Chapter 13 Vibrations and Waves Hooke s Law Reviewed F =!k When is positive, F is negative ; When at equilibrium (=0, F = 0 ; When is negative, F is positive ; 1 2 Sinusoidal Oscillation Graphing vs. t Pen traces a sine wave A A : amplitude (length, m : period (time, s 3 4 = Acos(!t " # = Acos(2$ ft " # = Acos 2$t ' " # ( * Some Vocabulary f = 1! = 2" f = 2" Phases Phase is related to starting time $ = Acos 2!t " # ' ( $ = Acos 2!(t " t 0 ' ( if # = 2!t 0 f = Frequency! = Angular Frequency = Period A = Amplitude " = phase 90-degrees changes cosine to sine $ cos!t " # ' 2 ( = sin!t 5 6
Velocity and Acceleration vs. time Velocity is 90 out of phase with : When is at ma, v is at min... Acceleration is 180 out of phase with a = F/m = - (k/m v a v and a vs. t = Acos!t v = "v ma sin!t a = "a ma cos!t Find v ma with E conservation 1 2 ka2 = 1 2 mv 2 ma v ma = A Find a ma using F=ma!k = ma k m!kacos"t =!ma ma cos"t a ma = A k m 7 8 What is!? Requires calculus. Since d Acos!t = "! Asin!t dt v ma =! A = A k m Formula Summary f = 1! = 2" f = 2"! = k m = Acos(!t " # v = "! Asin(!t " # a = "! 2 A(cos!t " #! = k m 9 10 Eample13.1 An block-spring system oscillates with an amplitude of 3.5 cm. If the spring constant is 250 N/m and the block has a mass of 0.50 kg, determine (a the mechanical energy of the system Eample 13.2 A 36-kg block is attached to a spring of constant k=600 N/m. he block is pulled 3.5 cm away from its equilibrium positions and released from rest at t=0. At t=0.75 seconds, (b the maimum speed of the block a what is the position of the block? a -3.489 cm (c the maimum acceleration. b what is the velocity of the block? b -1.138 cm/s a 0.153 J b 0.783 m/s c 17.5 m/s 2 11 12
Eample 13.3 A 36-kg block is attached to a spring of constant k=600 N/m. he block is pulled 3.5 cm away from its equilibrium position and is pushed so that is has an initial velocity of 5.0 cm/s at t=0. a What is the position of the block at t=0.75 seconds? a -3.39 cm Eample 13.4a epression, (t = 4 + 2cos[!(t " 3] Where will be in cm if t is in seconds he amplitude of the motion is: a 1 cm b 2 cm c 3 cm d 4 cm e -4 cm 13 14 Eample 13.4b epression, (t = 4 + 2cos[!(t " 3] Here, will be in cm if t is in seconds he period of the motion is: a 1/3 s b 1/2 s c 1 s d 2 s e 2/# s Eample 13.4c epression, (t = 4 + 2cos[!(t " 3] Here, will be in cm if t is in seconds he frequency of the motion is: a 1/3 Hz b 1/2 Hz c 1 Hz d 2 Hz e # Hz 15 16 Eample 13.4d epression, (t = 4 + 2cos[!(t " 3] Here, will be in cm if t is in seconds he angular frequency of the motion is: a 1/3 rad/s b 1/2 rad/s c 1 rad/s d 2 rad/s e # rad/s Eample 13.4e epression, (t = 4 + 2cos[!(t " 3] Here, will be in cm if t is in seconds he object will pass through the equilibrium position at the times, t = seconds a, -2, -1, 0, 1, 2 b, -1.5, -0.5, 0.5, 1.5, 2.5, c, -1.5, -1, -0.5, 0, 0.5, 1.0, 1.5, d, -4, -2, 0, 2, 4, e, -2.5, -0.5, 1.5, 3.5, 17 18
Simple Pendulum F =!mgsin" sin" = # 2 + L 2 L F #! mg L Looks like Hooke s law (k $ mg/l Simple Pendulum F =!mgsin" sin" = # 2 + L 2 L F #! mg L! = g L " = " ma cos(!t # $ 19 20 Simple pendulum Pendulum Demo! = g L Frequency independent of mass and amplitude! (for small amplitudes 21 22 Eample 13.5 A man enters a tall tower, needing to know its height h. He notes that a long pendulum etends from the roof almost to the ground and that its period is 15.5 s. In real systems, friction slows motion Damped Oscillations (a How tall is the tower? a 59.7 m (b If this pendulum is taken to the Moon, where the free-fall acceleration is 1.67 m/s 2, what is the period of the pendulum there? b 37.6 s 23 24
