Chapter 15, example problems:


 Ralf Roberts
 7 years ago
 Views:
Transcription
1 Chapter, example problems: (.0) Ultrasound imaging. (Frequenc > 0,000 Hz) v = 00 m/s. λ <.0 mm. Frequenc required > 00 m/s /.0 mm = Hz. (Smaller wave length implies larger frequenc, since their product, being equal to the sound velocit, can not change. (.) Speed of propagation vs. particle speed. Eq. (.): (x, t) = A cos[ω (x/v t)] = A cos [πf (x/v t)]. (a) Putting in ω = v k, k = π / λ, and using the fact that v (x/v t) = (x v t), we obtain: (x, t) = A cos[( π / λ) (x v t)]. (b) The transverse velocit of a particle in the string on which the wave travels: The transverse displacement of a particle at position x on that string as a function of t is (x, t) = A cos[( π / λ) (x v t)], so the transverse velocit of that particle as a function of t is: v (x, t) = {A cos[( π / λ) (x v t)] } / t x = A sin[( π / λ) (x v t)] ( π / λ) ( v) = ( π f) Α sin[( π / λ) (x v t)]. (c) The maximum transverse speed of a particle in the string, v max, is just the amplitude of the above velocit wave, namel π f Α (= ω Α). (This is the same expression as that for the maximum velocit of a harmonic oscillator. This is not a coincidence. Ever point in a wave is indeed executing a simple harmonic motion, onl with a phase which varies with x.) This maximum transverse velocit of a particle in the string is in general different from the wave velocit. For them to be equal, we would have to require ω Α = v, or, since ω = v k, we would have to require kα =, which just means Α = λ / π. For this maximum transverse velocit to be less or greater than the wave velocit, wee need A to be less or greater than λ / π. (In realit, A is practicall alwas much less than λ / π. Thus the maximum transverse velocit of a particle in the string is practicall alwas much less than the wave velocit.) (.6).0 m string, weight. N. Tied to the ceiling. Lower end supports a weight W. Plucked slightl. Wave traveling up the string obes: (x, t) = (8.0 mm) cos(7 m x 70 s t)] (thus x is measured upward). (a) Time for a pulse to travel the full length of the string: We need the wave velocit, which is 70 s / 7 m =.87 m/s. Then the travel time is.0 m /.87 m/s = 0.09 s. (b) Actuall, this problem is correct onl if the weight of the string,. N, is much less than the weight W, and therefore can be neglected with respect to the latter. Then the tension in the string, F, is due to W alone, and is therefore equal to W. The linear mass densit of the string, μ, is (. N / 9.8 m/s ) /.0 m = kg/m. Then using the formula for the velocit of a transverse wave on a string, v = (F / μ), we can find W = v μ = (.87 m/s) kg/m =. N, which is indeed much larger than. N, confirming the validit of the approximation we made in the beginning. (But the error is about %.)
2 (c) The number of wavelengths on the string at an instant of time =.0 m / (π / 7 m ) =.. (d) Equation for waves traveling down the string is: (x, t) = A cos [( π / λ) (x m/s t)]. (That is, A and λ can be different for different waves, but the velocit of the waves must all be the same 87. m/s, since it is determined b the tension and the linear mass densit of the string.) (.) Threshold of pain. Intensit of sound is 0. W/m at 7. m from a point source. At what distance from the point source will intensit reach the threshold of pain,.0 W/m? Denote the new distance D. Then.0 W/m πd = 0. W/m π (7. m) = power output of the point source. Hence D = 7. m (0. W/m /.0 W/m ) / =.87 m. (Note: πd is the area of a sphere of radius D.) Thus ou have to move 7. m .87 m =.0 m toward the source of sound to reach the threshold of sound. (.8) Interference of triangular pulses. v =.00 cm/s..00 cm.00 cm.00 cm.00 cm.00 cm.00 cm.00 cm.00 cm/s 0. s = 0. cm. The two pulses will barel touch after traveling for 0. s, producing:.00 cm.00 cm.00 cm.00 cm.00 cm.00 cm.00 cm/s 0.0 s =.0 cm. The two pulses now overlapped b.00 cm..00 cm.00 cm The superposed pulses now show a flat top for a width of.00 cm.
