Reading: Chapter 15 Siple Haronic Motion Siple Haronic Motion Frequency f Period T T 1. f Siple haronic otion x ( t) x cos( t ).
Aplitude x Phase Angular frequency Since the otion returns to its initial value after one period T, Thus x cos( t ) x cos[ ( t T) )], t ( t T), T. f. T
Velocity dx d v( t) [ x cos( t )], dt dt v( t) x sin( t ). Velocity aplitude v x. Acceleration dv d a( t) [ x sin( t )], dt dt a( t) x cos( t ). Acceleration aplitude a x. Note that a( t) x( t), d x x 0. dt This equation of otion will be very useful in identifying siple haronic otion and its frequency.
The Force Law for Siple Haronic Motion Consider the siple haronic otion of a block of ass subject to the elastic force of a spring. Newton s law: F kx a. d x kx 0. dt d x k x 0. dt Coparing with the equation of otion for siple haronic otion, k. Siple haronic otion is the otion executed by a particle of ass subject to a force that is proportional to the displaceent of the particle but opposite in sign. Angular frequency: Period: Since T, T k.. k
Exaples 15-1 A block whose ass is 680 g is fastened to a spring whose spring constant k is 65 N -1. The block is pulled a distance x = 11 c fro its equilibriu position at x = 0 on a frictionless surface and released fro rest at t = 0. (a) What are the angular frequency, the frequency, and the period of the resulting oscillation? (b) What is the aplitude of the oscillation? (c) What is the axiu speed of the oscillating block? (d) What is the agnitude of the axiu acceleration of the block? (e) What is the phase constant for the otion? (f) What is the displaceent function x(t)? k 1 (a) 65 9.78 rads (ans) 0.68 f 1.56 Hz (ans) 1 T 0.643s (ans) f (b) (c) x 11c (ans) 1 v x (9.78)(0.11) 1.08 s (ans) x (9.78) (0.11) 10.5 s (d) a (ans) (e) At t = 0, x ( 0) x cos 0.11 (1) v ( 0) x sin 0 () (): sin 0 0 (ans) (f) x( t) x cos( t ) 0.11cos(9.78t) (ans)
15- At t = 0, the displaceent of x(0) of the block in a linear oscillator is 8.50 c. Its velocity v(0) then is 0.90 s 1, and its acceleration a(0) is +47.0 s. (a) What are the angular frequency? (b) What is the phase constant and aplitude x? (a) x( t) x cos( t ) ( t) x sin( t ) v a( t) x cos( t ) At t = 0, x ( 0) x cos 0.085 (1) v ( 0) x sin 0.90 () x a (0) cos 47.0 (3) a(0) (3) (1): x(0) a( 0) 47.0 1 3.5 rads (ans) x(0) 0.0850 v(0) sin (c) () (1): tan x(0) cos v(0) 0.90 tan 0.4603 x(0) (3.51)( 0.085) o 4.7 or o o 180 4.7 o 155 x(0) (1): x cos If = 4.7 o 0.085, x o cos 4.7 0.094 9.4 c If = 155 o 0.085, x 0.094 9.4 c o cos155 Since x is positive, = 155 o and x = 9.4 c. (ans)
Energy in Siple Haronic Motion Potential energy: Since ( t) x cos( t ), x Kinetic energy: 1 1 U( t) kx kx cos Since ( t) x sin( t ), v Since = k/, Mechanical energy: 1 1 K( t) v x sin 1 K( t) kx sin ( t ). ( t ). ( t ). E U K 1 cos sin kx 1 ( t ) kx 1 kx [cos ( t ) sin ( t )]. ( t ) Since cos (t + ) + sin (t + ) = 1, E U K 1 kx.
The echanical energy is conserved.
15-3 Suppose the daper of a tall building has ass =.7 10 5 kg and is designed to oscillate at frequency f = 10 Hz and with aplitude x = 0 c. (a) What is the total echanical energy E of the daper? (b) What is the speed of the daper when it passes through the equilibriu point? See Youtube Discovery Channel Taipei 101 (3/5) and Taipei 101 Daper (a) k ( f ) 5 9 (.710 )(0 ) 1.07310 N The energy: E K U 1 v 1 kx 1 9 0 (1.07310 )(0.) 7.14710 J 15. MJ (ans) (b) Using the conservation of energy, 1 1 E K U v kx 7 1 5.147 10 (.710 ) v 0 1 v 1.6 s (ans)
An Angular Siple Haronic Oscillator When the suspension wire is twisted through an angle, the torsional pendulu produces a restoring torque given by. is called the torsion constant. Using Newton s law for angular otion, I, d I, 0. dt I Coparing with the equation of otion for siple haronic otion,. I Period: Since T, T I.
