Exam 1 Equation Sheet | Foundations of Physics I | PHYS 2306, Study notes of Physics

Download Exam 1 Equation Sheet | Foundations of Physics I | PHYS 2306 and more Study notes Physics in PDF only on Docsity! Waves  y(x, t) = Acos(kx – ωt) | A = amplitude| k = constant | ω = angular velocityt) | A = amplitude| k = constant | ωt) | A = amplitude| k = constant | ω = angular velocity = angular velocity  ωt) | A = amplitude| k = constant | ω = angular velocity = 2π T | T = Period  T= 1 f | f = frequency  ωt) | A = amplitude| k = constant | ω = angular velocity = 2πf  k = 2π λ  v=λf | λ = wavelength  v=√Tμ | T = Tension | μ = Linear Density ( mass length )  T = v2μ | T = Tension  Wave Speed - Use eqtns. and algebra  Transverse Velocity – Use derivative of wave eqtn.  fbeat=f2-f1  Doppler Effect – Source Receding o fobserved =( v+vobserver v+vsource )∗f source  Doppler Effect – Source Approaching o fobserved =( v+vobserver v−vsource )∗f source  In Doppler Effect, v = c = speed of light = 343 m/s  Open Pipes, Closed Pipes, and Strings Tied at Both Ends o n = 1, 2, 3, etc. o f = nv 2L | L = length of pipe  Open-Closed Pipes and Strings Tied at One End o n = 1, 3, 5, etc. o f = nv 4 L | L = length of pipe  Sound Intensity o I=10∗log I I 0 | I = Intensity that can be detected o I∝ I 0 r2 I0 = Intensity that is made Electricity  Electric field points in direction that protons move  They point towards electrons, away from protons  k = Coulomb’s constant = 1 4 π ϵ 0  F = Gm1m2 r2 | G = 6.011 x 10-11  ∫E∗dA=¿ qenc ϵ 0  E = Gm r2 | For point charge  E = λ 2π ϵ 0 r | For infinite line | λ=linear charge density  λ= charge length  E = σ 2 ϵ0 | For infinite sheet  σ = charge density = Q A  Flux = φ = EA(cos(θ)) = )) = ∫E∗dA | THERE IS NOT FLUX INSIDE A CHARGED SHAPE  φtotal = Q ϵ 0 Problems One of the harmonics of a column of air open at one end and closed at the other has a frequency of 448 Hz and the next higher resonance has a frequency of 576 Hz. What is the fundamental frequency of the air column? Since it is CLOSED at one end… n = 1, 3, 5, etc. and f = nv 4 L SO f= Δnv 4 L = 2v 4 L = v 2L = Δ f 2 = 128 2 = 64 Hz. Three point charges q, q and –q are placed at the corners of a square of side d as shown. What is the magnitude of the net electric field at the center of the square? q First, find the distance to the center. If the side is length d, q -q r2 = ( 1 2 d)2 + ( 1 2 d)2 | r = √ 12 d2 = d √2 The bottom left field is: E = kq r2 = 2 kq d2 The bottom right and top left field work together to create: E = 2 kq r2 = 4 kq d2 To find the net electric field, we need to get the magnitude by using Pyth. Theorem: E = √( 2kqd2 ) 2 +¿¿ = √ 4k 2q2 d4 + 16k2q2 d4 = √20∗kq d2 The figure shows two tiny 5.0 gram spheres suspended from two very thin 1.0 m long threads. The spheres repel each other after being charged and hang at rest as shown at angle θ=0.006π rad. What is the charge on each sphere? Fdown=mg= (.005 kg ) (9.8 )=.049N F side=F side∗tan ( .006 π )=9.24∗10 −4 N r = (1m)sin(.006π)*2 )*2 F side= kqq r2 = k q2 r2 = (9∗109)(q)2 .03762 q=√ F side∗r 2 k =√ 9.24∗10 −4 ∗.03762 9∗109 =12nC A line of uniform charge is extended from x = -5.0m to x = -4.0m, and has a linear charge density of 7.0 x 10-6 C/m. What is the magnitude of the resulting electric field at x = 0m? E=∫ kdq r =kλ∫ x1 x2 dr r2 =kλ( 1 x1 − 1 x2 )=(9∗10 9) (7∗10−6 )( 14− 1 5 )=3.2 kN /C A traveling wave on a horizontal vibrating string has wavelength 2.0m and vertical amplitude 5.0cm. The string has a wave speed of 5.0m/s. A light insect with mass 6.0mg holds tight to the top side of the string. What is the maximal normal force on the insect exerted upwards by the string (do not ignore gravity)? A = .05m | m = .006g | λ = 