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18-2-2015 FCI 1 Faculty of Computer and Information Basic Science department 2013/2014 Prof. Nabila.M.Hassan 18-2-2015 FCI 2 •Aims of Course: •The graduates have to know the nature of vibration wave motions with emphasis on their mathematical descriptions and superposition. •The fundamental ideas can be introduced with reference to mechanical systems which are easy to visualize. •The graduates have to know the nature of vibration and wave motions with emphasis on their mathematical description and superposition Developing the graduate's skills and creative thought needed to meet new trends in science. •Supplying graduates with basic attacks and strategies for solving problems. 18-2-2015 FCI 3 1- A particle oscillates with simple harmonic motion, so that its displacement varies according to the expression x = (5 cm)cos(2t + π/6) where x is in centimeters and t is in seconds. At t = 0 find (a) the displacement of the particle, (b) its velocity, and (c) its acceleration. (d) Find the period and amplitude of the motion. Solution: The displacement as a function of time is x(t) = A cos(ωt + φ). Here ω = 2/s, φ = π/6, and A = 5 cm. The displacement at t = 0 is x(0) = (5 cm)cos(π/6) = 4.33 cm. (b) The velocity at t = 0 is v(0) = -ω(5 cm)sin(π/6) = -5 cm/s. (c) The acceleration at t = 0 is a(0) = -ω2(5 cm)cos(π/6) = -17.3 cm/s2. (d) The period of the motion is T = π sec, and the amplitude is 5 cm. 18-2-2015 FCI 4 1- An oscillator consists of a block of mass 0.50 kg connected to a spring. When set into oscillation with amplitude 35 cm, it is observed to repeat its motion every 0.50 s. The maximum speed is : (a) 4.4 m/s ,(b) 44.0 m/s ,( c) 0.44 m/s 2- A particle executes linear harmonic motion about the point x = 0. At t = 0, it has displacement x = 0.37 cm and zero velocity. The frequency of the motion is 0.25 Hz. The max speed of the motion equal: (a) 0.59 cm/s ,(b) 5.9 cm/s ,( c) 0.059 cm/s 3- An oscillating block-spring system has a mechanical energy of 1.0 J, amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The force constant of the spring is, (a) 100 N/m ,(b) 200 N/m ,( c) 20 N/m 4- An oscillating block-spring system has a mechanical energy of 1.0 J, amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The mass of the block is, (a) 1.4 kg ,(b) 14.0 kg ,( c) .140 kg 18-2-2015 FCI 5 1- An oscillator consists of a block of mass 0.50 kg connected to a spring. When set into oscillation with amplitude 35 cm, it is observed to repeat its motion every 0.50 s. The maximum speed is : (a) 4.4 m/s ,(b) 44.0 m/s ,( c) 0.44 m/s 2- A particle executes linear harmonic motion about the point x = 0. At t = 0, it has displacement x = 0.37 cm and zero velocity. The frequency of the motion is 0.25 Hz. The max speed of the motion equal: (a) 0.59 cm/s ,(b) 5.9 cm/s ,( c) 0.059 cm/s 3- An oscillating block-spring system has a mechanical energy of 1.0 J, amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The force constant of the spring is, (a) 100 N/m ,(b) 200 N/m ,( c) 20 N/m 4- An oscillating block-spring system has a mechanical energy of 1.0 J, amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The mass of the block is, (a) 1.4 kg ,(b) 14.0 kg ,( c) .140 kg 18-2-2015 FCI 6 Content: Part II: Waves Chapter 1 Oscillation Motion - Motion of a spring: - Energy of the Simple Harmonic Oscillator: -Comparing SHM with uniform motion - The simple pendulum: - Damped Oscillations: - Forced Oscillation 18-2-2015 FCI 7 •Objectives: Student will be able to: -Calculate the periodic time of S H M. -- Define the damped motion - Define the resonance. -Compare between free, damped and derived oscillations 18-2-2015 FCI 8 The simple pendulum: A simple pendulum is a mechanical system that exhibits motion consists of a small mass, which is suspended from a light wire or string of length L. The displacement is defined by the arc S. The net force on the bob (mass) is tangent to the arc and equals ( F mgsin ) For small angles (sin θ ≈ θ ) , then the force equal F mgsin mg 18-2-2015 FCI 9 The displacement is directly related to θ S = L θ so that θ = S/ L , then F m g L Apply Newton's second law for motion in the tangential direction d 2 g F m 2 m L dt The equation of motion for simple pendulum A simple oscillation of the pendulum, the resulting force is F m gsin m g when θ is small max cos(t ) max is the max angular position and the angular frequency 18-2-2015 FCI d 2 g 2 L dt becomes g L T 2 L g 10 Example (1) a) How much energy is stored in the spring of a toy gun that has a force constant of 50 N /m and is compressed 0.150 m? Sol: The energy can be found directly from equation 1 1 2 PE el KX (50 N / m)( 0.150 m ) 2 2 2 0.563N.m 18-2-2015 FCI 0.563J 11 b)In absence of friction and neglecting the mass of the spring, at what speed will a 2.00 g plastic bullet be ejected from the gun? Sol: Since there is no friction, the potential energy is converted into kinetic energy. KEF PEel 1 1 mV 2 KX 2 .563 J 2 2 2 PEl v m 18-2-2015 FCI 1 2 2(0.563J ) 3 2 10 kg 1 2 23.7m / s 12 Problem: - A 20 g particle moves in simple harmonic motion with a frequency of 3 oscillations per second and an amplitude of 5cm. (a) Through what total distance does the particle move during one cycle of its motion? (b) What is its maximum speed? Where does that occur? (c) Find the maximum acceleration of the particle. Where in the motion does the maximum acceleration occur? 