Engineering Physics (1) BAS021 Chapter 3 Oscillation PDF

Engineering Physics(1) BAS021 Chapter 3 Oscillation Dr.H.Elhendawi 1 1 What Really Happened at the tacoma narrows bridge? Why do some buildings fall in earthquakes?...

Engineering Physics(1) BAS021 Chapter 3 Oscillation Dr.H.Elhendawi 1 1 What Really Happened at the tacoma narrows bridge? Why do some buildings fall in earthquakes? Dr.H.Elhendawi 2 2  Objectives Demonstrate your understanding of Oscillation definition , its 1 classification and types, motion function and energy function. Write and apply formulas for calculating Motion Function 2 and Energy Conservation Law. 3 Solve problems involving each of the parameters in the above objectives. Dr.H.Elhendawi 3 3 I-Concepts Dr.H.Elhendawi 4 4  1- Oscillation Oscillation is a periodic change in a physical quantity by increasing or decreasing some central value. ‫ التذبذب هو تغير دورى فى قيمة فيزيائية بالزيادة أو النقصاا واوق قيماة مةيناة تسامى القيماة‬.‫المركزية‬ Dr.H.Elhendawi 5 5  2-Classification of oscillation Mechanical oscillation: ‫تذبذب ميكانيكى‬ An object moves around equilibrium position..‫ حركة جسم حول موضع اتزان معين‬ As simple pendulum, spring and sound..‫ حركة البندول و السلك الزنبركى‬ Dr.H.Elhendawi 6 6 Non mechanical oscillation: Is a periodic change in a physical quantity..‫ تغير دوري في قيمة كمية فيزيائية‬ As Electromagnetic waves, voltage and alternating current. Dr.H.Elhendawi 7 7  ‫‪3- Types of Oscillation‬‬ ‫‪‬‬ ‫‪Simple Harmonic Oscillation‬‬ ‫إهتزازة توافقية بسيطة أو الوركة التوافقية البسيطة ويت ال يودت فقد فى الطاقة‪.‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪Damped Harmonic Oscillation‬‬ ‫إهتزازه مخمدة ويث يودث فقد فى الطاقة‬ ‫‪‬‬ ‫‪‬‬ ‫‪Forced Harmonic Oscillation.‬‬ ‫إهتزازة مجبرة ويث يضاف لها طاقة لتصبح إهتزازة توافقية بسيطة ‪.