3 .00 cm/s 0.7 s =. cm. The two pulses now totall overlap. Their superposition gives one triangular pulse of twice the height, but the same old width of.00 cm. But this shape is maintained for onl one instant of time, since part of it is moving to the right, and part of it is moving to the left..00 cm.00 cm.0 cm.00 cm.00 cm.0 cm.00 cm/s.00 s =.00 cm. The two pulses have now moved pass each other b.0 cm. The superposition again produces the trapezoidal shape with a flat top of.00 cm just like when it has onl moved for 0. s. But notice the directions of the two horizontal arrows, which indicate the directions of motion of the two constituent pulses..00 cm.00 cm.00 cm/s. s =.0 cm. The two pulses have now moved pass each other b their own width, so the no longer overlap. The situation appears just like when the have moved for onl 0. s, except for the directions of the two horizontal arrows, which indicate the directions of motion of the two constituent pulses..00 cm.00 cm.00 cm.00 cm.00 cm.00 cm
4 (.) Distance between adjacent antinodes of a standing wave is.0 cm. A particle at an antinode oscillates in SHM with amplitude 0.80 cm and period s. The string is along x and is fixed at x = 0. (a) Adjacent nodes are also.0 cm apart. (b) Recall that cos (A B) + cos (A + B) = cos A cos B, which implies that A cos (kx ωt) + A cos (kx + ωt) = A cos (kx) cos (ωt). Hence the amplitude of a standing wave is a factor two larger than the amplitude of either of the two constituent traveling waves, but their wavelengths are the same, and their frequencies are the same. Thus: Wavelength of each traveling wave λ = 0.0 cm; Amplitude of each traveling wave A = 0. cm. Speed of each traveling wave v = λ f = λ / Τ = 0.0 cm / 0.07 s = 00 cm/s = m/s. (c) Maximum transverse speed of a point at the antinode of the standing wave = Aω = A (π / T) = 0.80 cm (π / s) = 7. cm/s. Minimum speed = 0. (d) Shortest distance bweteena node and an antinode is 7.0 cm. (.) Weightless ant. An ant of mass m stands on top of a horizontal stretched rope. Rope has mass per unit length μ and tension F. A sinusoidal transverse wave of wave length λ propagates on the rope. Motion is in the vertical plane. Find the minimum wave amplitude which will make the ant momentaril weightless. The mass m is so small that it will not affect the propagating wave. The ant will becomes momentaril weightless when it momentaril needs no support from the rope. This happens when its mass times the maximum downward acceleration of the SHM of the ant is exactl equal to its weight. This happens when the ant is at the highest point of its SHM. [Then just before or after this moment, the acceleration will still be downward but with a magnitude less than this maximum value, and the weight of the ant will need to be partiall cancelled b an upward support force from the rope, so that the net downward force is still equal to (now smaller) ma. Thus the weightlessness feeling happens onl at that brief moment when the ant is at the highest point of the SHM. During the half ccle of the motion when the acceleration is pointing upward, the weight is pointing in the wrong direction to provide the needed force. So an upward support force from the rope larger than the weight of the ant will exist, so that the net force is upward, and is still equal to ma. Thus in this half ccle the ant actuall weights more than its actual weight, if ou put a scale between the ant and the rope to measure this apparent weight.] The magnitude of the maximum acceleration of the SHM is mω. But ω = v k = (F / μ) (π / λ). Thus we must require ma(g / μ) (π / λ) = mg, giving A = (μ g / F) (λ / π). Let us check the unit. The unit of μ g is N/m. The unit of (μ g / F) is therefore just /m. The unit of (λ / π) is just m. So their product has the unit of m, which is the right unit for A.
5 (.6) More general sinusoidal wave: (x, t) = A cos (k x ω t + φ ). (a) At t = 0. (x, t = 0) = A cos (k x + φ ). For φ = 0, (x, 0) = A cos (kx): 0 x For φ = π/, (x, 0) = A cos (kx + π/): That is, even at x = 0, is alread cos (π/). 0 x For φ = π/, (x, 0) = A cos (kx + π/): That is, even at x = 0, is alread cos (π/). From this figure, we can see that it is the same as: (x, 0) = A sin (kx). The wave is then given b: (x, 0) = A sin (kx ωt). For φ = π/ = π/ + π/, (x, 0) = A cos (kx +π/): That is, even at x = 0, is alread cos (π/). 0 0 x x For φ = π/ = π + π/, (x, 0) = A cos (kx + π/): That is, even at x = 0, is alread cos (π/). From this figure, we can see that it is the same as: (x, 0) = + A sin (kx). The wave is then given b: (x, 0) = + A sin (kx ωt). 0 x
