# Hooke s Law. Spring. Simple Harmonic Motion. Energy. 12/9/09 Physics 201, UW-Madison 1

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1 Hooke s Law Spring Simple Harmonic Motion Energy 12/9/09 Physics 201, UW-Madison 1

2 relaxed position F X = -kx > 0 F X = 0 x apple 0 x=0 x > 0 x=0 F X = - kx < 0 x 12/9/09 Physics 201, UW-Madison 2

3 We know that if we stretch a spring with a mass on the end and let it go, the mass will oscillate back and forth (if there is no friction). k m This oscillation is called Simple Harmonic Motion, and is actually easy to understand... k k m m 12/9/09 Physics 201, UW-Madison 3

4 k m m d 2 x dt 2 x d 2 x dt 2 = k m x a differential equation 12/9/09 Physics 201, UW-Madison 4

5 d 2 x define dt = k 2 m x ω = k m d 2 x dt 2 = ω 2 x Where ω is the angular frequency of motion Try the solution x = A cos(ωt) dx dt d 2 x dt 2 = ω Asin ( ωt ) = ω 2 Acos( ωt) = ω 2 x This works, so it must be a solution! 12/9/09 Physics 201, UW-Madison 5

6 d 2 x We just showed that (from F = ma) dt 2 = ω 2 x has the solution x = A cos(ωt) but x = A sin(ωt) is also a solution. The most general solution is a linear combination (sum) of these two solutions! x = Bsin( ωt) + C cos( ωt) = Acos( ωt + φ) dx dt d 2 x dt 2 = ω Bcos ωt ( ) ωc sin ( ωt ) = ω 2 Bsin( ωt) ω 2 C cos( ωt) = ω 2 x ok 12/9/09 Physics 201, UW-Madison 6

7 x = Acos( ωt + ϕ) 3 But wait a minute...what does angular frequency ω have to do with moving back & forth in a straight line?? 4 2 y x The spring s motion with amplitude A is identical to the x-component of a particle in uniform circular motion with radius A. 2 3 π 2 4 π 5 6 θ 12/9/09 Physics 201, UW-Madison 7

8 A mass m = 2 kg on a spring oscillates with amplitude A = 10 cm. At t = 0 its speed is maximum, and is v = +2 m/s. What is the angular frequency of oscillation ω? What is the spring constant k? v MAX = ωa ω = v MAX A = 2 m s 10cm = 20 s 1 ω = k Also: k = mω 2 m So k = (2 kg) x (20 s -1 ) 2 = 800 kg/s 2 = 800 N/m k m 12/9/09 Physics 201, UW-Madison 8 x

9 Use initial conditions to determine phase φ! Suppose we are told x(0) = 0, and x is initially increasing (i.e. v(0) = positive): x(t) = A cos(ωt + φ) v(t) = -ωa sin(ωt + φ) a(t) = -ω 2 A cos(ωt + φ) x(0) = 0 = A cos(φ) v(0) > 0 = -ωa sin(φ) φ < 0 φ = π/2 or -π/2 So φ = -π/2 π θ 2π k m cosθ sinθ 0 x 12/9/09 Physics 201, UW-Madison 9

10 So we find φ= -π/2 x(t) = A cos(ωt - π/2 ) v(t) = -ωa sin(ωt - π/2 ) a(t) = -ω 2 A cos(ωt - π/2 ) x(t) = A sin(ωt) v(t) = ωa cos(ωt) a(t) = -ω 2 A sin(ωt) A x(t) k m -A π 2π ωt 0 x 12/9/09 Physics 201, UW-Madison 10

11 We already know that for a vertical spring if y is measured from the equilibrium position The force of the spring is the negative derivative of this function: U = 1 2 ky2 So this will be just like the horizontal case: k ky = ma = m d 2 y dt 2 y = 0 Which has solution y = A cos(ωt + φ) where ω = F = -ky m 12/9/09 Physics 201, UW-Madison 11 k m

12 Recall that the torque due to gravity about the rotation (z) axis is τ = -mgd d = Lsin θ Lθ for small θ so τ = -mg Lθ τ = I d 2 θ dt 2 z But τ = Iα, where I = ml 2 d 2 θ where dt 2 = ω 2 θ ω = g L Differential equation for simple harmonic motion! θ = θ 0 cos( ωt + ϕ) θ d L m mg 12/9/09 Physics 201, UW-Madison 12

