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The Ideal Spring & Simple Harmonic Motion

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The Ideal Spring & Simple Harmonic Motion Simple Harmonic Motion http://www.youtube.com/watch?v=yVkdfJ9PkRQ What it shows: Fifteen uncoupled simple pendulums of increasing lengths dance together to produce visual traveling waves, standing waves, beating, and random motion. How it works: The period of one complete cycle of the dance is 60 seconds. The length of the longest pendulum has been adjusted so that it executes 51 oscillations in this 60 second period. The length of each successively shorter pendulum is carefully adjusted so that it executes one additional oscillation in this period. Thus, the 15th pendulum (shortest) undergoes 65 oscillations. When all 15 pendulums are started together, they quickly fall out of sync—their relative phases continuously change because of their different periods of oscillation. However, after 60 seconds they will all have executed an integral number of oscillations and be back in sync again at that instant, ready to repeat the dance. The Ideal Spring & Simple Harmonic Motion Hooke’s Law: The force (F) applied directly to a spring moves or causes a displacement (x) in the spring directly proportional to the amount of force. F x; F x and F x Hooke’s Law Equation: F = -k·x Ideal springs follow this relationship. F = Fs = restoring force = the force applied by the spring. Units: N. k = the spring constant = “stiffness” of spring (unique for each spring… depends on material and depends on number of coils) Units: N/m. x = displacement of spring from unstrained length. Units: m. F = -k·x; this is negative b/c restoring force always points in a direction opposite to the displacement of the spring. Unstrained length is when the spring is “at rest”—not stretched or compressed F = 0 Force applied by hand, FA, is opposite the force applied by spring, Fs FA = -Fs Fs = -k·x Force exerted by spring is opposite in direction to the displacement, x. Note: the restoring force will be different for each change of x, while k (spring constant) is constant. Which has a greater spring constant value, a 10-coil spring or the same spring cut in half, so there are two 5-coil springs? Compress the entire 10-coil spring and the 5-coil spring by 1.0 cm. x = F/k 10-coil spring: each coil moves 1.0cm/10 or 0.1 cm, but overall, x = 1.0 cm, 5-coil spring: each coil moves 1.0cm/5 or 0.2 cm, but overall, x = 1.0cm Since each coil in the 5-coil spring moves twice as far, the force must be twice as great when compared to the 10-coil spring… 10-coil spring: x = F/k = 1.0cm 5-coil spring: x = F/k = 2F/2k = 1.0cm the 5-coil spring has a spring constant twice as great as the 10-coil spring. Shorter springs are stiffer springs, all other factors being equal! The spring constant is inversely proportional to the # of coils in a spring. The shorter and less coils a spring has, the greater the force needed to compress it. In general: k 1 / (# of coils) Simple Harmonic Motion If a spring is stretched and then let go and no external forces act on it (such as friction) and no loss of energy, then the spring will vibrate back and forth indefinitely producing a pattern called Simple Harmonic Motion. Simple Harmonic Motion: motion which repeats itself with a specific frequency and its velocity changes throughout motion. Examples of SHM Recall: Period (T) = the time for one oscillation (back and forth motion); Frequency (f) = the number of cycles completed every second. f = 1/T The weight of an object on a vertical spring stretches the spring by an amount do. Simple harmonic motion of amplitude A occurs with respect to the equilibrium position of the object on the stretched spring. Harmonic motion graphically looks like a sine or cosine wave and is sinusoidal. The type of function that describes simple harmonic motion is A. linear B. exponential C. quadratic D. sinusoidal E. inverse Answer The type of function that describes simple harmonic motion is A. linear B. exponential C. quadratic D. sinusoidal E. inverse Energy and Simple Harmonic Motion Recall: An object possesses gravitational potential energy when it is at a location where gravity can do work on the object. PEg = m·g·h = Wg = Fg·d A spring also has another type of potential energy, elastic potential energy. Elastic potential energy: a stretched or compressed spring has Elastic PE and can do work on an object that is attached to the spring. Example: Garage doors w/springs do work on door to close or open it. Screen doors w/springs do work on the door to close/open it. Recall: W = F·d where F and d are parallel to one another. For a spring: F = +*k·x , but F is not constant. F is specific to an amount of x. As x changes, F changes, we will use the average force to determine work. * positive when work is done by spring to another object, negative for spring to restore itself. Average F = Favg = (Fi + Ff)/2 = (kxi + kxf)/2 Favg = ½ k (xi + xf) W = Fd = [ ½ k (xi + xf)]·d W = [½ k (xi + xf)]·(xi - xf) b/c d = (xi - xf) W = ½ k (xi + xf)(xi - xf) 2 2 W = ½ k (xi + xfxi – xfxi — xf ) 2 2 W = ½ k (xi – xf ) 2 2 W = ½ k xi – ½ k xf W = initial Elastic PE – final Elastic PE In general: 2 PEelastic = ½ k x Units: Joules k = spring constant x = distance stretched or compressed relative to unstrained length. Energy in Simple Harmonic Motion As a mass on a spring goes through its cycle of oscillation, energy is transformed from potential to kinetic and back to potential. Law of Conservation of Energy, E Efinal = Einitial Assuming no non-conservative forces acting externally… no friction, no air resistance, then… If a mass falls… 2 Etotal = ½ mv + mgh If a mass falls and rotates while falling… 2 2 Etotal = ½ mv + mgh + ½ I· (½ I·2=rotational energy; I=moment of inertia) If a mass falls and rotates and is on a spring while falling…such as bungee jumping… 2 2 2 Etotal = ½ mv + mgh + ½ I· + ½ k x No friction, no air resistance… Efinal = Einitial 2 2 2 2 2 2 ½mvf +mghf+½If +½kxf = ½mvi +mghi+½Ii +½kxi A 0.20 kg ball is attached to a vertical spring. The spring constant of the spring is 28 N/m. The ball, supported initially so that the spring is neither stretched nor compressed, is released from rest. In the absence of air resistance, how far does the ball fall before being brought to a momentary stop by the spring? And the solution is… 2 2 2 2 2 2 ½mvf +mghf+½If +½kxf = ½mvi +mghi+½Ii +½kxi Since the ball starts at rest, and reaches a final velocity of zero at the bottom of the fall, vi=0, vf=0; KEf =0 and KEi=0 Since the ball and spring system do not rotate during 2 2 this fall, f =0 and i=0; ½If =0 and ½Ii =0 hi = ho in the drawing, and hf = 0 m, PEf =0 Since the ball is unstrained to begin with, xi = 0, 2 ½kxi =0 2 2 2 2 2 2 ½mvf +mghf+½If +½kxf = ½mvi +mghi+½Ii +½kxi 2 So, 0 + 0 + 0 +½kxf = 0 + mghi + 0 + 0 Next slide… Given: k=28 N/m; m = 0.20 kg Find: hi (height) where vf = 0 2 ½kxf = mghi 2 2 ½ (28 N/m) xf = (0.20 kg) (9.8 m/s ) hi Since x = hi 2 2 ½ (28 N/m) hi = (0.20 kg) (9.8 m/s ) hi hi = 0.14 m Here is a good one from Phet http://phet.colorado.edu/simulations/sims.p hp?sim=Masses_and_Springs The Pendulum T = 2(L/g) (for small angles only 15o) T = period of pendulum L = length of pendulum g = acceleration due to gravity Recall T=1/f 1/f = 2 (L/g) 1 = f 2 (L/g) 2f = (g/L) 2f = = (g/L) small angles only! A Pendulum’s Period, T, depends upon the length, L, of the pendulum in what way??? T = 2(L/g) T L As length of the pendulum increases, the Period, T, increases by L Or… T2 L As period of the pendulum increases, length, L, increases by T2 If double T, L is ??? 2 2 2 L T (2To) = 4·To If double T, L increases by 22 or 4. Or, in order to double period, T, length, L, has to be quadrupled. A Pendulum’s Angular Frequency, , depends upon the length, L, of the pendulum in what way??? 2f = = (g/L) 1/L As length of the pendulum increases, the angular velocity or angular frequency (rotating rate) decreases by 1/L Or… 1/2 L If double , L is ??? 2 2 2 L 1/ 1/(2o) = 1/4·1/(o) If double , L is reduced by (½)2 or ¼ If you want angular velocity/frequency to double, then need to reduce L by (½)2 or ¼. Damped Harmonic Motion In a perfect world, damping does not occur, and the amplitude of the object experiencing SHM does not change… Damped Harmonic Motion is SHM in the real world… the amplitude decreases due to friction, air resistance, etc. (not gravity) as time goes by. Application of damping: Suspension system of automobiles… The suspension system is designed and built so that damped harmonic motion will occur when a car goes over a bump, for example. The opposite of Damped Harmonic Motion can occur too… it’s called resonance.
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