6 (b) Transverse velocit: v = / t = + Aω sin (k x ω t + φ ). (c) At t = 0, a particle at x = 0 has displacement = A /. Can one determine φ? No. There are several candidates: At t = x = 0, (0, 0) = + A cos (φ ). So we need to demand cos (φ ) = /. The answer is φ = π/ or 7π/. (Note: φ = 7π/ is the same as φ = π/, since their difference is π.) If we also know that a particle at x = 0 is moving toward = 0 at t = 0, determine φ. For x = 0, (0, t) = A cos ( ω t + φ ). We test φ = π/ and 7π/, and see which one works. We first calculate the transverse velocit v = / t = + Aω sin ( ω t + φ ), and evaluate it at t = 0, and get v (x = 0, t = 0) = + Aω sin (φ ). At φ = π/, this transverse velocit is positive, so it is not moving toward =0. So we reject it. At φ = 7π/, this transverse velocit is negative, so it is moving toward =0. So we accept this answer. Hence we conclude that φ = 7π/ (or, equivalentl, π/). Can one get this answer without doing so much math? es! First, one should realize that if (x, t) = A cos (kx ωt + φ), then the wave is moving toward positive x, or to the right in our plots. Then looking at our plot for φ = π/, and let the curve move to the right. One will find that will increase. at x = 0. So we should reject this answer. Then we do the same thing for φ = 7π/ or π/, and see that it can be accepted. (d) In general, to determine φ, we need to know: (i) (x = 0, t = 0); and (ii) the sign of v (x = 0, t = 0). (.68) Vibrating string. 0.0 cm long. Tension F =.00 N. Five stroboscopic pictures shown. Strobo rate is 000 flashes per minute. Maximum displacement occurs at flashes and, with no other maxima in between. P.0 cm.0 cm (a) Period: T = min = s (the time for 8 flashes). Frequenc f = / T = 0. Hz. Wavelength = 0.0 cm. All for either of the two the traveling wave on this string that are moving in opposite directions to form this standing wave. (b) The second harmonic (also known as the first overtone). (c) Speed of the traveling waves on this string v = λ / T = 0.0 cm / (0.096 s) = 0.8 cm/s =. m/s. (d) Point P. (i) In position it is moving with a transverse velocit of v = 0. (ii) In position it is moving with a transverse velocit of v = Aω = A(π f ) =
7 .0 cm (π 0. s ) =.0 cm 6.7 s = 96. cm/s =.96 m/s. (e) Mass of this string m = 0.0 m μ = 0.0 m [.00 N / (. m/s) ] = 0.08 kg = 8. g. (.7) String, Both ends held fixed. Vibrates in the third harmonic. Speed 9 m/s. Frequenc 0 Hz. Amplitude at an antinode is 0.00 cm. λ = 9 m/s / 0 Hz = 0.8 m. (a) Find amplitude of oscillation: (i) at x = 0.0 cm from the left end of the string. The general equation for the standing wave is : (x, t) = 0.00 m sin (π x / 0.8 m) sin (π 0 Hz t ) [We let x = 0 be the left end of the string. We should have the factor sin (π x / 0.8 m) because the left end is a node, and sin 0 = 0 Also, the wavelength is 0.8 m, and so at x = 0.8 m we should get sin (π) = 0.] So the amplitude of transverse oscillation at x = 0.0 cm is: 0.00 m sin (π 0 cm / 0.8 m) = 0. (ii) at x = 0.0 cm from the left end of the string, the amplitude is: 0.00 cm sin (π 0 cm / 0.8 m) = 0.00 cm. (iii) at x = 0.0 cm from the left end of the string, the amplitude is: 0.00 cm sin (π 0 cm / 0.8 m) = 0.8 cm/s. (b) At each point of part (a), find time to go from largest upward displacement to largest downward displacement. The answer is the same, and is equal to: T /, or / ( f ) = s, except at the nodes of the standing wave, i.e., at x = 0.0 cm in the three cases of part (a). (c) Maximum transverse velocit of the motion: v (x, t) = (x, t) / t = 0.00 cm (π 0 Hz) sin (π x / 0.8 m) cos (π 0 Hz t ) The amplitude of transverse velocit at distance x from the left end is 0.00 m (π 0 Hz) sin (π x / 0.8 m). So at x = 0.0 cm from the left end of the string, the maximum transverse velocit is: 0.00 m (π 0 Hz) sin (π 0.0 m / 0.8 m) = 0. at x = 0.0 cm from the left end of the string, the maximum transverse velocit is: 0.00 cm (π 0 Hz) sin (π 0.0 m / 0.8 m) = 60. cm/s = 6.0 m/s. at x = 0.0 cm from the left end of the string, the maximum transverse velocit is: 0.00 cm (π 0 Hz) sin (π 0.0 m / 0.8 m) = 6. cm/s =.7 m/s. Maximum transverse acceleration of the motion: a (x, t) = v (x, t) / t = 0.00 cm (π 0 Hz) sin (π x / 0.8 m) sin (π 0 Hz t ) The amplitude of transverse acceleration at distance x from the left end is 0.00 m (π 0 Hz) sin (π x / 0.8 m). So at x = 0.0 cm from the left end of the string, the maximum transverse acceleration is: 0.00 m (π 0 Hz) sin (π 0.0 m / 0.8 m) = 0. at x = 0.0 cm from the left end of the string, the maximum transverse acceleration is: 0.00 cm (π 0 Hz) sin (π 0.0 m / 0.8 m) =
8 60. cm/s = cm/s = 9096 m/s. at x = 0.0 cm from the left end of the string, the maximum transverse accelration is: 0.00 cm (π 0 Hz) sin (π 0.0 m / 0.8 m) = 67 cm/s = 6 m/s. Actuall, these answers are eas to see b looking at the figure: The second node from the left end is at 0.8 m, because ou see a whole wavelength between here and the left end. Then the first node from the left end must be at 0. m. The first antinode from the left end must be at 0. m. This is wh at x = 0. m we find the largest amplitudes of velocit and acceleration, and at x = 0. m we find vanishing amplitudes of velocit and acceleration. In addition, the amplitudes of velocit and acceleration at x = 0. m is down from those at x = 0. m b a factor sin (π/) = / = 0.707, as the xdependent factor is simpl a sine function to give sin (π/) = at x = 0. m, and to give sin (π) = 0 at x = 0. m. It will also give 0 at x = 0.8 m, and at x =. m, which is the right end of the string. The correspond to sin (π) and sin (π) from the xdependent factor.