13 You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T 1. Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T 2. Which of the following is true: (a) T 1 = T 2 (b) T 1 > T 2 (c) T 1 < T 2 12/9/09 Physics 201, UW-Madison 13

14 We have shown that for a simple pendulum ω = g L Recall the pendulum depended on the torque from gravity. Now the moment arm is from the center of mass and the oscillation point so L is shorter Since T = 2π ω T = 2π L g If we make a pendulum shorter, it oscillates faster (shorter period) 12/9/09 Physics 201, UW-Madison 14

15 Force: d 2 s dt 2 = ω 2 s ω = k k m k m m s 0 0 s Solution: s = A cos(ωt + φ) ω = g L s L 12/10/09 Physics 201, UW-Madison 15

16 U t x = Acos( ωt) U = 1 2 kx2 = 1 2 ka2 cos 2 ( ωt) K t v = dx dt = ω Asin ( ωt ) K = 1 2 mv2 = 1 2 mω 2 A 2 sin 2 ( ωt) t E = K +U = K Max = 1 2 mω2 A 2 = U Max = 1 2 k 2 A 2 12/10/09 16

17 U = 1 2 kx2 U = 1 2 kx2 K = 1 2 mv2 E = U + K = 1 2 ka2 = 1 2 mv2 max 12/10/09 Physics 201, UW-Madison 17

18 For both the spring and the pendulum, we can derive the SHM solution by using energy conservation. The total energy (K + U) of a system undergoing SHM will always be constant! This is not surprising since there are only conservative forces present, hence K+U energy is conserved. U E K U -A 0 A s 12/10/09 Physics 201, UW-Madison 18

19 m x=0 x E total = 1/2 Mv 2 + 1/2 kx 2 = constant KE PE KE max = Mv 2 max /2 = Mω2 A 2 /2 =ka 2 /2 PE max = ka 2 /2 E total = ka 2 /2 12/10/09 Physics 201, UW-Madison 19

20 U 1 2 kx2 for small x 12/10/09 Physics 201, UW-Madison 20

21 Suppose we have some arbitrarily shaped solid of mass M hung on a fixed axis, and that we know where the CM is located and what the moment of inertia I about the axis is. The torque about the rotation (z) axis for small θ is (using sin θ θ) τ = -Mgd -MgRθ MgRθ = I d 2 θ dt 2 d 2 θ dt = ω 2 θ 2 where ω = MgR I ( ) θ = θ 0 cos ωt + ϕ τ α z-axis R θ x CM d Mg 12/10/09 Physics 201, UW-Madison 21

22 Consider an object suspended by a wire attached at its CM. The wire defines the rotation axis, and the moment of inertia I about this axis is known. The wire acts like a rotational spring. When the object is rotated, the wire is twisted. This produces a torque that opposes the rotation. In analogy with a spring, the torque produced is proportional to the displacement: τ = -kθ τ wire θ I 12/10/09 Physics 201, UW-Madison 22

23 Since τ = -kθ, τ = Iα becomes kθ = I d 2 θ dt 2 d 2 θ dt = ω 2 θ where ω = k 2 I τ I wire θ This is similar to the mass on spring except I has taken the place of m (no surprise). 12/10/09 Physics 201, UW-Madison 23

24 In many real systems, nonconservative forces are present The system is no longer ideal Friction is a common nonconservative force In this case, the mechanical energy of the system diminishes in time, the motion is said to be damped The amplitude decreases with time The blue dashed lines on the graph represent the envelope of the motion 12/10/09 Physics 201, UW-Madison 24

25 One example of damped motion occurs when an object is attached to a spring and submerged in a viscous liquid The retarding force can be expressed as R = - b v where b is a constant b is called the damping coefficient The restoring force is kx From Newton s Second Law ΣF x = -k x bv x = ma x or, m d 2 x dt 2 + b dx dt + kx = 0 12/10/09 Physics 201, UW-Madison 25

26 m d 2 x dt 2 + b dx dt + kx = 0 When b is small enough, the solution to this equation is: with x = A 0 e (b/2m)t cos( ω t + δ ) ω = ω 0 (This solution applies when b<2mω 0.) b 1 2mω 0 2, ω 0 = k m 12/10/09 Physics 201, UW-Madison 26

27 The position can be described by x = A e (b/2m)t cos( ω t + δ ) 0 The angular frequency is ω = k m b 2m 2 When the retarding force is small, the oscillatory character of the motion is preserved, but the amplitude decreases exponentially with time The motion ultimately ceases 12/10/09 Physics 201, UW-Madison 27

