Physics 20 Lesson 28 Simple Harmonic Motion Dynamics & Energy

Transcription

1 Phyic 0 Leon 8 Siple Haronic Motion Dynaic & Energy Now that we hae learned about work and the Law of Coneration of Energy, we are able to look at how thee can be applied to the ae phenoena. In general, we ue the ethod of dynaic to ole for force and acceleration, and we ue energy coneration when we are aked to ole for peed, diplaceent and height. In thi leon we will reiit iple haronic otion (pendulu and ibrating horizontal pring) and will apply our knowledge of dynaic and energy coneration. I. The Pendulu Dynaic We can undertand the otion of a pendulu a an application of the dynaic or a an energy yte. Conider a pendulu a (alo called a bob ) that i initially hanging in the equilibriu poition. Note that F g equal F T and the net force i zero. In addition, the total echanical energy for the pendulu i alo zero (i.e. no potential and no kinetic energy). F T F g We then oe the a fro it equilibriu poition through an angle and then releae it. A the force diagra to the right indicate, F T no longer oppoe F g. Recall fro our work with inclined plane in Leon 17 that on an incline the noral force act at an angle to the force of graity. We broke up the force due to graity (F g ) into two coponent (F and F y ). We do the ae thing here and reole the force of graity into coponent. Note that the perpendicular coponent of graity (F y ) i equal and oppoite the tenion force (F T ). The reult i a retoring force (F ) that act toward the equilibriu point. F F in g Therefore, the bob accelerate toward the equilibriu point and the bob gain peed. But the acceleration i not unifor a the angle becoe aller the retoring force becoe aller until at the equilibriu point the acceleration i zero. At the equilibriu point the bob ha reached it aiu peed. A it pae through the equilibriu point, the retoring force now act to low the bob down. (For a detailed decription of the dynaic of a pendulu wing ee Pearon page 360.) F T F g F y F Dr. Ron Licht 8 1

2 It i ery iportant to note that ince the acceleration i not unifor one can not ue equation like f i a or f i a d t to calculate peed. Such equation aue unifor acceleration. To deal with nonunifor acceleration require the ue of either integral calculu, which you will learn about in Matheatic 31, or you can apply the principle of coneration of echanical energy. Eaple 1 Deterine the acceleration of a 0.50 kg pendulu bob a it pae through an angle of 16 o to the right of the equilibriu point. Firt calculate the retoring force. g F F in left F gin left F (0.50kg)(9.81 )in16 left F 0.676N left Calculate the acceleration uing Newton nd Law F T F g F y FNET a 0.676Nleft a 0.50kg a.70 left F Dr. Ron Licht 8

3 II. The Pendulu Mechanical energy We can alo undertand the otion of a pendulu a an eaple of the coneration of echanical energy. When a pendulu i pulled to the ide (poition A), the pendulu ha graitational potential energy i.e. point A i higher than B. When the pendulu i releaed, the graitational potential energy i tranfored into kinetic energy until all of the graitational potential energy i conerted into kinetic energy at the botto of the wing (poition B). The pendulu keep winging through the equilibriu poition, loing kinetic energy a it regain graitational potential energy. Auing that there i no lo to heat or friction, it will rie to a point equal to the original height aboe the equilibriu poition (poition C). At the top of the wing the pendulu i intantaneouly at ret (E k = 0) and all of the energy i in graitational potential for. Then it will begin to wing back toward the equilibriu poition. The figure below i a graph of the changing graitational potential and kinetic energie of a pendulu bob during one-half of an ocillation. Note that the u or total (E p + E k ) of the kinetic and graitational potential energie i contant. E p + E k E p E k Throughout it otion a pendulu ha a cobination of kinetic and graitational potential energie. Thu at any point the total echanical energy of the pendulu i equal to: E total = E p + E k E total = g h + ½ At the top of the wing (h a ) the pendulu ha no kinetic energy (E k = 0). It ha graitational potential energy only. Thu E total = E p a = g h a At the botto of a wing ( a ) the total echanical energy i kinetic only. Thu E total = E k a = ½ a (For a different decription of the echanical energy of a pendulu ee Pearon page 314 to 316.) Dr. Ron Licht 8 3