SOLUTIONS TO CONCEPTS CHAPTER 15
SOLUTIONS TO CONCEPTS CHAPTER 15 1. v = 40 cm/sec As velocity of a wave is constant location of maximum after 5 sec = 40 5 = 00 cm along negative xaxis. [(x / a) (t / T)]. Given y = Ae a) [A] = [M 0 L
More informationSolution: F = kx is Hooke s law for a mass and spring system. Angular frequency of this system is: k m therefore, k
Physics 1C Midterm 1 Summer Session II, 2011 Solutions 1. If F = kx, then k m is (a) A (b) ω (c) ω 2 (d) Aω (e) A 2 ω Solution: F = kx is Hooke s law for a mass and spring system. Angular frequency of
More informationphysics 1/12/2016 Chapter 20 Lecture Chapter 20 Traveling Waves
Chapter 20 Lecture physics FOR SCIENTISTS AND ENGINEERS a strategic approach THIRD EDITION randall d. knight Chapter 20 Traveling Waves Chapter Goal: To learn the basic properties of traveling waves. Slide
More informationStanding Waves on a String
1 of 6 Standing Waves on a String Summer 2004 Standing Waves on a String If a string is tied between two fixed supports, pulled tightly and sharply plucked at one end, a pulse will travel from one end
More information16.2 Periodic Waves Example:
16.2 Periodic Waves Example: A wave traveling in the positive x direction has a frequency of 25.0 Hz, as in the figure. Find the (a) amplitude, (b) wavelength, (c) period, and (d) speed of the wave. 1
More informationSpring Simple Harmonic Oscillator. Spring constant. Potential Energy stored in a Spring. Understanding oscillations. Understanding oscillations
Spring Simple Harmonic Oscillator Simple Harmonic Oscillations and Resonance We have an object attached to a spring. The object is on a horizontal frictionless surface. We move the object so the spring
More informationCopyright 2008 Pearson Education, Inc., publishing as Pearson AddisonWesley.
Chapter 20. Traveling Waves You may not realize it, but you are surrounded by waves. The waviness of a water wave is readily apparent, from the ripples on a pond to ocean waves large enough to surf. It
More information1) The time for one cycle of a periodic process is called the A) wavelength. B) period. C) frequency. D) amplitude.
practice wave test.. Name Use the text to make use of any equations you might need (e.g., to determine the velocity of waves in a given material) MULTIPLE CHOICE. Choose the one alternative that best completes
More informationThe Physics of Guitar Strings
The Physics of Guitar Strings R. R. McNeil 1. Introduction The guitar makes a wonderful device to demonstrate the physics of waves on a stretched string. This is because almost every student has seen a
More informationPhysics 231 Lecture 15
Physics 31 ecture 15 Main points of today s lecture: Simple harmonic motion Mass and Spring Pendulum Circular motion T 1/f; f 1/ T; ω πf for mass and spring ω x Acos( ωt) v ωasin( ωt) x ax ω Acos( ωt)
More informationAP Physics C. Oscillations/SHM Review Packet
AP Physics C Oscillations/SHM Review Packet 1. A 0.5 kg mass on a spring has a displacement as a function of time given by the equation x(t) = 0.8Cos(πt). Find the following: a. The time for one complete
More informationCambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level
Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level *0123456789* PHYSICS 9702/02 Paper 2 AS Level Structured Questions For Examination from 2016 SPECIMEN
More informationPractice Test SHM with Answers
Practice Test SHM with Answers MPC 1) If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system are true? (There could be more than one
More information226 Chapter 15: OSCILLATIONS
Chapter 15: OSCILLATIONS 1. In simple harmonic motion, the restoring force must be proportional to the: A. amplitude B. frequency C. velocity D. displacement E. displacement squared 2. An oscillatory motion
More informationPhysics 1120: Simple Harmonic Motion Solutions
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Physics 1120: Simple Harmonic Motion Solutions 1. A 1.75 kg particle moves as function of time as follows: x = 4cos(1.33t+π/5) where distance is measured
More informationANALYTICAL METHODS FOR ENGINEERS
UNIT 1: Unit code: QCF Level: 4 Credit value: 15 ANALYTICAL METHODS FOR ENGINEERS A/601/1401 OUTCOME  TRIGONOMETRIC METHODS TUTORIAL 1 SINUSOIDAL FUNCTION Be able to analyse and model engineering situations
More informationLesson 11. Luis Anchordoqui. Physics 168. Tuesday, December 8, 15
Lesson 11 Physics 168 1 Oscillations and Waves 2 Simple harmonic motion If an object vibrates or oscillates back and forth over same path each cycle taking same amount of time motion is called periodic
More informationDetermination of Acceleration due to Gravity
Experiment 2 24 Kuwait University Physics 105 Physics Department Determination of Acceleration due to Gravity Introduction In this experiment the acceleration due to gravity (g) is determined using two
More informationPHYS 1014M, Fall 2005 Exam #3. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
PHYS 1014M, Fall 2005 Exam #3 Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A bicycle wheel rotates uniformly through 2.0 revolutions in
More informationPhysics 41 HW Set 1 Chapter 15
Physics 4 HW Set Chapter 5 Serway 8 th OC:, 4, 7 CQ: 4, 8 P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59, 67, 74 OC CQ P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59,
More informationSimple Harmonic Motion
Simple Harmonic Motion 1 Object To determine the period of motion of objects that are executing simple harmonic motion and to check the theoretical prediction of such periods. 2 Apparatus Assorted weights
More informationAP1 Oscillations. 1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false?
