Physics 231 Topic 7: Oscillations Alex Brown October 14-19 2015 MSU Physics 231 Fall 2015 1 Key Concepts: Springs and Oscillations Springs Periodic Motion Frequency & Period Simple Harmonic Motion (SHM) Amplitude & Period Angular Frequency SHM Concepts Energy & SHM Uniform Circular Motion Simple Pendulum Damped & Driven Oscillations Covers chapter 7 in Rex & Wolfson MSU Physics 231 Fall 2015 2 Springs xo=0 MSU Physics 231 Fall 2015 3 Hooke’s Law xo=0 Fs = -kx Hooke’s law k = spring constant in units of N/m. x = displacement from the equilibrium point xo=0 MSU Physics 231 Fall 2015 4 How to find the spring constant k When the object hanging from the spring is not moving: ∑F = ma = 0 x Fspring + Fgravity = 0 x=0 -k(d) + mg = 0 Fspring = -kx x=d k = mg/d Fgravity = mg MSU Physics 231 Fall 2015 5 Springs Hooke’s Law: Fs = -kx k: spring constant (N/m) x: distance the spring is stretched from equilibrium W = area under F vs x plot = ½kx2 The energy stored in a spring depends on how far the spring has been stretched: elastic potential energy PEspring = ½ k x2 MSU Physics 231 Fall 2015 6 PINBALL! The ball-launcher spring has a constant k = 120 N/m. A player pulls the handle 0.05 m. The mass of the ball is 0.1 kg. What is the launching speed? PEspring = ½ k x2 PEgravity = mgh KE = ½ mv2 ME = PE + KE = Constant (PEgravity+PEspring+KEball)pull = (PEgravity+PEspring+KEball)launch mghpull + ½kxpull2 + ½mvpull2 = mghlaunch + ½kxlaunch2 + ½mvlaunch2 ½120(0.05)2 = 0.1 (9.81) ( 0.05*sin(10o) ) + ½(0.1)vlaunch2 0.15 = 8.5x10-3 + 0.05v2 v =1.7 m/s MSU Physics 231 Fall 2015 7 Momentum p = mv Impulse p = pf - pi = Ft Conservation of momentum (closed system) p1i + p2i = p1f + p2f Collisions in one dimension given given m1 and m2 or c = m2 /m1 v1i and v2i Inelastic (stick together, KE lost) vf = [v1i + c v2i] / (1+c) Elastic (KE conserved) v1f = [(1-c) v1i + 2c v2i] / (1+c) v2f = [(c-1) v2i + 2v1i] / (1+c) MSU Physics 231 Fall 2015 8 Collisions in one dimension given given m1 = m2 or c = 1 v1i and v2i Inelastic (stick together, KE lost) vf = [v1i + v2i] / 2 Elastic (KE conserved) v1f = v2i v2f = v1i MSU Physics 231 Fall 2015 9 Carts on a spring track m1 v spring m2 ||||||||||| k = 50 N/m v = 5.0 m/s m1 = m2 = 0.25 kg What is the maximum compression of the spring if the carts collide a) elastically and b) perfectly inelastic? A) Conservation of momentum and KE with c = 1 v1f = v v2f = 0 :: v1f = 0 v2f = v = 5 m/s Conservation of energy: ½mv2 = ½kx2 (0.5) (0.25) (5)2 = (0.5)(50)x2 x = 0.35 m B) Conservation of momentum only with c = 1 vf = (v1f + v2f)/2 = (v + 0)/2= 2.5 m/s Conservation of energy: ½mvf2 = ½kx2 (0.5)(0.5)(2.5)2 = (0.5)(50)x2 x=0.25 m MSU Physics 231 Fall 2015 10 Hooke’s Law Fs = -kx Hooke’s law If there is no friction, the mass continues to oscillate back and forth. If a force is proportional to the displacement x, but opposite in direction, the resulting motion of the object is called: simple harmonic oscillation MSU Physics 231 Fall 2015 11 Simple harmonic motion displacement x A c) b) time (s) a) (b) Amplitude (A): maximum distance from equilibrium (unit: m) Period (T): Time to complete one full oscillation (unit: s) Frequency (f): Number of completed oscillations per second. f=1/T (unit: 1/s = 1 Herz [Hz]) (a) Angular frequency = 2πf = 2π/T (c) MSU Physics 231 Fall 2015 12 Concept Quiz displacement x 5m 2 4 6 -5 m 8 10 time (s) a) what is the amplitude of the harmonic oscillation? b) what is the period of the harmonic oscillation? c) what is the frequency of the harmonic oscillation? a) Amplitude: 5 m b) period: time to complete one full oscillation: T = 4s c) frequency: number of oscillations per second f = 1/T = 0.25/s = 0.25 Hz MSU Physics 231 Fall 2015 13 What the equation? displacement x 5m 2 4 6 8 -5 m 10 time (s) x = A cos(t) A=5 T = 2π = 2π/T = 2π/4 = 1.57 rad/s MSU Physics 231 Fall 2015 14 Clicker Quiz! A mass on a spring in SHM has amplitude A and period T. What is the total distance traveled after a time interval of 2T? A) B) C) D) E) 0 A/2 A 4A 8A In one cycle covering period T, the mass moves from +A to -A and back: 2A+2A = 4A In two periods, the mass moves twice this distance: 8A MSU Physics 231 Fall 2015 