4 Eaple A.0 kg a on a tring i pulled o that the a i 1.74 higher than at the equilibriu point. When it i releaed: A. What i the aiu peed that the pendulu will hae during it wing? E E E 1 1 total k() p() gh gh gh (9.81 )(1.74) 5.0 B. What i the peed of the pendulu when it i half way down fro it aiu height? Etotal Ep() Ep Ek 1.74 h gh gh 1 1 gh gh (gh gh ) (gh gh ) ( ) Dr. Ron Licht 8 4

5 III. Horizontal Vibrating Spring Dynaic Like the pendulu that we dicued aboe, we can undertand the otion of a apring yte a an application dynaic. Conider a horizontal a-pring yte a hown to the right. If the a-pring yte i copreed to the left a retoring force i produced to the right reulting in an acceleration toward the equilibriu point. The acceleration ay be calculated uing a cobination of Hooke Law and Newton nd Law of otion. F and a are (+) i (-) equilibriu point F NET F a k k a A the a oe toward the equilibriu point, the retoring force and acceleration decreae while the peed increae. A the a oe through the equilibriu point, no copreion or tenion eit in the pring. Therefore, the retoring force and acceleration are zero when the peed i at a aiu. F and a are 0 i 0 equilibriu point Once the a pae beyond the equilibriu point the pring begin to tretch reulting in a negatie retoring force and acceleration toward the equilibriu point. The a low down until it coe to a top and then oe toward the left. (For a detailed decription of the dynaic of a a-pring yte ee Pearon page 354 and 355.) Note the negatie ign in the acceleration equation aboe. If the a i gien a poitie diplaceent (+) the reulting equilibriu point F and a are (-) acceleration i negatie, and if the a i gien a negatie diplaceent ( ) the reulting acceleration i poitie. In other word, the retoring force and therefore the reulting acceleration, i alway toward the equilibriu point of the yte. i (+) Like the acceleration of a pendulu, the acceleration of a a-pring yte i not unifor. Thu, in order to calculate the peed of the a at any gien point we need to apply the principle of coneration of echanical energy. Dr. Ron Licht 8 5

6 Eaple 3 A 0.74 kg a i ocillating on a horizontal frictionle urface attached to a pring (k = 8.1 N/). What i the a' diplaceent when it intantaneou acceleration i 4.11 / [left]? F F NET a k a k 0.74kg( N 0.36[right] [left]) IV. Horizontal Vibrating Spring Mechanical energy A ibrating horizontal a on a pring i ery iilar to a pendulu. When the pring i at ret at it equilibriu poition it will hae no energy. Howeer, diplacing the a fro the equilibriu point gie the pring potential energy. When the a i releaed, the pring potential i conerted into kinetic energy. In turn the kinetic energy i conerted into pring potential, which in turn i conerted into kinetic energy and o on. Like a pendulu, a weighted ibrating pring ha contant total energy. At any point the total energy i gien by: E total = E p + E k E total = ½ k + ½ At the end of it ibration the diplaceent i at a aiu ( a ) and the yte i intantaneouly at ret. Thu E total = E p a = ½ k a At the equilibriu point ( = 0) the total echanical energy i kinetic only ( a ). Thu Eaple 4 E total = E k a = ½ a A 0.40 kg a ibrate at the end of a horizontal pring along a frictionle urface. If the pring force contant i 55 N/ and the aiu diplaceent (aplitude) i 15 c, what i the aiu peed of the a a it pae through the equilibriu point? E E E total k() p() 1 1 k 1.76 k 55 N (0.15 ) 0.40kg Dr. Ron Licht 8 6

