Physics 231. Topic 7: Oscillations. Alex Brown October MSU Physics 231 Fall

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1 Physics 231 Topic 7: Oscillations Alex Brown October MSU Physics 231 Fall

2 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 7in Rex & Wolfson MSU Physics 231 Fall

3 Springs x o =0 MSU Physics 231 Fall

4 Hooke s Law x o =0 F s = -kx Hooke s law k = spring constant in units of N/m. x = displacement from the equilibrium point x o =0 MSU Physics 231 Fall

5 How to find the spring constant k When the object hanging from the spring is not moving: x F = ma = 0 F spring + F gravity = 0 x= 0 x= d F spring = kx F gravity = mg -k(d) + mg = 0 k = mg/d MSU Physics 231 Fall

6 Springs Hooke s Law: F s = kx k: spring constant (N/m) x: distance the spring is stretched from equilibrium W = area under F vs x plot = ½kx 2 The energy stored in a spring depends on how far the spring has been stretched: elastic potential energy PE spring = ½ k x 2 MSU Physics 231 Fall

7 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? PE spring = ½ k x 2 PE gravity = mgh KE = ½ mv 2 ME = PE + KE = Constant (PE gravity +PE spring +KE ball ) pull = (PE gravity +PE spring +KE ball ) launch mgh pull + ½kx pull 2 + ½mv pull2 = mgh launch + ½kx launch 2 + ½mv launch 2 ½120(0.05) 2 = 0.1 (9.81) ( 0.05*sin(10 o ) ) + ½(0.1)v 2 launch 0.15 = 8.5x v 2 v =1.7 m/s MSU Physics 231 Fall

8 Momentum p = mv Impulse p = p f -p i = F t Conservation of momentum (closed system) p 1i + p 2i = p 1f + p 2f Collisions in one dimension given given m 1 and m 2 or c = m 2 /m 1 v 1i and v 2i Inelastic (stick together, KE lost) v f = [v 1i + c v 2i ] / (1+c) Elastic (KE conserved) v 1f = [(1-c) v 1i + 2c v 2i ] / (1+c) v 2f = [(c-1) v 2i + 2v 1i ] / (1+c) MSU Physics 231 Fall

9 Collisions in one dimension given given m 1 = m 2 or c = 1 v 1i and v 2i Inelastic (stick together, KE lost) v f = [v 1i + v 2i ] / 2 Elastic (KE conserved) v 1f = v 2i v 2f = v 1i MSU Physics 231 Fall

10 Carts on a spring track v m 1 m 2 spring What is the maximum compression of the spring if the carts collide a) elastically and b) perfectly inelastic? k = 50 N/m v = 5.0 m/s m 1 = m 2 = 0.25 kg A) Conservation of momentum and KE with c = 1 v 1f = v v 2f = 0 :: v 1f = 0 v 2f = v = 5 m/s Conservation of energy: ½mv 2 = ½kx 2 (0.5) (0.25) (5) 2 = (0.5)(50)x 2 x = 0.35 m B) Conservation of momentum only with c = 1 v f = (v 1f + v 2f )/2 = (v + 0)/2= 2.5 m/s Conservation of energy: ½mv f 2 = ½kx 2 (0.5)(0.5)(2.5) 2 = (0.5)(50)x 2 x=0.25 m MSU Physics 231 Fall

11 Hooke s Law F s = -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

12 Simple harmonic motion displacement x c) A b) a) time (s) (c) (b) (a) 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]) Angular frequency = 2πf = 2π/T MSU Physics 231 Fall

13 displacement x 5 m Concept Quiz -5 m 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

14 displacement x 5 m What the equation? -5 m time (s) x= A cos( t) A = 5 T= 2π = 2π/T = 2π/4 = 1.57 rad/s MSU Physics 231 Fall

15 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) 0 B) A/2 C) A D) 4A E) 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

16 Energy and Velocity x=+a E kin (½mv 2 ) E pot,spring (½kx 2 ) Sum 0 ½kA 2 ½kA 2 x=0 ½mv 2 max 0 ½mv 2 max x=-a 0 ½k(-A) 2 ½kA 2 conservation of ME: ½mv 2 max = ½kA 2 so v max = ±A (k/m) MSU Physics 231 Fall

17 Velocity and acceleration in general Total ME at any displacement x: ½mv 2 + ½kx 2 Total ME at max. displacement A: ½kA 2 Conservation of ME: ½kA 2 = ½mv 2 + ½kx 2 So: v = ± [(A 2 -x 2 )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

18 An Example The maximum velocity and acceleration of mass connected to a spring are v max =0.95 m/s and a max =1.56 m/s 2, respectively. Find the oscillation amplitude. MSU Physics 231 Fall