1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false? (A) The displacement is directly related to the acceleration. (B) The
More informationOscillations. Vern Lindberg. June 10, 2010
Oscillations Vern Lindberg June 10, 2010 You have discussed oscillations in Vibs and Waves: we will therefore touch lightly on Chapter 3, mainly trying to refresh your memory and extend the concepts. 1
More informationwww.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x
Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity
More informationPhysics 9e/Cutnell. correlated to the. College Board AP Physics 1 Course Objectives
Physics 9e/Cutnell correlated to the College Board AP Physics 1 Course Objectives Big Idea 1: Objects and systems have properties such as mass and charge. Systems may have internal structure. Enduring
More informationAnswer, Key Homework 3 David McIntyre 1
Answer, Key Homewor 3 Daid McIntyre 1 This printout should hae 26 questions, chec that it is complete Multiplechoice questions may continue on the next column or page: find all choices before maing your
More informationC B A T 3 T 2 T 1. 1. What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N
Three boxes are connected by massless strings and are resting on a frictionless table. Each box has a mass of 15 kg, and the tension T 1 in the right string is accelerating the boxes to the right at a
More informationWaves and Sound. AP Physics B
Waves and Sound AP Physics B What is a wave A WAVE is a vibration or disturbance in space. A MEDIUM is the substance that all SOUND WAVES travel through and need to have in order to move. Two types of
More information18 Q0 a speed of 45.0 m/s away from a moving car. If the car is 8 Q0 moving towards the ambulance with a speed of 15.0 m/s, what Q0 frequency does a
First Major T042 1 A transverse sinusoidal wave is traveling on a string with a 17 speed of 300 m/s. If the wave has a frequency of 100 Hz, what 9 is the phase difference between two particles on the
More informationLecture 6. Weight. Tension. Normal Force. Static Friction. Cutnell+Johnson: 4.84.12, second half of section 4.7
Lecture 6 Weight Tension Normal Force Static Friction Cutnell+Johnson: 4.84.12, second half of section 4.7 In this lecture, I m going to discuss four different kinds of forces: weight, tension, the normal
More informationPHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 17. February 13, 2013
PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 17 February 13, 2013 0.1 A 2.00kg object undergoes an acceleration given by a = (6.00î + 4.00ĵ)m/s 2 a) Find the resultatnt force acting on the object
More informationMatter Waves. Home Work Solutions
Chapter 5 Matter Waves. Home Work s 5.1 Problem 5.10 (In the text book) An electron has a de Broglie wavelength equal to the diameter of the hydrogen atom. What is the kinetic energy of the electron? How
More informationTennessee State University
Tennessee State University Dept. of Physics & Mathematics PHYS 2010 CF SU 2009 Name 30% Time is 2 hours. Cheating will give you an Fgrade. Other instructions will be given in the Hall. MULTIPLE CHOICE.
More informationSimple Harmonic Motion Experiment. 1 f
Simple Harmonic Motion Experiment In this experiment, a motion sensor is used to measure the position of an oscillating mass as a function of time. The frequency of oscillations will be obtained by measuring
More informationColumbia University Department of Physics QUALIFYING EXAMINATION
Columbia University Department of Physics QUALIFYING EXAMINATION Monday, January 13, 2014 1:00PM to 3:00PM Classical Physics Section 1. Classical Mechanics Two hours are permitted for the completion of
More informationUnit  6 Vibrations of Two Degree of Freedom Systems
Unit  6 Vibrations of Two Degree of Freedom Systems Dr. T. Jagadish. Professor for Post Graduation, Department of Mechanical Engineering, Bangalore Institute of Technology, Bangalore Introduction A two
More informationNotice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case.