7 V. Practice proble 1. A pendulu bob ( = 50.0 g) eperience a retoring force of N. Through what angle i it diplaced? (11.0 o ). A 0.50 kg pendulu i pulled back and releaed. When the height i 0.60 aboe it equilibriu poition it peed i 1.9 /. What i the aiu height of the pendulu? (0.78 ) 3. A.60 g a eperience an acceleration of 0.0 / at a diplaceent of on a pring. What i k for thi pring? ( N/) 4. A a ibrate at the end of a horizontal pring (k = 15 N/) along a frictionle urface reaching a aiu peed of 7.0 /. If the aiu diplaceent of the a i 85.0 c, what i the a? (1.84 kg) Dr. Ron Licht 8 7

8 VI. Hand-in aignent 1. Why i the acceleration of a iple haronic ocillator not unifor?. Decribe the poition that a a-pring yte and pendulu are in when: (a) acceleration i a aiu (b) elocity i a aiu (c) retoring force i aiu 3. Deterine the retoring force of a pendulu that i pulled to an angle of 1.0 left of the ertical. The a of the bob i g. (0.61 N [right]) 4. At what angle ut a pendulu be diplaced to create a retoring force of 4.00 N on a bob with a a of g? (54.6 ) 5. A 1.35 kg pendulu bob i hung fro a 3.0 tring. When the pendulu i in otion it peed i.40 / when it height i 85.0 c aboe the equilibriu point. What i the aiu peed of the pendulu? What i the period of ibration for the pendulu? (4.74 /, 3.59 ) 6. A.50 kg pendulu bob i hung fro a.75 tring. The pendulu i pulled back and releaed fro a height h aboe the equilibriu point. If it aiu peed i 6.6 / what i h? What i the period of ibration for the pendulu? (.00, 3.33 ) 7. Tarzan wing on a 30.0 long ine initially inclined at an angle of 37.0 fro the ertical. What i hi peed at the botto of the wing if he doe the following? a. tart fro ret (10.9 /) b. puhe off with a peed of 4.00 / (11.6 /) 8. For each ituation below, draw accurate free-body diagra howing all force acting on the rock. a. Supended fro a pring. Pulled downward lightly and releaed. No friction. b. Supended fro a pring. Intantaneouly at ret at the top of it trael. Dr. Ron Licht 8 8

9 c. Supended fro a pring. Moing downward through the equilibriu poition. No friction. d. Supended fro a pring. Moing upward through the equilibriu poition. No friction. 9. A a of 3.08 kg ocillate on the end of a horizontal pring with a period of What acceleration doe the a eperience when it diplaceent i.85 to the right? ( / [left]) 10. A 4.0 kg a ibrate at the end of a horizontal pring along a frictionle urface reaching a aiu peed of 5.0 /. If the aiu diplaceent of the a i 11.0 c, what i the pring contant? ( N/) 11. An object ibrate at the end of a horizontal pring, pring contant = 18 N/, along a frictionle horizontal urface. If the aiu peed of the object i 0.35 / and it aiu diplaceent i 0.9, what i the peed of the object when the diplaceent i 0.0? (0.5 /) 1. A kg object ibrate at the end of a horizontal pring (k = 995 N/) along a frictionle urface. The peed of the object i.50 / when it diplaceent i What i the aiu diplaceent and peed of the object? (0.160, 5.84 /) 13. A a of.50 kg i attached to a horizontal pring and ocillate with an aplitude of The pring contant i 40.0 N/. Deterine: a) the acceleration of the a when it i at a diplaceent of ( 4.80 / ) b) the aiu peed (3.0 /) c) the period (1.57 ) 14. A horizontal a-pring yte ha a a of 0.00 kg, a aiu peed of /, and an aplitude of What i the a' poition when it acceleration i 3.58 / to the wet? ( [eat]) 15. An object ibrate at the end of a horizontal pring, k = 18 N/, along a frictionle horizontal urface. If the aiu peed of the object i 0.35 / and it aiu diplaceent i 0.9, what i the peed of the object when the diplaceent i 0.0? (0.5 /) Dr. Ron Licht 8 9