HO 25 Solutions. = s. = 296 kg s 2. = ( kg) s. = 2π m k and T = 2π ω. kg m = m kg. = 2π. = ω 2 A = 2πf

Transcription

1 HO 5 Soution 1.) haronic ociator = g with an idea pring T = 0.00 T = π T π π o = = ( g) 0.00 = 96 g = 96 N.) haronic ociator = 0.00 g and idea pring = 140 N F = x = a = d x dt o the dipaceent i a oution to the differentia equation d x dt = x = ω x the oution i x( t) = Aco( ωt) where ω = or uing T = π and T = π ω π or ω = T = π π = ω = N 140 = 0.00 g = 6.5 N g = 6.5 g g = 6.5 T = π ω = π 6.5 = 0.4 aternativey T = π = π 0.00 g 140 N = ) ipe haronic otion with apitude A = 0.18 and frequency f = 6.00 Hz x( t) = Aco( ωt) and ω = πf v( t) = dx dt = d dt a( t) = dv dt = d dt ( Aco( ωt) ) = ωain( ωt) o v ax = ωa = πfa = π ( )( ) = 6.79 ( ωain( ωt) ) = ω Aco( ωt) o a ax = ω A = πf ( ( )) ( ) A = π ) haronic ociator apitude A and anguar frequency ω, x( t) = Aco( ωt) and v( t) = ωain( ωt) ( ) = 56 inetic energy i K = 1 v = 1 ( ωain ( ωt )) = 1 ω A in ( ωt) potentia energy i U = 1 x = 1 ( Aco ( ωt )) = 1 A co ( ωt) K = U when 1 ω A in ( ωt) = 1 A co ( ωt) ao ω = o ω = or = ω o 1 A in ( ωt) = 1 A in ( ωt) = 1 A co ( ωt) are equa when ωt = ( n +1) π 4 (odd utipe of π 4 ) o x = Aco n +1 ( ) π 4 = ± A and v = ωain n +1 ( ) π 4 = ωa

2 4.) (continued) occur four tie each cyce when ωt = π 4, 3π 4, 5π 4, and 7π 4 tie between occurrence i when ωδt = π π π or Δt = and ince ω = ω T Δt = π π = T 4 T 5.) boc = 3.00 g upended fro idea pring tretching it Δx = 0.00 F = Δx and the force i provided by the weight of the boc o F g = g = Δx o = g 3.00 g Δx = 0.00 ( ) 9.8 = 147 N for ipe haronic otion T = π = π 3.00 g 147 N = ) ipe penduu ae 100 copete wing in 55.0 where g = 9.8 T = = 0.55 and T p = π g gt o = 4π = 4π ( ) = ) on Earth ipe penduu T E = 1.60, and paced on Moon where g Moon = 1.6 on Earth T E = π g E o on Moon T Moon = π o = g T E E 4π g Moon = π g E T E 4π g Moon = T E g E g Moon = ( 1.60 ) = ) ipe penduu = 0.55 defected 7 highet peed i at the botto of it wing or one-fourth of it period or when ωt = π and t = π ω = T = π 0.55 = π g 9.8 = 1.49 and t = T 4 = 1.49 = π π = T 4 T 9.) = g ipe haronic otion on a horizonta pring = 400 N when x = 0.01, v = E = K +U = 1 v + 1 x = g ( ) N 0.01 dipaceent i a axiu when v = 0 and a of the energy i potentia ( ) = J E = U = 1 x ax and the axiu dipaceent i x ax = E = A = ( J) = N

3 9.) (continued) v( t) = ωain( ωt) o v ax = ωa = A = ( ) 400 N g = aternativey when the peed i a axiu a energy i inetic E = K = 1 v ax and v ax = E = ( J) g = ) ipe penduu = 0.50 g and = 1.00 dipace θ = 15 A = inθ = ( 1.00 )in15 = 0.6 θ T = π g and ω = π T = g A v( t) = ωain( ωt) o v ax = ωa = A g = ( 0.6 ) = 0.81 a( t) = ω Aco( ωt) o a ax = ω A = A g = ( 0.6 ) 9.8 = and α ax = a ax = =.55 uing Newton nd Law F = a = 0.50 g ( ).55 = 0.64 N 11.) phyica penduu f = Hz, =.0 g, d = τ = α = d θ dt and ince τ = rfinθ = dginθ cobining thee dginθ = d θ dt for a ange inθ θ o dgθ = d θ dt or d θ dt = gd θ = ω θ and the otion i ipe haronic therefore for a phyica penduu ω = gd and T = π ω = π gd 1 ince ω = πf it foow that = gd =.0 g πf ( ) ( ) = g π ( ) 1.) phyica penduu thin rod L = 1.0, M = 0.40 g, axi i d = 0.0 fro end for a thin rod c = 1 1 ML ditance fro end to center of a d 1 = 0.5 ditance fro axi to center of a d = 0.3 uing parae axi theore = c + Md = 1 1 ML + Md = M 1 1 L + d d d d 1