‬‬ ‫‪‬‬ ‫‪Dr.H.Elhendawi‬‬ ‫‪8‬‬ ‫‪8‬‬  1- Simple Harmonic Motion Spring Oscillations An object of mass m oscillating at the end of a spring. Dr.H.Elhendawi 99  Hook’s Law The restoring force F(‫ )قوة اإلسترجاع‬is directly proportional to the displacement x )‫ (اإلزاحة‬the spring has been stretched or compressed from the equilibrium position. 𝑭=−𝒌𝒙 Where F restoring force (N) »‫قوة اإلسترجاع «القوة التى تعيد الجسم لوضع اإلتزان دائما‬ k spring constant (N/m) x displacement (m) Minus sign indicates that the restoring force is in the opposite direction to the displacement. 𝑭𝒆𝒙𝒕 = −𝑭 = 𝒌 𝒙 Dr.H.Elhendawi 1 10 0  II- Motion Function 1- Displacement Equation 𝑭=−𝒌𝒙 but 𝒎𝒂 = − 𝒌 𝒙 where m mass , a acceleration 𝒌 𝒂=− 𝒙 𝒎 𝒅𝟐 𝒙 𝒌 𝟐 + 𝒙=𝟎 𝒅𝒕 𝒎 𝒅𝟐 𝒙 𝒌 + 𝝎𝟐 𝒙 =𝟎 (𝝎𝟐 = ) 𝒅𝒕𝟐 𝒎 2nd order difference equation it’s solution is 𝒙 𝒕 = 𝑨 𝒔𝒊𝒏 (𝝎𝒕 + ∅) 1 Dr.H.Elhendawi 11 1 1- Displacement Equation 𝒙 𝒕 = 𝑨 𝒔𝒊𝒏 𝝎𝒕 + ∅ Where 𝒙 𝒕 Displacement at any time t (sec) (t) ‫اإلزاحة عند أى لحظة زمنية‬ A Amplitude ( max displacement) (m) )‫السعة (أقصى إزاحة يصل إليه االجسم‬ 𝝎 Angular frequency ‫التردد الزاوى‬ 2π 𝒌 𝟏 𝝎 = 2πf = (rad/sec) Or (𝛚 = ,𝑻 = ), T 𝒎 𝒇 ∅ Phase constant (rad) )‫ثابت الطور و منها تحدد اإلزاحة اإلبتدائية (االزاحة عند زمن صفر‬ 𝒙𝒎𝒂𝒙 = ±𝑨 Dr.H.Elhendawi 1 12 2 2- Velocity Equation 𝒙 𝒕 = 𝑨𝒔𝒊𝒏 𝝎𝒕 + ∅or 𝒙 𝒕 = 𝑨𝒄𝒐𝒔 𝝎𝒕 + ∅ 𝒅𝒙 𝒅𝒙 𝝑 𝒕 = = 𝑨 𝝎 𝒄𝒐𝒔 𝝎𝒕 + ∅ or 𝝑 𝒕 = = −𝑨 𝝎 𝒔𝒊𝒏 𝝎𝒕 + ∅ 𝒅𝒕 𝒅𝒕 𝝑𝒎𝒂𝒙 = ±𝑨𝝎 3- Acceleration Equation 𝒅𝟐 𝒙 𝒂 𝒕 = 𝟐 = −𝑨 𝝎𝟐 𝒔𝒊𝒏 𝝎𝒕 + ∅ = −𝝎𝟐 𝒙 𝒕 𝒅𝒕 Or 𝒂 𝒕 = −𝑨 𝝎𝟐 𝒄𝒐𝒔 𝝎𝒕 + ∅ = −𝝎𝟐 𝒙 𝒕 𝒂𝒎𝒂𝒙 = ±𝑨𝝎𝟐 Dr.H.Elhendawi 1 13 3  SHM Function displacement - velocity - acceleration Dr.H.Elhendawi 1 14 4  ‫‪III- Conservation Energy of SHM‬‬ ‫‪ The total mechanical energy is constant.‬‬ ‫‪ ‬فى الوركة التوافقية البسيطة و تبةا لقانو بقاء الطاقة فإ مجموع طاقتى الوضع و الوركة يظق ثابت عند اى لوظة‬ ‫زمنية‪.‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪‬‬ ‫)‪The Potential Energy(U‬‬ ‫طافة الوضع )∅ ‪𝑈 = 𝑘𝑥 2 = 𝑘𝐴2 𝑠𝑖𝑛2 (𝜔𝑡 +‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪‬‬ ‫= 𝐸 ‪The Kinetic Energy(K.E) 𝐾.‬‬ ‫‪𝑚𝑣 2‬‬ ‫=‬ ‫‪𝑚𝜔2 𝐴2 𝑐𝑜𝑠 2‬‬ ‫∅ ‪𝜔𝑡 + ∅ = 𝑘𝐴2 𝑐𝑜𝑠 2 𝜔𝑡 +‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫) ∅ ‪𝐸 = 𝑈 + 𝐾. 𝐸 = 2 𝑘𝐴2 (𝑠𝑖𝑛2 𝜔𝑡 + ∅ + 𝑐𝑜𝑠 2 𝜔𝑡 +‬‬ ‫‪‬‬ ‫)‪The Total Energy (E‬‬ ‫𝟐 𝟏‬ ‫𝑨𝒌 = 𝑬‬ ‫𝟐‬ ‫الوظ أ الطاقة الكلية مقدار ثابت و يتناسب مع مربع السةة‬ ‫‪‬‬ ‫‪1‬‬ ‫‪15‬‬ ‫‪Dr.H.Elhendawi‬‬ ‫‪5‬‬  Conservation Energy of SHM 𝟏 Potential Energy 𝑼 = 𝒌𝒙𝟐 𝟐 at x=0 U=0 1 at x= ±A 𝑈 = 𝑘𝐴2 (Maximum) 2 𝟏 Kinetic Energy 𝑲. 𝑬 = 𝒎𝒗𝟐 𝟐 at x= 0 𝝑 = 𝝑𝒎𝒂𝒙 = ±𝑨𝝎 1 1 𝐾. 𝐸 = 𝑚𝑣 2 = 𝐴2 𝜔2 (Maximum) 2 2 at x= ±A 𝝑=𝟎 𝐾. 𝐸 = 0 Total Energy 𝟏 𝟐 1 2 1 𝑬 = 𝒌𝑨 = 𝑘𝑥 + 𝑚𝑣 2 𝟐 2 2 𝑘 𝑣 2 = (𝑨𝟐 − 𝑥 2 ) 𝑚 so 𝜗 = ±𝜔 𝑨𝟐 − 𝑥 2 1 16 Dr.H.Elhendawi 6  ‫خطوات حل المسائل‬ ‫‪.1‬كتابة معادلة اإلزاحة و مقارنتها بالمعادلة المعطاة بالمسألة‬ ‫‪.2‬من المقارنه نوجد قيمة ‪ A‬و𝝎 و ∅‬ ‫‪.3‬إذا كانت هناك رسم نستخرج منه الثوابت السابقة‬ ‫𝝅𝟐‬ ‫= 𝒇𝝅𝟐 = 𝝎‬ ‫‪.4‬لحساب التردد و الطول الزمن الدورى‬ ‫𝑻‬ ‫‪ released from rest.5‬معناها أنه بدأ من السكون ‪ t=0‬و عندها ‪x=A‬‬ ‫و لهذا إذا كان الجسم المرتبط بالسلك الزنبركى مشدود‬ ‫𝝅‬ ‫‪1) Stretched Position x = A‬‬ ‫𝒐𝟎𝟗 = = ∅ 𝐨𝐬 𝟏 = ∅𝒏𝒊𝒔 ‪𝒙 = 𝑨 𝒔𝒊𝒏∅ so‬‬ ‫𝟐‬ ‫𝟎 = ∅ 𝐨𝐬 𝟏 = ∅𝒔𝒐𝒄 ‪𝒙 = 𝑨 𝒄𝒐𝒔∅ so‬‬ ‫𝒐‬ ‫لهذا إذا كان الجسم المرتبط بالسلك الزنبركى مضغوط‬ ‫𝝅𝟑‬ ‫𝒐𝟎𝟕𝟐 = 𝟐 = ∅ 𝒐𝒔 𝟏‪2) Compression Position x = -A 𝒙 = 𝑨 𝒔𝒊𝒏∅ = −𝑨 so 𝒔𝒊𝒏∅ = −‬‬ ‫𝑨‪𝒙 = 𝑨 𝒄𝒐𝒔∅ = −‬‬ ‫𝒐𝟎𝟖𝟏 = 𝝅 = ∅ 𝐨𝐬 𝟏‪so 𝒄𝒐𝒔∅ = −‬‬ ‫‪1‬‬ ‫‪17‬‬ ‫‪Dr.H.Elhendawi‬‬ ‫‪7‬‬ Ex(1) The position of a particle moving along the x axis is given by 𝒙(𝒕) = 𝟎. 𝟖𝒔𝒊𝒏 𝟏𝟐𝒕 + 𝟎. 𝟑 where x is in meter and t is in sec. (a) what are the amplitude and the period of the motion? (b) Determine the position, velocity and acceleration when t=0.6s Solution 𝒙 𝒕 = 𝑨𝒔𝒊𝒏 𝝎𝒕 + ∅ 𝒙 = 𝟎. 𝟖𝒎𝒔𝒊𝒏 𝟏𝟐𝒕 + 𝟎. 