15 Energy and Velocity Ekin(½mv2) x=+A x=0 x=-A conservation of ME: Epot,spring(½kx2) 0 ½kA2 ½kA2 0 ½mv2max ½mv2max 0 ½mv2max = ½kA2 MSU Physics 231 Fall 2015 Sum ½k(-A)2 so ½kA2 vmax = ±A(k/m) 16 Velocity and acceleration in general Total ME at any displacement x: ½mv2 + ½kx2 Total ME at max. displacement A: ½kA2 Conservation of ME: ½kA2 = ½mv2 + ½kx2 So: v = ±[(A2-x2)k/m] also F = ma = -kx so a = -kx/m position x velocity v Acceleration a +A 0 -kA/m 0 ±A(k/m) 0 -A 0 kA/m MSU Physics 231 Fall 2015 17 An Example The maximum velocity and acceleration of mass connected to a spring are vmax=0.95 m/s and amax=1.56 m/s2, respectively. Find the oscillation amplitude. MSU Physics 231 Fall 2015 18 An Example The maximum velocity and acceleration of mass connected to a spring are vmax=0.95 m/s and amax=1.56 m/s2, respectively. Find the oscillation amplitude. We do not know the mass of the object or the spring constant. So we must find a way to eliminate these variables! so vmax = ±A(k/m) so (vmax )2 = A2 (k/m) so amax = A (k/m) A = vmax2/amax = (0.95)2/(1.56) = 0.58 m MSU Physics 231 Fall 2015 19 Harmonic oscillations vs circular motion x(t) = A cos The projection of the position of the circulating object on the x-axis as a function of time is the same as the position of the oscillating spring. MSU Physics 231 Fall 2015 r=A 20 Harmonic oscillations vs circular motion vx(t) = -v sin v vx The vx projection of the linear velocity of the rotating object is the velocity of the mass on the spring. MSU Physics 231 Fall 2015 r=A 21 x=+A But remember for circular motion with constant speed around the circle = t and v = r = A where is the “angular frequency” x=0 x=-A So x(t) = A cos = A cos(t) vx(t) = -v sin = - A sin(t) r=A time to complete one circle = T = one period so T = 2 MSU Physics 231 Fall 2015 or = 2/T 22 Harmonic oscillations vs circular motion The simple harmonic motion can be described by the projection of circular motion on the horizontal axis. x(t) = A cos(t) v(t) = -A sin(t) • • • • A = 2/T = 2f T f = 1/T amplitude of the oscillation angular frequency period frequency. MSU Physics 231 Fall 2015 23 x = A cos(t) A x time (s) -A A velocity v = -A sin(t) v time (s) -A Remember vmax k A m thus k m The angular frequency does not depend on A MSU Physics 231 Fall 2015 24 Displacement vs Acceleration displacement x A time (s) -A Newton’s second law: F=ma -kx=ma a=-kx/m acceleration(a) kA/m -kA/m MSU Physics 231 Fall 2015 25 x = A cos(t) A x time (s) -A A v = -A sin(t) v time (s) -A A 2 acceleration a = -A 2 cos(t) time (s) a -A 2 MSU Physics 231 Fall 2015 26 Period of a Spring 2 m T 2 k Increase m: increase T MSU Physics 231 Fall 2015 Increase k: decrease T 27 Clicker Quiz! C A mass is oscillating horizontally while attached to a spring with spring constant k. Which of the following is true? a) When the magnitude of the displacement is largest, the magnitude of the acceleration is also largest. b) When the displacement is positive, the acceleration is also positive c) When the displacement is zero, the acceleration is non-zero MSU Physics 231 Fall 2015 28 The pendulum x=+A x=-A x=0 MSU Physics 231 Fall 2015 29 The pendulum = max = A y=h Conservation of Energy! At x=+A: PE = mgh KE = ½mv2 = 0 At x=0: PE = mg(0) = 0 KE = ½mv2 ½mv2 = mgh v2 = 2gh velocity at the bottom MSU Physics 231 Fall 2015 30 The pendulum Tension at max angle = max T = mg cosmax L Tension at bottom = 0 T = mg + mv2/L T s mg sin mg cos MSU Physics 231 Fall 2015 31 The pendulum Restoring force: F=-mg sin The force pushes the mass m back to the central position. L T s mg sin mg cos MSU Physics 231 Fall 2015 32 The pendulum Restoring force: F=-mg sin The force pushes the mass m back to the central position. L sin (radians) if is small T F = - mg s mg sin so: also s = L F=-(mg/L)s mg cos MSU Physics 231 Fall 2015 33 pendulum vs spring The pendulum k mg / L m m spring g L Does not depend on mass pendulum F kx F (mg / L) s k m g L MSU Physics 231 Fall 2015 34 Damped Oscillations Newton’s 1st Law: An object will stay in motion unless acted upon by a force. SHM is the same: oscillations will last forever in the absence of additional forces. Common force of friction