Exaple 15-4 A thin rod whose length L is 1.4 c and whose ass is 135 g is suspended at its idpoint fro a long wire. Its period T a of angular SHM is easured to be.53 s. An irregularly shaped object, which we call X, is then hung fro the sae wire, and its period T b is found to be 4.76 s. What is the rotational inertia of object X about its suspension axis? Rotational inertia of the rod about the center 1 Ia ML 1 1 (0.135)(0.14) 1 1.79810 4 kg Ia Since Ta and Ta I a Tb Ib Therefore, T b Ib, we have T b Ib Ia Ta 4.76 4 4 (1.7310 ) 6.110 kg (ans).53
The Siple Pendulu The restoring torque about the point of suspension is = g sin L. Using Newton s law for angular otion, = I, g sin L L, d g sin 0. dt L When the pendulu swings through a sall angle, sin. Therefore d g 0. dt L Coparing with the equation of otion for siple haronic otion, g. L Period: Since T, T L. g
The Physical Pendulu The restoring torque about the point of suspension is = g sin h. Using Newton s law for angular otion, = I, d gh gsin h I, sin 0. dt I When the pendulu swings through a sall angle, sin. Therefore d gh 0. dt I Coparing with the equation of otion for siple haronic otion, gh. I Period: Since T, T I. gh
If the ass is concentrated at the center of ass C, such as in the siple pendulu, then T I gh L gl L. g We recover the result for the siple pendulu. Exaples 15-5 A eter stick, suspended fro one end, swings as a physical pendulu. (a) What is its period of oscillation T? (b) A siple pendulu oscillates with the sae period as the stick. What is the length L 0 of the siple pendulu? Rotational inertia of a rod about one end 1 ML 3 Period T L / 3 gl / I gh L (ans) 3g (b) For a siple pendulu of length L 0, L0 T g L0 L L 0 L 66.7 c (ans) g 3g 3
15-6 A diver steps on the diving board and akes it ove downwards. As the board rebounds back through the horizontal, she leaps upward and lands on the free end just as the board has copleted.5 oscillations during the leap. (With such tiing, the diver lands when the free end is oving downward with greatest speed. The landing then drives the free end down substantially, and the rebound catapults the diver high into the air.) Modeling the spring board as the rod-spring syste (Fig. 15-1(d)), what is the required spring constant k? Given = 0 kg, diver s leaping tie t fl = 0.6 s. See Youtube Guo Jingjing.
Daped Siple Haronic Motion The liquid exerts a daping force proportional to the velocity. Then, F d bv, b = daping constant. Using Newton s second law, bv kx a. d x dx b kx 0. dt dt Solution: bt/ x( t) x e cos( ' t ), where k b 4 ' If b = 0, reduces to k / of the undaped oscillator. If b k, then. bt/ The aplitude, x( t) x e, gradually decreases with tie. The echanical energy decreases exponentially with tie. 1 bt/ E( t) kx e..
Exaple 15-7 For the daped oscillator with = 50 g, k = 85 N 1, and b = 70 gs 1. (a) What is the period of the otion? (b) How long does it take for the aplitude of the daped oscillations to drop to half its initial value? (c) How long foes it take for the echanical energy to drop to half its initial value? 0.5 (a) T 0.34 s (ans) k 85 (b) When the aplitude drops by half, bt/ 1 x e x e bt/ 1 bt 1 Taking logarith, ln ln ln ()(0.5)(ln ) t 4.95 s (ans) b 0.07 (c) When the energy drops by half, 1 bt/ 1 1 kx e kx bt/ 1 e bt 1 Taking logarith, ln ln ln (0.5)(ln ) t.48 s (ans) b 0.07
Forced Oscillations and Resonance When a siple haronic oscillator is driven by a periodic external force, we have forced oscillations or driven oscillations. Its behavior is deterined by two angular frequencies: (1) the natural angular frequency () the angular frequency d of the external driving force. The otion of the forced oscillator is given by x ( t ) x cos( t ). (1) It oscillates at the angular frequency d of the external driving force. () Its aplitude x is greatest when. d This is called resonance. See Youtube Tacoa Bridge Disaster. d