𝟑 (a) The amplitude (A) =0.8m 𝟐𝝅 𝟐𝝅 𝟐𝒙𝟑.𝟏𝟒 The period of (T) 𝝎 = 𝟏𝟐 = so 𝑇= = = 𝟎. 𝟓𝟐𝟑 𝒔𝒆𝒄 𝑻 𝝎 𝟏𝟐 (b) the position at t=0.6s 𝒙(𝒕) = 𝟎. 𝟖𝒔𝒊𝒏 𝟏𝟐(𝟎. 𝟔𝒔) + 𝟎. 𝟑 = 𝟎. 𝟎𝟕𝟓𝒎 𝑑𝑥 the velocity 𝜗 𝑡 = = 0.8 12 𝑐𝑜𝑠 12𝑡 + 0.3 𝑎𝑡 𝑡 = 0.6𝑠 𝜗 = 0.333𝑚/𝑠 𝑑𝑡 𝑑2 𝑥 the acceleration 𝑎 𝑡 = = −(0.8) (122 )𝑠𝑖𝑛 12𝑡 + 0.3 𝑑𝑡 2 𝒂𝒕 𝒕 = 𝟎. 𝟔𝒔 𝒂 = −𝟏𝟎. 𝟖 𝒎/𝒔𝟐 181 Dr.H.Elhendawi 8 Ex(2) A 2 kg block is attached to a spring for which k=200N/m. It is held at an extension of 5 cm and then released at t=0.Find (a) The displacement as a function of time. (b) The velocity and the acceleration when x=A/2 Solution m=2kg , k=200N/m , A=0.05m (a) The displacement as a function of time, we need A, 𝛚, ∅ 𝑘 200𝑁/𝑚 𝛚= = = 10 𝑟𝑎𝑑/𝑠 𝑚 2𝑘𝑔 𝝅 To find ∅ at t=0 x = A 𝑨 = 𝑨𝒔𝒊𝒏∅ , ∅= 𝟐 𝝅 𝒙(𝒕) = 𝟎. 𝟎𝟓𝒔𝒊𝒏 𝟏𝟎𝒕 + 𝒎 𝟐 Dr.H.Elhendawi 1 19 9 (b) to find the velocity at x=A/2 we have to find the time t 𝝅 𝝅 𝝅 𝟓𝝅 𝒔𝒊𝒏 𝟏𝟎𝒕 + = 𝟎. 𝟓 so 𝟏𝟎𝒕 + = 𝒐𝒓 𝟐 𝟐 𝟔 𝟔 dx π ϑ t = = 0.05 10 cos 10t + dt 2 π 5π = 0.5cos or 0.5cos( ) 6 6 m = 0.43 or -0.43m/s s 𝑘 10𝑟𝑎𝑑 2 0.05 acceleration when x=A/2 𝑎 = − 𝑥 = −𝜔2 𝑥 =− 𝑚 = −2.5 𝑚/𝑠 2 𝑚 𝑠 2 Dr.H.Elhendawi 2 20 0  𝝅 Ex(3) In Ex(2) the displacement equation is 𝒙 = 𝟎. 𝟎𝟓𝒔𝒊𝒏 𝟏𝟎𝒕 + 𝟐 Find (a) K.E and U and E at t= π/15 s (b) the speed at x=A/2. Solution m=2kg , k=200N/m , A=0.05m 𝟏 1 200𝑁 (a) The Total Energy E 𝑬 = 𝒌𝑨𝟐 = 0.05 2 = 0.25𝐽 𝟐 2 𝑚 1 1 10𝜋 𝝅 3 The Kinetic Energy K.E 𝐾. 𝐸 = 𝑚𝑣 2 = (2 𝑘𝑔) [cos 𝑠 + ]2 = 𝐽 2 2 15 𝟐 16 2 1 1 10𝜋 𝜋 1 Potential Energy U 𝑈 = 𝑘𝑥 2 = (200𝑚 𝑁 ) 𝑠𝑖 𝑛 𝑠 + = 𝐽 2 2 15 2 16 𝟏 1 𝐴 1 (b) the speed at x=A/2 𝑘𝐴2 = 𝑘( )2 + 𝑚𝑣 2 𝟐 2 2 2 3𝑘𝐴2 So ϑ= = 0.43 𝑚/𝑠 4𝑚 2 21 Dr.H.Elhendawi 1  Ex(4) For the mass-spring system m = 200g, k = 5.0 N/m if the block were released from the same initial position, xi=5.0 cm, but with an initial velocity of vi= −0.1 m/s. Express the position as a function of time in SI units. Solution Let : x)t( = A