PHY 140Y FOUNDATIONS OF PHYSICS Tutorial Questions #10 Solutions November 19/20

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1 PHY 40Y FOUNDTIONS OF PHYSICS Tutrial Questins #0 Slutins Nveer 9/0 Dape an Driven Harnic Mtin, Resnance. ass f 0 g is cnnecte t a light spring having frce cnstant 5.4 N/. It is free t scillate n a hrizntal frictinless surface. (a) If the ass is isplace 7.0 c fr the equiliriu psitin an release fr t, what are the natural frequency an the peri f its tin? () What are the axiu spee an acceleratin f the ass? (c) What is the ttal energy f the syste? What are the kinetic an ptential energies f the syste when the ass has a isplaceent f.0 c fr equiliriu? () If the surface is n lnger frictinless, ut instea gives rise t a aping frce with aping cnstant.7 kg/s, is the tin f the ass unerape, critically ape, r verape? Justify yur answer. Slutin: k equiliriu psitin k initial psitin 0 L 7.0 c (a) Given: 0 g 0. kg L 7.0 c k 5.4 N/ k 5.4 N/ natural angular frequency: 4.9(5) ra/s 0. kg 4.95 ra/s natural frequency: f 0.79 Hz π π π π peri: T.3 secns 4.95 ra/s x PHY 40Y Funatins f Physics (K. Strng) Tutrial 0 Slutins, page

2 () The axiu spee ccurs at x 0, while the axiu acceleratin is at x L (the axiu extensin). pply Cnservatin f Mechanical Energy t fin the axiu spee: K + U K + U v 0 ax v v 0 ax ax L kl kl L L 4.95 ra/s /s L pply Newtn s Secn Law t fin the axiu acceleratin: F a kl a ax ax kl L (4.95 ra/s) /s (c) The ttal energy f the syste is: E kl K ax U ax (5.4 kg/s) (0.070 ) 0.03 J When the ass has a isplaceent f.0 c fr equiliriu, U(0.00 c) kx (5.4 kg/s) (0.00 ) 0.00 J K(0.00 c) [ v ] 0.03 J J 0.0 J E U(0.00 c) () In rer t eterine if the tin f the ass is unerape, critically ape, r verape, nee t cpare with..7 kg/s 0. kg 4.95 ra/s. kg/s Thus: < an s the tin is unerape, with the aplitue f the scillatin ecreasing with tie.. 50-g ass is unte n a spring with a spring cnstant f k 3.3 N/. The aping cnstant fr this syste is kg/s. Hw any scillatins will the syste unerg uring the tie it takes the aplitue t ecay t /e f its riginal value? Slutin: First, nee t check whether the syste is unerape, critically ape, r verape, y cparing with. PHY 40Y Funatins f Physics (K. Strng) Tutrial 0 Slutins, page

3 k k 0.5 kg 3.3 N/.8 kg/s With kg/s, <<, an s the tin is unerape an will scillate with: x(t) t cs( t + δ) 4 Slve fr the tie t at which this happens: ske fr the nuer f scillatins efre the aplitue ecays t /e f its riginal value i.e., t exp( ) t t ( kg) kg/s s ls nee t calculate the peri fr ne scillatin: π π π T k 3.3 N/ kg π 3 (8.4 0 kg/s) 4(0.5 kg).7(3) secns Thus, the nuer f scillatins is 59.5/ penulu f length.00 is release fr an initial angle f 5.0 (a) What is the nant (angular) frequency f this penulu? () fter 000. secns, its aplitue is reuce y frictin t What is the value f /? Slutin: (a) The nant angular frequency is the sae as the natural angular frequency, an s is: g l 9.8 /s ra/s () The penulu will scillate fllwing: θ ( t) θ t cs ( t + δ) (assuing sall angles θ; nte that θ ra, while sin ra) PHY 40Y Funatins f Physics (K. Strng) Tutrial 0 Slutins, page 3

4 t t 0, the aplitue is: t t 000. secns, the aplitue is: Thus: 5.0 t t ln t 5.0 θ ( 0) θ θ ( 000.) θ t t θ ln s s 4. ass-spring syste has / /5, where is the aping cnstant an is the natural frequency. When this syste is riven at frequencies 0% ave an elw, hw es its aplitue cpare with its aplitue at? Slutin: In general, the aplitue is: t nance ( F, an s the aplitue is: The rati f the aplitue t the nant aplitue is thus: F F ( ) + F ) + F + If 5 an. (0% ave nance), then % 5 +. (. ) If 0. 9 (0% elw nance), then % 76% ( ) PHY 40Y Funatins f Physics (K. Strng) Tutrial 0 Slutins, page 4

5 5..00 kg ass attache t a spring is riven y an external frce F(t) (3.00N) cs(πt). If the frce cnstant f the spring is 0.0 N/ an aping is negligile, eterine: (a) the peri, an () the aplitue f the tin. Slutin: k 0.0 N/ (a) natural angular frequency: 3.6 ra/s.00 kg angular frequency f tin: (aping negligile) 4 angular frequency f riving frce: π ra/s The peri f the tin is calculate using the riving angular frequency, as the spring will scillate at this frequency after the transients ie ut an it reaches a steay state tin. π π peri: T.00 secns π ra/s () The aplitue f the tin is: F F (aping negligile) ( ) + ( ) Thus: F 3.00 N [( π ra/s) (3.6 ra/s) ] ( ) (.00 kg ) PHY 40Y Funatins f Physics (K. Strng) Tutrial 0 Slutins, page 5