HW1 Possible Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. Tipler 14.P.003 An object attached to a spring has simple
More informationHOOKE S LAW AND OSCILLATIONS
9 HOOKE S LAW AND OSCILLATIONS OBJECTIVE To measure the effect of amplitude, mass, and spring constant on the period of a springmass oscillator. INTRODUCTION The force which restores a spring to its equilibrium
More informationGiant Slinky: Quantitative Exhibit Activity
Name: Giant Slinky: Quantitative Exhibit Activity Materials: Tape Measure, Stopwatch, & Calculator. In this activity, we will explore wave properties using the Giant Slinky. Let s start by describing the
More informationAcceleration due to Gravity
Acceleration due to Gravity 1 Object To determine the acceleration due to gravity by different methods. 2 Apparatus Balance, ball bearing, clamps, electric timers, meter stick, paper strips, precision
More informationDetermination of g using a spring
INTRODUCTION UNIVERSITY OF SURREY DEPARTMENT OF PHYSICS Level 1 Laboratory: Introduction Experiment Determination of g using a spring This experiment is designed to get you confident in using the quantitative
More informationCHAPTER 15 FORCE, MASS AND ACCELERATION
CHAPTER 5 FORCE, MASS AND ACCELERATION EXERCISE 83, Page 9. A car initially at rest accelerates uniformly to a speed of 55 km/h in 4 s. Determine the accelerating force required if the mass of the car
More informationW i f(x i ) x. i=1. f(x i ) x = i=1
Work Force If an object is moving in a straight line with position function s(t), then the force F on the object at time t is the product of the mass of the object times its acceleration. F = m d2 s dt
More informationHOOKE S LAW AND SIMPLE HARMONIC MOTION
HOOKE S LAW AND SIMPLE HARMONIC MOTION Alexander Sapozhnikov, Brooklyn College CUNY, New York, alexs@brooklyn.cuny.edu Objectives Study Hooke s Law and measure the spring constant. Study Simple Harmonic
More informationPHYSICS HIGHER SECONDARY FIRST YEAR VOLUME  II. Revised based on the recommendation of the Textbook Development Committee. Untouchability is a sin
PHYSICS HIGHER SECONDARY FIRST YEAR VOLUME  II Revised based on the recommendation of the Textbook Development Committee Untouchability is a sin Untouchability is a crime Untouchability is inhuman TAMILNADU
More informationThe Sonometer The Resonant String and Timbre Change after plucking
The Sonometer The Resonant String and Timbre Change after plucking EQUIPMENT Pasco sonometers (pick up 5 from teaching lab) and 5 kits to go with them BK Precision function generators and Tenma oscilloscopes
More informationv = fλ PROGRESSIVE WAVES 1 Candidates should be able to :
PROGRESSIVE WAVES 1 Candidates should be able to : Describe and distinguish between progressive longitudinal and transverse waves. With the exception of electromagnetic waves, which do not need a material
More informationSOLID MECHANICS DYNAMICS TUTORIAL NATURAL VIBRATIONS ONE DEGREE OF FREEDOM
SOLID MECHANICS DYNAMICS TUTORIAL NATURAL VIBRATIONS ONE DEGREE OF FREEDOM This work covers elements of the syllabus for the Engineering Council Exam D5 Dynamics of Mechanical Systems, C05 Mechanical and
More informationAP1 Waves. (A) frequency (B) wavelength (C) speed (D) intensity. Answer: (A) and (D) frequency and intensity.
1. A fire truck is moving at a fairly high speed, with its siren emitting sound at a specific pitch. As the fire truck recedes from you which of the following characteristics of the sound wave from the
More informationMidterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m
Midterm Solutions I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel a solid disc) of mass M, radius R, anchored at its center but free to rotate i) Which of
More informationPHYA2. General Certificate of Education Advanced Subsidiary Examination June 2010. Mechanics, Materials and Waves
Centre Number Surname Candidate Number For Examiner s Use Other Names Candidate Signature Examiner s Initials Physics A Unit 2 For this paper you must have: a ruler a calculator a Data and Formulae Booklet.
More informationLecture L222D Rigid Body Dynamics: Work and Energy
J. Peraire, S. Widnall 6.07 Dynamics Fall 008 Version.0 Lecture L  D Rigid Body Dynamics: Work and Energy In this lecture, we will revisit the principle of work and energy introduced in lecture L3 for
More informationPhysics 53. Kinematics 2. Our nature consists in movement; absolute rest is death. Pascal
Phsics 53 Kinematics 2 Our nature consists in movement; absolute rest is death. Pascal Velocit and Acceleration in 3D We have defined the velocit and acceleration of a particle as the first and second
More informationSimple Harmonic Motion(SHM) Period and Frequency. Period and Frequency. Cosines and Sines
Simple Harmonic Motion(SHM) Vibration (oscillation) Equilibrium position position of the natural length of a spring Amplitude maximum displacement Period and Frequency Period (T) Time for one complete
More informationExam 4 Review Questions PHY 2425  Exam 4
Exam 4 Review Questions PHY 2425  Exam 4 Section: 12 2 Topic: The Center of Gravity Type: Conceptual 8. After a shell explodes at the top of its trajectory, the center of gravity of the fragments has
More informationPHYS 211 FINAL FALL 2004 Form A
1. Two boys with masses of 40 kg and 60 kg are holding onto either end of a 10 m long massless pole which is initially at rest and floating in still water. They pull themselves along the pole toward each
More information4.4 WAVE CHARACTERISTICS 4.5 WAVE PROPERTIES HW/Study Packet
4.4 WAVE CHARACTERISTICS 4.5 WAVE PROPERTIES HW/Study Packet Required: READ Hamper pp 115134 SL/HL Supplemental: Cutnell and Johnson, pp 473477, 507513 Tsokos, pp 216242 REMEMBER TO. Work through all
More informationSolution Derivations for Capa #13