2m | v = 5m/s F = ma = FN – mg | a= d2 y d t2 =−ω2 Asin (kx−ωt ) |amax|=ω 2 A=(2πf )2∗A=(2π vλ ) 2 ∗A F=m (a+g )=.006 ( 4 π 2 ∗52 4 ∗( .05 )+9.8)=.13N Raptors, having a much more advanced ability to hear, can detect sound with an intensity level as low as -4.0 dB. When you walk, the resulting sound has an intensity level of 40 dB at a distance of 0.5 m. How close to the raptor can you walk? 10 log I 1 I 2 =−40−4=−44 | log I 1 I 2 =−4.4 Then raise both sides to the 10th power and make ratio of distances: I 1 I 2 =10−4.4=( R2 R1 ) 2 R2 = .5m | .5 R1 =√10−4.4 | R1= .5 √10−4.4 =79m A standard demonstration charges a conducting sphere with small aluminum pie plates on top… Approximately what charge has to be on the sphere for the last one to fly off? Assume the pie plates had a radius of 5 cm and mass of 1 g, while the sphere has a radius of 15 cm. m = .001kg | R = .15m | r = .05m | F=qE | q=( π r 2 4 π R2 )∗Q E= kQ R2 | mg=F=qE=( r2 4 R2 )(k Q2 R2 ) | Q=√ 4mgR 4 k∗r2 =1∗10−6C Two 10cm x 10cm conducting flat plates have a total charge of +1 μC and -1 μC C and -1 μC and -1 μC C respectively. When they are very close to each other, separated by just an insulating oxide layer about 100 nm thick, what is the force pulling them together? For Plates: E= σ 2 ϵ 0 | σ = 10−6(Thecharge) 10−2(Thearea) =10−4 F=qE= q∗σ 2 ϵ 0 = (10−6 ) (10−4 ) 2 (8.8∗10−12) =5.7N A long cylindrical insulator has a uniform charge density, p, throughout its volume. What is the electric field as a function of r, the radial distance away from the center (r<1)? ∫E∗dA= qenc ϵ 0 | E (2πrl )= π r2 lp ϵ 0 | E= pr 2 ϵ 0 Two waves are present on the same string: y1 = cos(2x + 3t) and y2 = cos(2x – 3t). What is the distance between adjacent nodes on the string? k= 2π λ =2 so λ=π Therefore, since 1 wavelength includes 2 nodes, the distance between nodes is λ 2 = π 2 . A conducting sphere has a charge +2Q on it, and is located inside a conducting shell with net charge +Q on it. Rank the outgoing electric flux from largest to smallest through centered spherical surfaces: A) 0 < r < R1 B) R1 < r < R2 C) R2 < r < R3 and D) r > R3 ∫E∗dA= qenc ϵ 0 =Flux Flux(A) = 0; Flux(B) = 2Q ϵ 0 ; Flux(C) = 0; Flux(D) = 3Q ϵ 0 ; NOTE: NO FLUX IN A CHARGE OBJECT A charge Q and a charge -2Q are located at A and B along a straight line. Which graph correctly shows the magnitude of EX(x) along this line? The answer is A because only A and C represent two different charges and only the charge at a large positive x must be negative (like in A). A line of uniform charge exists between x = 0.70 and x = 1.00, and has a linear charge density of 7.0 μC and -1 μC C/m. What is the magnitude of the resulting electric field at x = 0.00? E=k∫ dq r2 =k∫ 0.7 1.0 λdr r2 =−kλ( 11− 1 0.7 )=(9∗10 9 ) (7∗10−6 )( 10.7− 1 1 )=2.7∗10 4 N C A sphere has a volume charge density given by p(r) = D r for r < R. What is the resulting electric field for r < R? ∫E∗dr= qenc ϵ 0 | E (4π r2 )= 1 ϵ 0 ∫ 0 r p (4 π r2 )dr E r2= D ϵ 0 ∫ 0 r r∗dr= D ϵ 0 ∗r2 2 so dividing by r2… E= D 2 ϵ 0 A point charge, Q, is located a distance, d, from the center of a circle of radius d, along the axis of the circle. What is the electric flux to the right through the plane bounded by this circle, due to the charge? Flux total= Q ϵ 0 Since the shape is a ring: 2π∫ 0 π 4 sin (θ )dθ 4 π = −cos ( π4 )−(−cos (0 )) 2 = 1 2 − √2 4 =.146 Q ϵ 0 If you are 4 meters from a spherical sound source, how much further must you go to change the sound level by -6 dB? −6=10 log 42 (4+x)2  10−0.6=( 4 4+x ) 2  4+x= 4 √10−0.6  x=3.98m A sound wave is traveling to the right, and the displacement of air molecules is show in the graph. Rank the delta-pressure at points A, B, C, and D from most pos to most neg.