often “damps” oscillations and brings them to a stop. In homework A is reduced by some factor f after one period After one oscillation A1 = f A So for n oscillations An = fn A mechanical energy ME = (1/2) k A2 MSU Physics 231 Fall 2015 35 Driven Oscillations With driven oscillation, the amplitude grows! The same can be applied in reverse: An object NOT in motion can be put into oscillation by applying a force This is known as driven oscillations. If the force acts with the same frequency as the as that of the object this is known as resonance. MSU Physics 231 Fall 2015 36 Spring problem A bungee jumper with height h=2 m and mass m = 80 kg leaps from a bridge with height H = 30 m. A bungee cord with spring constant k = 100 N/m attached to his legs. What is the maximum length the cord needs to be if he is to avoid hitting the water below? Define A = extended “amplitude” of the bungee cord Total extension = jumper’s height + nominal length of the cord + extended length of the cord = h+L+A Must stop before h+L+A = H = 30m, or A = H-L-h Use conservation of ME: ½mv2 + ½kx2 + mgH On top of the bridge: ME = mgH Maximum extension of the cord: ME = ½ kA2 = ½ k(H-L-h)2 ME(bridge) = ME(ground) mgH = ½k(H-L-h) 2 (80 kg)(9.81 m/s2)(30m) = ½(100 N/m)(30m-L-2m) 2 23544 = 50 (28m-L) 2 (28m-L) = sqrt(23544/50) = 21.7 m L = 6.3 m MSU Physics 231 Fall 2015 37 Spring problem A h=2m tall, m=80 kg bungee jumper leaps from a H=30m bridge with a bungee cord with spring constant k = 100 N/m attached to his legs. What is the frequency of his subsequent SHM? 𝐓𝐬𝐩𝐫𝐢𝐧𝐠 = 𝟐𝛑 Frequency 𝐦 = 𝟐𝛑 𝐤 𝟖𝟎 𝟏𝟎𝟎 = 5.62 sec f = 1/T = 1/5.62s = 0.18 Hz MSU Physics 231 Fall 2015 38 Spring problem A mass of 0.2 kg is attached to a spring with k=100 N/m. The spring is stretched over 0.1 m and released. a) What is the angular frequency () of the corresponding motion? b) What is the period (T) of the harmonic motion? c) What is the frequency (f)? d) What are the functions for x,v and t of the mass as a function of time? Make a sketch of these. a) =(k/m) = (100/0.2) = 22.4 rad/s b) = 2/T T= 2/ = 0.28 s c) = 2f f=/2 = 3.55 Hz (=1/T) d) x(t) = Acos(t) = 0.1 cos(22.4t) v(t) = -Asin(t) = -2.24 sin(22.4t) a(t) = -2Acos(t) = -50.2 cos(22.4t) MSU Physics 231 Fall 2015 39 0.1 x -0.1 0.28 0.56 0.28 0.56 0.28 0.56 time (s) velocity v 2.24 -2.24 50.2 a -50.2 MSU Physics 231 Fall 2015 40 pendulum problem The machinery in a pendulum clock is kept in motion by the swinging pendulum. Does the clock run faster, at the same speed, or slower if: a) The mass is hung higher b) The mass is replaced by a heavier mass c) The clock is brought to the moon d) The clock is put in an upward accelerating elevator? L g L m Moon Elevator Faster(A) Same(B) Slower(C) MSU Physics 231 Fall 2015 41 Pendulum Quiz A pendulum with length L of 1 meter and a mass of 1 kg, is oscillating harmonically. The period of oscillation is T. If L is increased to 4 m, and the mass replaced by one of 0.5 kg, the period becomes: A) 0.5 T L T 2 B) 1 T g C) 2 T 4L L T 2 4 2T D) 4 T g g E) 8 T Mass does not matter MSU Physics 231 Fall 2015 42 Spring problem A mass of 1 kg is hung from a spring. The spring stretches by 0.5 m. Next, the spring is placed horizontally and fixed on one side to the wall. The same mass is attached and the spring stretched by 0.2 m and then released. What is the acceleration upon release? 1st step: find the spring constant k Fspring =-Fgravity or -kd =-mg k = mg/d = (1)(9.8)/0.5 = 19.6 N/m 2nd step: find the acceleration upon release Newton’s second law: F=ma -kx = ma a = -kx/m a = -(19.6)(0.2)/1 = -3.92 m/s2 MSU Physics 231 Fall 2015 43 Sprng problem A block with mass of 200 g is placed over an opening. A spring is placed under the opening and compressed by a distance d=5 cm. Its spring constant is 250 N/m. The spring is released and launches the block. How high will it go relative to its rest position (h)? (the spring can be assumed massless) x d ME = ½mv2 Initial: v = 0 + ½kd2 + mgx d = -0.05 m must be conserved. x=0 m = 0.2 kg 0 + (0.5)(250)(0.05)2 + 0 Final: v = 0 0 + d=0 0 Conservation of ME: = 0.3125 x=h + (0.2)(9.8)(h) = 1.96 h 0.3125 = 1.96 h so h = 0.16 m MSU Physics 231 Fall 2015 44