Physics 31 ecture 15 Main points of today s lecture: Simple harmonic motion Mass and Spring Pendulum Circular motion T 1/f; f 1/ T; ω πf for mass and spring ω x Acos( ωt) v ωasin( ωt) x ax ω Acos( ωt) Ai is a constant t k / m for pendulum ( ωt ) θ θ cos t max ω g /
Graphical Representation of Simple Harmonic Motion x T x Acos t When does x resume its maximum? cos(0)1; cos(π)1; cos(4π)1, etc ωtπ; T(π)/ ω; f1/t; fπ/ω ωπf is the angular frequency. T is the period. It is the time for the motion to repeat itself fi is the frequency. It is the number of times the motion repeats itself per second. Units are Hertz (Hz) or s -1. The period does not depend on the amplitude A a a kx/m When x is a maximum or minimum, velocity is zero When x is zero, the speed is a maximum When x achieves its most positive value, a is at its most negative value. v ω v ωω Asin ω t A ω cosωt
Conceptual question A mass attached to a spring oscillates back and forth as indicated in the position vs. time plot below. At point P, the mass has a. positive velocity and positive acceleration. b. positive velocity and negative acceleration. c. positive velocity and zero acceleration. d. negative velocity and positive acceleration. e. negative velocity and negative acceleration. x is increasing with time therefore the velocity is positive x is increasing with time, therefore the velocity is positive. x is positive, a-kx/m, therefore the accleration is negative.
Example The motion of an object is described by the equation x (0.30 m) cos(πt/3), where t is assumed to be in seconds. Find (a) the position, (b) velocity and (c) acceleration of the object at t 0 and t 0.60 s, (d) the amplitude of the motion, (e) the frequency of the motion, and (f) the period of the motion. d) general form: x A cos( ωt); A 0.3m; e) & f ) ω π /3Hz 105Hz 1.05Hz, f ω /( π).17hz; ( ) ω π ( ) x Asin( ωt) 0.3m cos(1.05t) (t is in sec.) v - ω Asin( ω t) ( 105Hz 1.05Hz ) ( ) a ω x 1.1x 03m 0.3m sin(105t) sin(1.05t) a) c)just plug the time into the equations above at t 0 : x 0.3 m; v 0 ; a 0.33m/s at t 0.6s : x 0.4 m; v 0.37m/s ; a.6m/s T 1/f 6s
Example A 50 coil spring has a spring constant t of 860 N/m. One end of fthe 50- coil spring is attached to a wall. An object of mass 45 kg is attached to the other end of the spring and the system is set in horizontal oscillation. What is the angular frequency of the motion? a).39 Hz b) 4.37 Hz k 860N / m c) 5.1 Hz ω 4.37Hz m 45kg d) 6.85 Hz e) 9. Hz testing recall If mass spring system was arranged vertically with the mass suspended from the 50 coil spring, how would the frequency change? a) it would be smaller because gravity subtracts from the spring force at the bottom of its motion. b) it would be larger because gravity adds to the spring force at the top of its motion. c) it would be exactly the same. Gravity only displaces the equilibrium point so that the equilibrium length is greater.
Example If a mass of 0.4 kg is suspended vertically by a spring, it stretches the spring by m. Assume the spring is stretched further and released, and the mass plus spring system undergoes vertical oscillations. Calculate the angular frequency of the oscillatory motion.(hint: Solve the static equilibrium ilibi to get tk and dthen solve for ω) ) a) 0.7 rad/s 0.7 Hz b) 4.5 rad/s4.5 Hz c).9 rad/s.9 Hz d) 13.8 rad/s13.8 Hz e). rad/s. Hz 0 F total mgg ky y k mg y ( )( 0.4kg 9.8m / s ) 1.96N / m m 0.4 kg d m ω k m 1.96N / m.hz 0.4kg
Verification of Sinusoidal Nature This experiment shows the sinusoidal nature of simple harmonic motion The spring mass system oscillates in simple harmonic motion The attached pen traces out the sinusoidal motion With no friction or viscosity, the amplitude of the oscillation remains the same. With damping, the amplitude decreases:
Simple Pendulum The simple pendulum is another example of simple harmonic motion The torque is given by the gravitational force times the moment arm dsinθ τ -mgd- m g sin θ Newton s second law states: note: Im τ Iα - m g sin θmθ α α - g/ sin θ For small angles < 15 : sin θ θ sin θ θ (in radians) α - g/ θ This is similar to the equation a x - k/m x which describes the motion of mass plus spring. We therefore expect simple harmonic motion with: ω f 1 g ; T 1/f ππ π π θ θ cos( π ft); s r θ; max g t max T depends on and g not on θ max v rπθ f sin(πft)
Comparison of simple pendulum to a spring-mass system
Checking Understanding A series of pendulums with different length strings and different masses is shown below. Each pendulum is pulled to the side by the same (small) angle, the pendulums are released, and they begin to swing from side to side. 0 cm 1 g f ; π Rank the frequencies of the five pendulums, from highest to lowest. A. A E > B D > C B. D > A C > B E C. A B C D E D. B > E > C > A > D Slide 14-17
Conceptual question A person swings on a swing.when the person sits still, the swing oscillates back and forth at its natural frequency. If, instead, two people sit on the swing, the natural frequency of the swing is a. greater. b. the same. c. smaller.
Quiz Two playground swings start t out together. th During the time that t swing 1 makes 10 complete cycles, swing makes only 8.5 cycles. What is the ratios 1 / of the lengths of the swings? (Hint: use ratio technique) a).3 b).4 c).5 d).6 e).7 Ncycles,1 f1δt; cycles, cycles,1 Ncycles, fδt N /N 8.5/10 0.85 1 1 f /f1.7 ( π ) ( π ) g/ / π Δ t g/ / Δt 1
Example The period of a simple pendulum is 0.% longer at location A than it is a location B. Find the ratio g A /g B of the acceleration due to gravity at these two locations. T B π A gb T π g A T /T B A TB π 1 gb g 1.00 g TA π g A A B g 1 A g B 1.00 0.996
Example An object tis attached dto the lower end of a 100 coil spring that tis hanging from the ceiling. The spring stretches by 0.16 m. The spring is then cut into two identical springs of 50 coils each. As the drawing shows, each spring is attached between the ceiling and the object. By how much does each string stretch? at first : 0 ( mg) ( 0. m) F total -ky mg k 16 k 100 0.16m second : k50 k100 keff k50 0 Ftotal -keff y mg mg mg y k mg eff 4 0.16m 1 100 4k 100 0.16m 4 mg 0.04m