28 The energy of a weakly damped harmonic oscillator decays exponentially in time. The potential energy is ½kx 2 : x 2 = ( A e (b/2m)t cos( ω t + δ )) 2 0 = A 2 e (b/m)t cos 2 ( ω t + δ ) 0 Average over a period when (b/m)<<ω : Average of cos 2 = ½, so averaged over a period, x 2 A 0 2 e (b/m)t. 12/10/09 Physics 201, UW-Madison 28

29 The energy of a weakly damped harmonic oscillator decays exponentially in time. The kinetic energy is ½mv 2 : v 2 = d dt ( A 0 e (b/2m)t cos( ω t + δ )) 2 ( ) 2 = (b / 2m)A 0 e (b/2m)t cos( ω t + δ ) A 0 e (b/2m)t sin( ω t + δ ) Average energy over a period when (b/m)<<ω : Average of cos 2 = average of sin 2 = ½, and average of sinx cosx=0, so averaged over a period, v 2 e (b/m)t. 12/10/09 Physics 201, UW-Madison 29

30 The energy of a weakly damped harmonic oscillator decays exponentially in time. Both kinetic and potential energy (averaged over a period) decay as e -(b/m)t. So total energy is proportional to e -(b/m)t. 12/10/09 Physics 201, UW-Madison 30

31 ω 0 is also called the natural frequency of the system If R max = bv max < ka, the system is said to be underdamped When b reaches a critical value b c such that b c / 2 m = ω 0, the system will not oscillate The system is said to be critically damped If R max = bv max > ka and b/2m > ω 0, the system is said to be overdamped For critically damped and overdamped there is no angular frequency 12/10/09 Physics 201, UW-Madison 31

32 It is possible to compensate for the loss of energy in a damped system by applying an external sinusoidal force (with angular frequency ω) The amplitude of the motion remains constant if the energy input per cycle exactly equals the decrease in mechanical energy in each cycle that results from resistive forces After a driving force on an initially stationary object begins to act, the amplitude of the oscillation will increase After a sufficiently long period of time, E driving = E lost to internal Then a steady-state condition is reached The oscillations will proceed with constant amplitude The amplitude of a driven oscillation is ω 0 is the natural frequency of the undamped oscillator 12/10/09 Physics 201, UW-Madison 32

33 When ω ω 0 an increase in amplitude occurs This dramatic increase in the amplitude is called resonance The natural frequency 0 is also called the resonance frequency At resonance, the applied force is in phase with the velocity and the power transferred to the oscillator is a maximum The applied force and v are both proportional to sin (ωt + φ) The power delivered is F. v» This is a maximum when F and v are in phase Resonance (maximum peak) occurs when driving frequency equals the natural frequency The amplitude increases with decreased damping The curve broadens as the damping increases The shape of the resonance curve depends on b 12/10/09 Physics 201, UW-Madison 33

34 The amplitude of a system moving with simple harmonic motion is doubled. The total energy will then be 4 times larger 2 times larger the same as it was half as much quarter as much U = 1 2 kx mv2 at x = A,v = 0 U = 1 2 ka2 12/10/09 Physics 201, UW-Madison 34

35 A glider of mass m is attached to springs on both ends, which are attached to the ends of a frictionless track. The glider moved by 0.2 m to the right, and let go to oscillate. If m = 2 kg, and spring constants are k 1 = 800 N/m and k 2 = 500 N/m, the frequency of oscillation (in Hz) is approximately 6 Hz 2 Hz 4 Hz 8 Hz 10 Hz f = ω 2π = k / m 2π = π = 4 Hz 12/10/09 Physics 201, Fall 2006, UW-Madison 35

36 A 810-g block oscillates on the end of a spring whose force constant is 60 N/m. The mass moves in a fluid which offers a resistive force proportional to its speed the proportionality constant is N.s/m. Write the equation of motion and solution. F s + F R = ma m d 2 x dt 2 dx = kx b dt Solution: x(t) = Ae b 2m t cos(ωt + φ) What is the period of motion? s ω = k m b 2m 2 What is the fractional decrease in amplitude per cycle? Write the displacement as a function of time if at t = 0, x = 0 and at t = 1 s, x = m. Solution: x(t) = 0.182e 0.100t sin(8.61t + φ) 12/10/09 Physics 201, Fall 2006, UW-Madison 36

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