4 1.) (continued) T = π T = 1 f Mgd = π M 1 1 L + d = π Mgd o f = 1 T = 1 = 0.65 Hz L + d = π gd ( ) + ( 0.3 ) ( ) = ) ipe penduu =.3, = 6.74 g given an initia peed v ax =.06 at it equiibriu point T = π g = π = 3.00 at equiibriu point a energy i inetic o E = K = 1 v ax = g ( ).06 = 14.3 J a( t) = ω Aco( ωt) o a ax = ω A and v( t) = ωain( ωt) o v ax = ωa T = π ω π o ω = T = g and a = ω g ax A = ( ωa)ω = v ax ω = v ax o a ax = = 4.3 v( t) = ωain( ωt) o v ax = ωa o A = v ax ω = v ax g θ ax = in 1 A = in 1 v ax g = v ax g and A = inθ ax = in 1 v ax g = 6 14.) vertica pring ength = when boc = 0.0 g upended fro it pring ength i = o the pring i tretched Δy = = = F g = g = Δy o = g Δy = ( 0.0 g) 9.8 = 196 N paced on horizonta urface and dipaced o the pring i = 0.10 ong and reeaed undergoing ipe haronic otion with an apitude A = = = ω = πf o f = ω π = π = 196 N 0.0 g π = 4.98 Hz

5 14.) (continued) when pring i = ong it i at it equiibriu point and the veocity v = v ax v( t) = ωain( ωt) o v ax = ωa = A = ( ) 196 N 0.0 g = a( t) = ω Aco( ωt) o a ax = ω A = A = ( ) 196 N = g energy i conerved and i a potentia energy when dipaceent i a axiu E = U = 1 x ax = N ( ) = 0.45 J aternativey, the energy i a inetic energy when veocity i a axiu E = K = 1 v ax = g ( ) = 0.45 J 15.) vertica pring tretche Δy = 0.0 when a = 0.40 g i hung fro it F g = g = Δy o = g Δy = ( 0.40 g) 9.8 = 196 N 0.00 a a = 0.0 g i attached and hed at ret when pring i not tretched and then reeaed and ociate with ipe haronic otion pring i untretched a i at axiu dipaceent uing Conervation of Energy ( v 1 = v = 0) K 1 +U g 1 +U e1 = K +U g +U e o 0 +U g1 + 0 = U e y 1 A y = 0 U g1 = U e and gy 1 = 1 Δy = 1 ( y y 1 ) gy 1 = 1 y o y 1 1 = g = A A = g ( 0.0 g) 9.8 = 196 N = o the apitude i the aount that the a woud tretch the pring when hung fro it v( t) = ωain( ωt) o v ax = ωa = A = ( ) 196 N 0.0 g = 0.313

6 16.) = 0.4 g ha poition x = (0.4 )co ( 3 π t) where t i in econd by inpection of the equation for poition and recaing that for ipe haronic otion x t A = 0.4 and ω = π 3 v( t) = ωain( ωt) o v ax = ωa = π ( ) = 0.4 ( ) = Aco( ωt) the energy i a inetic energy when veocity i a axiu E = K = 1 v ax = g ( ) 0.4 = J T = π ω = π = 6 π 3 a( t) = ω Aco( ωt) o a( 1.5 ) = π 3 ( 0.4 )co π 3 ( 1.5 ) = 0 one coud ao notice that when t = 1.5, t = T 4 where v = v ax and a = 0 and the a i oving through the equiibriu point 17.) = 1.5 g attached to a vertica pring = 300 N boc i reeaed and coe to ret and then pued down a ditance d = 0.0 and reeaed pring i untretched a i at axiu dipaceent at ret a pued down ditance d a reache ax height when v = 0 y o pring F g = Δy 1 Δy 1 = g Δy 1 y Δy A y 1 = 0 v 1 = 0 Δy 3 Δy = Δy 1 + d Δy 3 = Δy y uing Energy Conervation between the tie when the boc i reeaed and when it reache it axiu height K 1 +U g1 +U e1 = K +U g +U e and U e1 = 0 +U g +U e or U e1 = U g +U e 1 Δy = gy + 1 Δy and 1 ( Δy + d 1 ) = gy + 1 Δy y g + g d + d = gy + g ( ) or 1 g + gd + d = gy + 1 g + d y gy dy + d + y

7 g + gd + d = gy + g + gd gy dy + d + y 17.) (continued) 0 = dy + y o 0 = y ( d + y ) and the root are y = 0 and y = d o y = d = A and A = d = 0.00 the aount the boc wa pued down for ipe haronic otion y( t) = Aco( ωt) and for pring ociator ω = o ω = 300 N 1.5 g = 14.1 and y( t) = 0.00 ( ) ( )co ( )t axiu aount of tretch i when boc i at owet point and pring i tretched Δy = Δy 1 + d or Δy = g + d = ( 1.5 g) N = = iniu aount of tretch i when boc i at highet point and pring i tretched Δy 3 = Δy y or Δy 3 = Δy 1 + d d = g d = = 0.09 o the boc ociate ±A fro it equiibriu point which the aount the boc tretche the pring after coing to ret Soe genera coent on the two pecia cae for vertica pring ociation (1) n probe 15, the a i reeaed fro an untretched pring and the apitude of the ociation wa found to be A = g which i the aount the a tretche the pring when at ret. The range of vaue for the pring i bounded by Δy = 0 and Δy = g during the ociatory otion. () n probe 17, the a wa pued down a ditance d fro the point in which the a tretche the pring when at ret. The apitude of the ociation in thi cae wa the ditance d that boc wa pued down and the equiibriu point i when the boc return to point where the a tretche the pring when at ret. The aount of tretch in the pring in thi cae i bounded by Δy = g ± d during the ociatory otion.