Solution Derivations for Capa #13 1 Identify the following waves as TTransverse, or LLongitudinal. If the first is T and the rets L, enter TLLL. QUESTION: A The WAVE made by fans at sports events. B
More informationSample Questions for the AP Physics 1 Exam
Sample Questions for the AP Physics 1 Exam Sample Questions for the AP Physics 1 Exam Multiplechoice Questions Note: To simplify calculations, you may use g 5 10 m/s 2 in all problems. Directions: Each
More informationWeight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N)
Gravitational Field A gravitational field as a region in which an object experiences a force due to gravitational attraction Gravitational Field Strength The gravitational field strength at a point in
More informationAcoustics. Lecture 2: EE E6820: Speech & Audio Processing & Recognition. Spherical waves & room acoustics. Oscillations & musical acoustics
EE E6820: Speech & Audio Processing & Recognition Lecture 2: Acoustics 1 The wave equation 2 Acoustic tubes: reflections & resonance 3 Oscillations & musical acoustics 4 Spherical waves & room acoustics
More informationSOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS  VELOCITY AND ACCELERATION DIAGRAMS
SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS  VELOCITY AND ACCELERATION DIAGRAMS This work covers elements of the syllabus for the Engineering Council exams C105 Mechanical and Structural Engineering
More informationSound and stringed instruments
Sound and stringed instruments Lecture 14: Sound and strings Reminders/Updates: HW 6 due Monday, 10pm. Exam 2, a week today! 1 Sound so far: Sound is a pressure or density fluctuation carried (usually)
More informationPhysics Notes Class 11 CHAPTER 3 MOTION IN A STRAIGHT LINE
1 P a g e Motion Physics Notes Class 11 CHAPTER 3 MOTION IN A STRAIGHT LINE If an object changes its position with respect to its surroundings with time, then it is called in motion. Rest If an object
More informationChapter 4. Forces and Newton s Laws of Motion. continued
Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting
More informationALEVEL PHYSICS A. PHYA2 mechanics, materials and waves Mark scheme. 2450 June 2014. Version: 1.0 Final
ALEVEL PHYSICS A PHYA2 mechanics, materials and waves Mark scheme 2450 June 2014 Version: 1.0 Final Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions,
More informationPHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true?
1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always
More informationFRICTION, WORK, AND THE INCLINED PLANE
FRICTION, WORK, AND THE INCLINED PLANE Objective: To measure the coefficient of static and inetic friction between a bloc and an inclined plane and to examine the relationship between the plane s angle
More informationMECHANICS OF SOLIDS  BEAMS TUTORIAL TUTORIAL 4  COMPLEMENTARY SHEAR STRESS
MECHANICS OF SOLIDS  BEAMS TUTORIAL TUTORIAL 4  COMPLEMENTARY SHEAR STRESS This the fourth and final tutorial on bending of beams. You should judge our progress b completing the self assessment exercises.
More informationMechanics 1: Conservation of Energy and Momentum
Mechanics : Conservation of Energy and Momentum If a certain quantity associated with a system does not change in time. We say that it is conserved, and the system possesses a conservation law. Conservation
More informationEDEXCEL NATIONAL CERTIFICATE/DIPLOMA UNIT 5  ELECTRICAL AND ELECTRONIC PRINCIPLES NQF LEVEL 3 OUTCOME 4  ALTERNATING CURRENT
EDEXCEL NATIONAL CERTIFICATE/DIPLOMA UNIT 5  ELECTRICAL AND ELECTRONIC PRINCIPLES NQF LEVEL 3 OUTCOME 4  ALTERNATING CURRENT 4 Understand singlephase alternating current (ac) theory Single phase AC
More informationOscillations: Mass on a Spring and Pendulums
Chapter 3 Oscillations: Mass on a Spring and Pendulums 3.1 Purpose 3.2 Introduction Galileo is said to have been sitting in church watching the large chandelier swinging to and fro when he decided that
More informationLecture 16. Newton s Second Law for Rotation. Moment of Inertia. Angular momentum. Cutnell+Johnson: 9.4, 9.6
Lecture 16 Newton s Second Law for Rotation Moment of Inertia Angular momentum Cutnell+Johnson: 9.4, 9.6 Newton s Second Law for Rotation Newton s second law says how a net force causes an acceleration.
More informationLecture 2: Acoustics
EE E6820: Speech & Audio Processing & Recognition Lecture 2: Acoustics 1 The wave equation Dan Ellis & Mike Mandel Columbia University Dept. of Electrical Engineering http://www.ee.columbia.edu/ dpwe/e6820
More informationInfrared Spectroscopy: Theory
u Chapter 15 Infrared Spectroscopy: Theory An important tool of the organic chemist is Infrared Spectroscopy, or IR. IR spectra are acquired on a special instrument, called an IR spectrometer. IR is used
More informationwhile the force of kinetic friction is fk = µ
19. REASONING AND SOLUION We know that µ s =2.0µ k for a crate in contact with a MAX cement floor. he maximum force of static friction is fs = µ sfn while the force of kinetic friction is fk = µ kfn. As
More informationExperiment 9. The Pendulum
Experiment 9 The Pendulum 9.1 Objectives Investigate the functional dependence of the period (τ) 1 of a pendulum on its length (L), the mass of its bob (m), and the starting angle (θ 0 ). Use a pendulum
More informationTEACHER S CLUB EXAMS GRADE 11. PHYSICAL SCIENCES: PHYSICS Paper 1
TEACHER S CLUB EXAMS GRADE 11 PHYSICAL SCIENCES: PHYSICS Paper 1 MARKS: 150 TIME: 3 hours INSTRUCTIONS AND INFORMATION 1. This question paper consists of 12 pages, two data sheets and a sheet of graph
More informationINTERFERENCE OF SOUND WAVES
1/2016 Sound 1/8 INTERFERENCE OF SOUND WAVES PURPOSE: To measure the wavelength, frequency, and propagation speed of ultrasonic sound waves and to observe interference phenomena with ultrasonic sound waves.