sin ) ω t+ φ ( v)t ( = ω A cos ) ω t+ φ ( At t = 0 → x = 0.05 m and v = −0.1 m/s 𝒌 5.0 N/m 𝛚= = = 5𝑟𝑎𝑑/𝑠 𝒎 200x10−3 kg At t= 0→ x( 0 ) = 0.05 m A sin φ = 0.05 → 𝑒𝑞(1) At t= 0→ v( 0 ) = −0.1 m/s 5 A cos φ = − 0.1 A cos φ = −0.02 → 𝑒𝑞(2) [𝑒𝑞𝑛 1 ]2 +[𝑒𝑞𝑛 2 ]2 → 𝐴2 = (0.05)2 +(0.02)2 → A = 0.0538 m 2 22 Dr.H.Elhendawi 2 𝑒𝑞𝑛(1) = 𝑡𝑎𝑛∅ = −𝟐. 𝟓 𝑒𝑞𝑛 3 𝑒𝑞𝑛(2) − 𝑣𝑒 𝑣𝑎𝑙𝑢𝑒 𝑠𝑜 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑖𝑛 𝑠𝑒𝑐𝑜𝑛𝑑 or fourth quarter ∅ = 𝑡𝑎𝑛−1 2.5 = 1.19 𝑟𝑎𝑑 ∅ = 𝜋 − 1.19 𝑟𝑎𝑑 = 𝟏. 𝟗𝟓 𝒓𝒂𝒅 𝑥(𝑡) = 0.0538𝑚 sin(5𝑡 + 1.95) Dr.H.Elhendawi 2 23 3  2- Damped oscillation  There is energy loss due to friction or air resistance, so it’s amplitude decreases ‫ يوجد فقد في الطاقة نتيجة لوجود قوى تحاول إخماد هذه الحركة (قوة االحتكاك أو مقاومة المائع) وبالتالي تقل سعة‬.‫االهتزازة بالتدريج مع الوقت‬  The frictional force (𝐹𝑓 ): 𝐹𝑓 = −𝑏𝑣 Where 𝑣 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 m/s 𝑏 damping constant (kg/s) 2 24 Dr.H.Elhendawi 4 Appling newton’s 2𝑛𝑑 law on damped oscillations 𝐹 = 𝑚𝑎 −𝑘𝑥 − 𝑏𝑣 = 𝑚𝑎 𝑑𝑥 𝑑2𝑥 −𝑘𝑥 − 𝑏 =𝑚 2 𝑑𝑡 𝑑𝑡 𝑑 2 𝑥 𝑏 𝑑𝑥 𝑘 𝑘 + + 𝑥=0 𝑏𝑢𝑡 𝜔𝑜 = 𝑢𝑛𝑑𝑎𝑚𝑝𝑒𝑑 𝑆𝐻𝑂 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑑𝑡 2 𝑚 𝑑𝑡 𝑚 𝑚 𝑑2𝑥 𝑑𝑥 𝑘 𝑏 + 2𝜇𝜔𝑜 + 𝑥=0 𝑤ℎ𝑒𝑟𝑒 2𝜇𝜔𝑜 = 𝑑𝑡 2 𝑑𝑡 𝑚 𝑚 𝑏 𝜇= 𝑤ℎ𝑒𝑟𝑒 𝜇 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 𝑟𝑎𝑡𝑖𝑜, 𝑏 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 , 2𝑚𝜔𝑜 There are 3 solutions depending on the value of 𝝁 2 25 Dr.H.Elhendawi 5 There are 3 solutions depending on the value of 𝝁 𝝁 ‫يوجد ثالث أنواع م االهتزارة المجمدة تبةا لقيمة‬ 1- Overdamped (𝝁 > 1 )  The displacement is exponentially decay. ‫تقق اإلزاوة في صورة دالة أسية‬  No oscillations. ‫ال يوجد تذبذب‬  The system returned slowly to its equilibrium position. ‫يةود الجسم لموضع اتزانه ببطء شديد‬ 2- Critically damped 𝝁 = 1  The displacement is exponentially decay. ‫تقق اإلزاوة في صورة دالة أسية‬  No oscillations. ‫ال يوجد تذبذب‬  The system returned as quickly as possible to its equilibrium position..