19 An Example The maximum velocity and acceleration of mass connected to a spring are v max =0.95 m/s and a max =1.56 m/s 2, 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 v max = ±A (k/m) so (v max ) 2 = A 2 (k/m) so a max = A (k/m) A = v max2 /a max = (0.95) 2 /(1.56) = 0.58 m MSU Physics 231 Fall

20 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. r=a MSU Physics 231 Fall

21 Harmonic oscillations vs circular motion v x (t) = -v sin v v x The v x projection of the linear velocity of the rotating object is the velocity of the mass on the spring. r=a MSU Physics 231 Fall

22 x=-a x=+a x=0 But remember for circular motion with constant speed around the circle = t and v = r = A where is the angular frequency So x(t) = A cos = A cos( t) v x (t) = -v sin = - A sin( t) r=a time to complete one circle = T = one period so T = 2 or = 2 /T MSU Physics 231 Fall

23 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 amplitude of the oscillation = 2 /T = 2 f angular frequency T period f = 1/T frequency. MSU Physics 231 Fall

24 A x = A cos( t) x -A time (s) A velocity v = -A sin( t) v -A time (s) k Remember vmax A thus m The angular frequency does not depend on A k m MSU Physics 231 Fall

25 displacement x A Displacement vs Acceleration -A time (s) Newton s second law: F=ma -kx=ma a=-kx/m acceleration(a) ka/m -ka/m MSU Physics 231 Fall

26 A x = A cos( t) x -A time (s) A v = -A sin( t) v -A A 2 a acceleration a = -A 2 cos( t) time (s) time (s) -A 2 MSU Physics 231 Fall

27 Period of a Spring T 2 2 m k Increase m: increase T Increase k: decrease T MSU Physics 231 Fall

28 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

29 The pendulum x= A x=+a x=0 MSU Physics 231 Fall

30 The pendulum = max = A y=h Conservation of Energy! At x=+a: PE = mgh KE = ½mv 2 = 0 At x=0: PE = mg(0) = 0 KE = ½mv 2 ½mv 2 = mgh v 2 = 2gh velocity at the bottom MSU Physics 231 Fall

31 The pendulum Tension at max angle = max L T T = mg cos max Tension at bottom = 0 T = mg + mv 2 /L mg sin s mg cos MSU Physics 231 Fall

32 The pendulum L T Restoring force: F=-mg sin The force pushes the mass m back to the central position. mg sin s mg cos MSU Physics 231 Fall

33 The pendulum L Restoring force: F=-mg sin The force pushes the mass m back to the central position. sin (radians) if is small T F = - mg also s = L mg sin s mg cos so: F=-(mg/L)s MSU Physics 231 Fall

34 pendulum The pendulum vs spring k m mg / m L g L Does not depend on mass spring pendulum F kx F ( mg / L) s k m g L MSU Physics 231 Fall

35 Damped Oscillations Newton s 1 st 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 A 1 = f A So for n oscillations A n = f n A mechanical energy ME = (1/2) k A 2 MSU Physics 231 Fall

36 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

37 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: ½mv 2 + ½kx 2 + mgh On top of the bridge: ME = mgh Maximum extension of the cord: ME = ½ ka 2 = 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) = 50 (28m L) 2 (28m L) = sqrt(23544/50) = 21.7 m L = 6.3 m MSU Physics 231 Fall

38 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? = = 5.62 sec Frequency f = 1/T = 1/5.62s = 0.18 Hz MSU Physics 231 Fall

39 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) = 2 f f= /2 = 3.55 Hz (=1/T) d) x(t) = Acos( t) = 0.1 cos(22.4t) v(t) = - Asin( t) = sin(22.4t) a(t) = - 2 Acos( t) = cos(22.4t) MSU Physics 231 Fall

40 0.1 x -0.1 velocity v time (s) a MSU Physics 231 Fall

41 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 m Moon Elevator g L Faster(A) Same(B) Slower(C) MSU Physics 231 Fall

42 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 B) 1 T C) 2 T D) 4 T E) 8 T T 2 L g 4L L T 2 4 2T g g Mass does not matter MSU Physics 231 Fall

43 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? 1 st step: find the spring constant k F spring =-F gravity or -kd =-mg k = mg/d = (1)(9.8)/0.5 = 19.6 N/m 2 nd step: find the acceleration upon release Newton s second law: F=ma -kx = ma a = -kx/m a = -(19.6)(0.2)/1 = m/s 2 MSU Physics 231 Fall

44 d ME = ½mv 2 + ½kd 2 + mgx x 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) Initial: v = 0 d = m x = 0 m = 0.2 kg 0 + (0.5)(250)(0.05) = Final: v = 0 d = 0 x = h (0.2)(9.8)(h) = 1.96 h must be conserved. Conservation of ME: = 1.96 h so h = 0.16 m MSU Physics 231 Fall

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