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411_PartUniversity Physics Solution

411_PartUniversity Physics Solution - Periodic Motion...

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Periodic Motion 13-3 (b) Eq.(13.18): 0 0 arctan( / ) x v x φ ω = / (300 N/m)/2.00 kg 12.25 rad/s k m ω = = = ( 4.00 m/s) arctan arctan( 1.633) 58.5 (12.25 rad/s)(0.200 m) φ = = + = ° (or 1.02 rad) (c) cos( ) x A t ω φ = + gives (0.383 m)cos([12.2rad/s] 1.02 rad) x t = + E VALUATE : At 0 t = the block is displaced 0.200 m from equilibrium but is moving, so 0.200 m. A > According to Eq.(13.15), a phase angle φ in the range 0 90 φ < < ° gives 0 0. x v < 13.13. I DENTIFY : For SHM, 2 2 (2 ) x a ω x π f x = − = − . Apply Eqs.(13.13), (13.15) and (13.16), with A and φ from Eqs.(13.18) and (13.19). S ET U P : 1.1 cm x = , 0 15 cm/s x v = − . 2 f ω π = , with 2.5 Hz f = . E XECUTE : (a) ( ) 2 2 2 2 (2.5 Hz) (1.1 10 m) 2.71 m/s . x a π = − × = − (b) From Eq. (13.19) the amplitude is 1.46 cm, and from Eq. (13.18) the phase angle is 0.715 rad. The angular frequency is 2 15.7 rad/s, π f = so (1.46 cm) cos ((15.7 rad/s) 0.715 rad) x t = + , ( 22.9 cm s) sin ((15.7 rad/s) 0.715 rad) x v t = − + and 2 ( 359 cm/s ) cos ((15.7 rad/s) 0.715 rad) . x a t = − + E VALUATE : We can verify that our equations for x , x v and x a give the specified values at 0 t = . 13.14. I DENTIFY and S ET U P : Calculate x using Eq.(13.13). Use T to calculate ω and 0 x to calculate . φ E XECUTE : 0 x = at 0 t = implies that /2 rad φ π = ± Thus cos( /2). x A t ω π = ± 2 / T π ω = so 2 / 2 /1.20 s 5.236 rad/s T ω π π = = = (0.600 m)cos([5.236 rad/s][0.480 s] /2)= 0.353 m. x π = ± The distance of the object from the equilibrium position is 0.353 m. E VALUATE : The problem doesn’t specify whether the object is moving in the x + or x direction at 0. t = 13.15. I DENTIFY : Apply 2 m T k π = . Use the information about the empty chair to calculate k . S ET U P : When 42.5 kg m = , 1.30 s T = . E XECUTE : Empty chair: 2 m T π k = gives 2 2 2 2 4 4 (42.5 kg) 993 N/m (1.30 s) π m π k T = = = With person in chair: 2 m T k π = gives 2 2 2 2 (2.54 s) (993 N/m) 162 kg 4 4 T k m π π = = = and person 162 kg 42.5 kg 120 kg m = = . E VALUATE : For the same spring, when the mass increases, the period increases. 13.16. I DENTIFY and S ET U P : Use Eq.(13.12) for T and Eq.(13.4) to relate x a and k . E XECUTE : 2 / , T m k π = 0.400 kg m = Use 2 2.70 m/s x a = − to calculate k : x kx ma = gives 2 (0.400 kg)( 2.70 m/s ) 3.60 N/m 0.300 m x ma k x = − = − = + 2 / 2.09 s T m k π = = E VALUATE : a x is negative when x is positive. / x ma x has units of N/m and / m k has units of s. 13.17. I DENTIFY : 2 m T k π = . x k a x m = − so max k a A m = . F kx = − . S ET U P : x a is proportional to x so x a goes through one cycle when the displacement goes through one cycle. From the graph, one cycle of a x extends from 0.10 s t = to 0.30 s, t = so the period is 0.20 s. T = 2.50 N/cm 250 N/m. k = = From the graph the maximum acceleration is 2 12.0 m/s . E XECUTE : (a) 2 m T k π = gives 2 2 0.20 s (250 N/m) 0.253 kg 2 2 T m k π π = = = (b) 2 max (0.253 kg)(12.0 m/s ) 0.0121 m 1.21 cm 250 N/m ma A k = = = = (c) max (250 N/m)(0.0121 m) 3.03 N F kA = = =
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13-4 Chapter 13 E VALUATE : We can also calculate the maximum force from the maximum acceleration: 2 max max (0.253 kg)(12.0 m/s ) 3.04 N, F ma = = = which agrees with our previous results. 13.18. I DENTIFY : The general expression for ( ) x v t is ( ) sin( ) x v t A t ω ω φ = − + . We can determine ω and A by comparing the equation in the problem to the general form.
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