‫يةود الجسم لموضع اتزانه أسرع نسبيا م الوالة السابقة‬ 2 26 Dr.H.Elhendawi 6 3- Underdamping  The system oscillates with it’s amplitude exponentially decay. ‫ يتذبذب الجسم ووق موضع االتزا و تقق سةة االهتزازة في صورة دالة أسية‬  The displacement of under damped oscillations  𝒙 𝒕 = 𝑨𝒔𝒊𝒏 𝝎𝟏 𝒕 + ∅ where A Underdamped amplitude 𝒃𝒕 −𝟐𝒎  𝑨 = 𝑨𝒐 𝒆 where Aₒ initial amplitude 2 27 Dr.H.Elhendawi 7 https://math24.net/oscillations-electrical-circuits.html 2 28 Dr.H.Elhendawi 8  Energy of damped OSC  The energy of underdamped oscillations: 𝒃𝒕 𝟏 𝟏 − 𝑬= 𝒌𝑨𝟐 = 𝒌𝑨𝒐 𝒆 𝟐𝒎 𝟐 𝟐 Where 𝟏 𝒌𝑨𝟐𝒐 is initial energy 𝟐 Dr.H.Elhendawi 2 29 9  3- Forced oscillation  Energy (force) is added continually to replace the energy lost in friction.‫ يضاف طاقة باستمرار لتةويض الفقد في الطاقة نتيجة االوتكاك‬  This driving force has a frequency “ f ”.  The natural frequency of the system “ fₒ ”.  f= fₒ the driving force is most effective and it is in “resonance” with the system and “fₒ” is the “resonance frequency ”. 3 30 Dr.H.Elhendawi 0 https://courses.lumenlearning.com/suny-physics/chapter/16-8-forced-oscillations-and-resonance/ 3 31 Dr.H.Elhendawi 1 Now, can you explain What happened at the Tacoma narrows bridge or what happens to building during earthquake ? https://www.youtube.com/watch?v=mXTSnZgrfxM https://www.youtube.com/watch?v=uWoiMMLIvco https://www.youtube.com/watch?v=Il0U9_WPGPk https://www.youtube.com/watch?v=DtpckvIBrqo 3 32 Dr.H.Elhendawi 2 Recourses Serway, Raymond A., and John W. Jewett. Physics for scientists and engineers. Cengage learning, 2018. https://www.iris.edu/hq/inclass/animation/building_resonance_the_resonant_frequency_of_different_s eismic_waves 3 33 Dr.H.Elhendawi 3  3 34 Dr.H.Elhendawi 4