18-2-2015 FCI 13 Damped Oscillations: Where the force is proportional to the speed of the moving object and acts in the direction opposite the motion. The retarding force can be expressed as: R = - bv ( where b is a constant called damping coefficient) and the restoring force of the system is – kx, then we can write Newton's second law as dx d 2x kx b m dt dt 2 F kx bv ma x x x When the retarding force is small compared with the max restoring force that is, b is small the solution is, x(t ) Ae 18-2-2015 FCI b t 2m cos(t ) k b 2 ( ) m 2m 14 represent the position vs time for a damped oscillation with decreasing amplitude with time The fig. shows the position as a function in time of the object oscillation in the presence of a retarding force, the amplitude decreases in time, this system is know as a damped oscillator. The dashed line which defined the envelope of the oscillator curve, represent the exponential factor 18-2-2015 FCI 15 The fig. represent position versus time: •A - under damped oscillator • b- critical damped oscillator • c- Overdamped oscillator. as the value of "b" increase the amplitude of the oscillations decreases more and more rapidly. When b reaches a critical value bc (bc / 2m o ), the system does not oscillate and is said to be critically damped. And whenbc / 2m o 18-2-2015 FCI the system is overdamped. 16 Forced Oscillation: For the forced oscillator is a damped oscillator driven by an external force that varies periodically Where F (t ) Fo sin t where ω is the angular frequency of the driving force and Fo is a constant From the Newton's second law dx d 2x F m a Fo sin t b dt kx m dt 2 Fo / m A ( 18-2-2015 x A cos(t ) 2 ) 2 o 2 FCI b m 2 17 o k m is the natural frequency of the un-damped oscillator (b=0). The last two equations show the driving force and the amplitude of the oscillator which is constant for a given driving force. For small damping the amplitude is large when the frequency of the driving force is near the natural frequency of oscillation, or when ω͌ ≈ ωo the is called the resonance and the natural frequency is called the resonance frequency. 18-2-2015 FCI 18 Amplitude versus the frequency, when the frequency of the driving force equals the natural force of the oscillator, resonance occurs. Note the depends of the curve as the value of the damping coefficient b. 18-2-2015 FCI 19 Note that: The applied force F is in phase with the velocity. The rate at which work is done on the oscillator by F equals the dot product F. v ; this rate is the power delivered to the oscillator. Because the product F. v is a maximum when F and v are in phase, we conclude that at resonance the applied force is in phase with the velocity and the power transferred to the oscillator is a maximum. 18-2-2015 FCI 20 Summary of the chapter 1: 1- The acceleration of the oscillator object is proportional to its position and is in the direction opposite the displacement from equilibrium, the object moves with SHM. The position x varies with time according to, x(t ) A cos(t ) 2- The time for full cycle oscillation is defined as the period, T 2 / . For block spring moves as SHM on the frictionless surface with a period T 2 18-2-2015 2 k m FCI 21 and 3- The frequency is defined as the number of oscillation per second, is the inverse of the period 1 f 1 T 2 k m 4- The velocity and the acceleration of SHM as a function of time are dx v A sin(t ) dt d 2x a 2 2 A cos(t ) dt We not that the max speed is Aω , and the max acceleration is Aω2 . The speed is zero when the oscillator is at position of x=± A , and is a max when the oscillator is at the equilibrium position at the equilibrium position x=0. 18-2-2015 FCI 22 5- The kinetic energy and potential energy for simple harmonic oscillator are given by, K 1 1 mv 2 m 2 A 2 sin 2 (t ) 2 2 U 1 2 1 kx kA 2 cos 2 (t ) 2 2 The total energy of the SHM is constant of the motion and is given by E 1 kA 2 2 6- A simple pendulum of length L moves in SHM for small angular displacement from the vertical, its period is 18-2-2015 FCI T 2 L g 23 7- For the damping force R = - bv, its position for small damping is described by x(t ) Ae b t 2m cos(t ) k b 2 ( ) m 2m 8 - If an oscillator is driving with a force F (t ) Fo sin t it exhibits resonance, in which the amplitude is largest when driving frequency matches the natural frequency of the oscillator. 18-2-2015 FCI 24 What is the effect on the period of a pendulum of doubling its length? L 2L TL 2 L / g T2 L 2 2L / g 18-2-2015 T2 L 2TL 1.414TL FCI 25 Useful website http://cnx.org/content/m15880/latest/ http://www.acs.psu.edu/drussell/Demos/SHO/massforce.html 18-2-2015 FCI 26