 1/0.11 q2= 0.062 Thus the final imge is 0.062m right of the second lens M=M1M2=(-q1/p1)(-q2/p2) =(-0.030/0.15)(-0.062/-0.14)= -0.89 so it is inverted 0.89 times the object size. STUDY GUIDE 4 Diffaction is easiest to interpret when parallel wave fronts of light impinge on a narrow slit. Waves of light which pass through the slit are bent and interfere to produce alternating bands, bright and dark, on a distant screen. This is called the Fraunhofer diffraction pattern. This pattern results from the superposition of waves from new point sources. The first minimum on each side occurs when the difference in distance is lambda/2. It doesn’t represent the diffraction in our eyes because the pupil in our eyes is round rather than a slit. The fraunhofer diffraction pattern from a circular aperature such as a pupil consists of a bright spot surrounded by concentrict dark and bright circles. The half angle of the diffraction cone in the first dark ring is given by o=1.22lambda/a. Optical Defects of the Eye and Their Correction Three most common are 1. Hypermetropia (far-sighted in which the sharp image of a distant object calls beyond the retina) 2. Myopia (nearsightedness which the sharp image of the distant object falls in front of the retina) 3. Astigmatism (eye cannot focus clearly on horizontal and vertical lines simultaneously because the cornea is not spherical but more shaprly curved) In the hypermetropic eye, the curvature of the cornea is insufficient (power too small) to set the image on the back surface of the eye. In myopia, the cornea has too much curve (power too large) with respect to the shape of the eye. Thin specta
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