Physics 231 Lecture 18

Transcription

1 Physics 31 ecture 18 τ = Fd;d is the lever arm Main points of today s lecture: Energy Pendulum T = π g ( ) θ = θmax cos πft + ϑ0 Damped Oscillations x x equibrium = Ae bt/(m) cos(ω damped t) ω damped = k / m (b / (m)) Resonance Angular acceleration and kinematics: α average α = Δω Δ θ = ωt ω = ω + αt 1 1 ω = ( ω+ ω0) Δ θ = ( ω+ ω0) t 1 Δ θ = ω0t + αt ω = ω0 + αδθ Torques: τ Frsinθ 0

2 Energy and Velocity x=+a E kin (½mv ) E pot,spring (½kx ) Sum 0 ½kA ½kA x=0 ½mv max 0 ½mv max x=-a 0 ½k(-A) ½kA conserva3on of ME: ½mv max = ½kA so v max = A(k/m) 1/ v max = Aω

3 Velocity and acceleration in general Total E mech at any displacement x: ½mv + ½kx Total E mech at max. displacement A: ½kA also F = ma = -kx so a = -kx/m ConservaAon of E mech : ½kA = ½mv + ½kx So: v = ± [(A -x )k/m] 1/ =± [(A -x )k/m, if x=acos(ωt), then v=± (A -A cos(ωt) k/m)=±aωsin(ωt), using cos(ωt) + sin(ωt) =1 and k/m=ω posi3on x velocity v Accelera3on a +A 0 -ka/m 0 ±A (k/m) 0 -A 0 ka/m

4 Simple Pendulum The simple pendulum is another example of simple harmonic motion The Force along the tangential direction t is given by the component of the gravitational force in that direction F s = - m g sin θ Newton s second law states: F s = ma s =- m g sin θ a s =- g sin θ For small angles < 15 : sin θ s (in radians)è sinθ=s/ a s = -gs/= - g/ s 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: max f = 1 π g ; T = 1/f = π g d=sinθ T depends on and g not on θ max θ = θ cos( π ft + ϑ ); s = θ; v t = πfθ max sin(πft + ϑ)

5 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.

6 Comparison of simple pendulum to a spring-mass system

7 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

8 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 = π B g B T = π A g A T /T B A = T B T A = π π g B g A = g g A B g 1 A = = gb

9 Quiz Two playground swings start out together. During the time that 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 = f1; cycles, cycles,1 Ncycles, = f N / N 8.5 / = 1 = f /f1 1 = ( ) ( ) = = g / / π g / 1 / π.7

10 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 to get k and then solve for ω) a) 0.7 rad/s= 0.7 Hz b) 4.5 rad/s=4.5 Hz c).9 rad/s=.9 Hz d) 13.8 rad/s=13.8 Hz e). rad/s=. Hz 0= Ftotal = mg ky mg ( )( 0.4kg 9.8m / s ) k = = = 1.96N / m y m 0.4 kg d= m ω = k m 1.96N / m = =.Hz 0.4kg

11 This figure shows the sinusoidal nature of simple harmonic motion With no friction or viscosity, the solid blue amplitude of the oscillation remains the same. With damping, the dashed amplitude decreases. If the damping forces is from air resistance. F air = b! v Damping of Simple Harmonic Motion x x equibrium = Ae bt/(m) cos(ω damped t) ω damped = k / m (b / (m)) When (b/(m)) k/m, the amplitude doesn t oscillater. Instead, it decays exponentially. The quickest decay is for (b/(m)) = k/m, which is called critically damped. Auto spring are critically damped by the auto shocks.

12 Resonance One way to keep oscillation going when there is damping is to keep pushing it with a periodic force. F driver = F 0 cos( ω driver t) After some time the motion will be x = A( F 0,ω driver,ω 0 )cos( ω driver t) A(F 0, ω driver, ω 0 ) When the driving frequency equals the frequency without damping, the amplitude is largest. We call that resonance. ω driver

13 Review: Angular velocity The rate of change of the angular displacement is defined to be the angular velocity ω. In term of the angular displacement, the average angular velocity over a time interval is: ω = average ω = Δθ Δ t The instantaneous angular velocity can be obtained by making the time interval very short: Δθ ω = limδ t 0 Note: if Δθ measured in radians, Δs=r Δθ is the arc length covered in time. Thus: Δs rδθ v t = = = rω is the average tangential speed and Δs rδθ vt = limδ t 0 = limδ t 0 = rω is the instantaneous tangential speed of the object. Fig. 7.4, p. 189 Slide 5

14 Angular acceleration Many times, the angular velocity changes with time. We quantify this by the angular acceleration α. In term of the angular velocity, the average angular acceleration over a time interval is: Δω αaverage α = ω= ω0 + αt for constant α Note: if Δω measured in rad/s, Δv t =rδω is the change in tangential speed during time. Thus: Δvt rδω at,average at = = = r α ā t is the average tangential acceleration. The instantaneous angular acceleration can be obtained by making the time interval very short: Δω α = lim 0 And a t =rα is the instantaneous tangential acceleration of the object. Δvt rδω at = limδ t 0 = limδ t 0 = rα 13

15 Analogy between linear and angular kinematics for constant acceleration If the acceleration is constant, we obtained a set of equation on the left side of the table relating Δx, v, a and t. If the angular acceleration is constant and we go through the same steps, we obtain the analogous set of equation on the right side of the table relating Δθ, ω, α, and t. setting =t Δ x = vt; Δx Δ t = v v v= v + at 0 1 v= ( v+ vo ) 1 Δ x = ( v+ vo ) t 1 Δ x = v0t+ at v v = aδx 0 ave Δ θ = ωt; Δθ Δ t = ω ω ω = ω + αt 0 1 ω = ( ω+ ω0 ) 1 Δ θ = ( ω+ ω0 ) t 1 Δ θ = ω0t + αt ω = ω + αδθ 0 ave 14