RAVELING WAVES Longitudinal (Compression Waves Sound Surface of a liquid Vibration of strings Electromagnetic Radio waves Microwaves Infrared Visible Ultraviolet X-rays Gamma-rays Gravity Sound waves are longitudinal waves 25 Compression and ransverse Waves Demo 26 ransverse Waves Elements move perpendicular to wave motion Elements move parallel to wave motion 27 Snapshot of a ransverse Wave 28 Snapshot of Longitudinal Wave ( y = A cos ' 2! # $ * " wavelength ( y = A cos ' 2! # $ * " y could refer to pressure or density 29 30
Moving Wave y = Acos 2! " vt ( " $ ' # * moves to right with velocity v Replace with -vt if wave moves to the right. Replace with +vt if wave should move to left. Moving Wave: Formula Summary + # y = Acos 2! $ "! ft. - ' * 0, / Fiing =0, y = Acos!2" v ' # t! $ ( * f = v!, v = f! v = f! 31 32 Eample 13.6a A wave traveling in the positive direction has a frequency of f = 25.0 Hz as shown in the figure. he wavelength is: a 5 cm b 9 cm c 10 cm d 18 cm e 20 cm Eample 13.6b A wave traveling in the positive direction has a frequency of f = 25.0 Hz as shown in the figure. he amplitude is: a 5 cm b 9 cm c 10 cm d 18 cm e 20 cm 33 34 Eample 13.6c A wave traveling in the positive direction has a frequency of f = 25.0 Hz as shown in the figure. he speed of the wave is: a 25 cm/s b 50 cm/s c 100 cm/s d 250 cm/s e 500 cm/s Eample 13.7a Consider the following epression for a pressure wave, where it is assumed that is in cm,t is in seconds and P will be given in N/m 2. What is the amplitude? a 1.5 N/m 2 b 3 N/m 2 c 30 N/m 2 d 60 N/m 2 e 120 N/m 2 P = 60! cos 2 " 3t 35 36
Eample 13.7b Consider the following epression for a pressure wave, P = 60! cos 2 " 3t where it is assumed that is in cm,t is in seconds and P will be given in N/m 2. What is the wavelength? a 0.5 cm b 1 cm c 1.5 cm d # cm e 2# cm Eample 13.7c Consider the following epression for a pressure wave, where it is assumed that is in cm,t is in seconds and P will be given in N/m 2. What is the frequency? a 1.5 Hz b 3 Hz c 3/# Hz d 3/(2# Hz e 3#/2 Hz P = 60! cos 2 " 3t 37 38 Eample 13.7d Consider the following epression for a pressure wave, P = 60! cos 2 " 3t where it is assumed that is in cm,t is in seconds and P will be given in N/m 2. What is the speed of the wave? a 1.5 cm/s b 6 cm/s c 2/3 cm/s d 3#/2 cm/s e 2/# cm/s 39 Eample 13.8 Which of these waves move in the positive direction? 1 y =!21.3" cos(3.4 + 2.5t 2 y =!21.3" cos(3.4! 2.5t 3y =!21.3" cos(!3.4 + 2.5t 4y =!21.3" cos(!3.4! 2.5t 5 y = 21.3" cos(3.4 + 2.5t 6 y = 21.3" cos(3.4! 2.5t 7y = 21.3" cos(!3.4 + 2.5t 8y = 21.3" cos(!3.4! 2.5t a 5 and 6 b 1 and 4 c 5,6,7 and 8 d 1,4,5 and 8 e 2,3,6 and 7 40 Speed of a Wave in a Vibrating String Eample 13.9 v = µ where µ = m L A string is tied tightly between points A and B as a communication device. If one wants to double the wave speed, one could: For different kinds of waves: (e.g. sound Always a square root Numerator related to restoring force Denominator is some sort of mass density a Double the tension b Quadruple the tension c Use a string with half the mass d Use a string with double the mass e Use a string with quadruple the mass 41 42
Superposition Principle raveling waves can pass through each other without being altered. Reflection Fied End Reflected wave is inverted y(,t = y 1 (,t + y 2 (,t 43 44 Reflection Free End Reflected pulse not inverted 45