More informationState Newton's second law of motion for a particle, defining carefully each term used.
5 Question 1. [Marks 28] An unmarked police car P is, travelling at the legal speed limit, v P, on a straight section of highway. At time t = 0, the police car is overtaken by a car C, which is speeding
More informationChapter 18 Static Equilibrium
Chapter 8 Static Equilibrium 8. Introduction Static Equilibrium... 8. Lever Law... Example 8. Lever Law... 4 8.3 Generalized Lever Law... 5 8.4 Worked Examples... 7 Example 8. Suspended Rod... 7 Example
More informationCandidate Number. General Certificate of Education Advanced Level Examination June 2014
entre Number andidate Number Surname Other Names andidate Signature General ertificate of Education dvanced Level Examination June 214 Physics PHY4/1 Unit 4 Fields and Further Mechanics Section Wednesday
More informationResonance in a Closed End Pipe
Experiment 12 Resonance in a Closed End Pipe 12.1 Objectives Determine the relationship between frequency and wavelength for sound waves. Verify the relationship between the frequency of the sound, the
More information3 Work, Power and Energy
3 Work, Power and Energy At the end of this section you should be able to: a. describe potential energy as energy due to position and derive potential energy as mgh b. describe kinetic energy as energy
More informationWavesWave Characteristics
1. What is the wavelength of a 256hertz sound wave in air at STP? 1. 1.17 10 6 m 2. 1.29 m 3. 0.773 m 4. 8.53 107 m 2. The graph below represents the relationship between wavelength and frequency of
More informationState Newton's second law of motion for a particle, defining carefully each term used.
5 Question 1. [Marks 20] An unmarked police car P is, travelling at the legal speed limit, v P, on a straight section of highway. At time t = 0, the police car is overtaken by a car C, which is speeding
More information! n. Problems and Solutions Section 1.5 (1.66 through 1.74)
Problems and Solutions Section.5 (.66 through.74).66 A helicopter landing gear consists of a metal framework rather than the coil spring based suspension system used in a fixedwing aircraft. The vibration
More informationv v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )
Week 3 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution
More informationLecture 17. Last time we saw that the rotational analog of Newton s 2nd Law is
Lecture 17 Rotational Dynamics Rotational Kinetic Energy Stress and Strain and Springs Cutnell+Johnson: 9.49.6, 10.110.2 Rotational Dynamics (some more) Last time we saw that the rotational analog of
More informationSolving Quadratic Equations by Graphing. Consider an equation of the form. y ax 2 bx c a 0. In an equation of the form
SECTION 11.3 Solving Quadratic Equations b Graphing 11.3 OBJECTIVES 1. Find an ais of smmetr 2. Find a verte 3. Graph a parabola 4. Solve quadratic equations b graphing 5. Solve an application involving
More informationPhysics 201 Homework 8
Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 Nm is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kgm 2. What is the
More informationExamples of Uniform EM Plane Waves
Examples of Uniform EM Plane Waves Outline Reminder of Wave Equation Reminder of Relation Between E & H Energy Transported by EM Waves (Poynting Vector) Examples of Energy Transport by EM Waves 1 Coupling
More informationELASTIC FORCES and HOOKE S LAW
PHYS101 LAB03 ELASTIC FORCES and HOOKE S LAW 1. Objective The objective of this lab is to show that the response of a spring when an external agent changes its equilibrium length by x can be described
More informationWaves  Transverse and Longitudinal Waves
Waves  Transverse and Longitudinal Waves wave may be defined as a periodic disturbance in a medium that carries energy from one point to another. ll waves require a source and a medium of propagation.
More informationv = λ f this is the Golden Rule for waves transverse & longitudinal waves Harmonic waves The golden rule for waves Example: wave on a string Review
L 23 Vibrations and Waves [3] resonance clocks pendulum springs harmonic motion mechanical waves sound waves golden rule for waves musical instruments The Doppler effect Doppler radar radar guns Review
More informationUnderstanding Poles and Zeros
MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.14 Analysis and Design of Feedback Control Systems Understanding Poles and Zeros 1 System Poles and Zeros The transfer function
More informationSecond Order Systems
Second Order Systems Second Order Equations Standard Form G () s = τ s K + ζτs + 1 K = Gain τ = Natural Period of Oscillation ζ = Damping Factor (zeta) Note: this has to be 1.0!!! Corresponding Differential
More information