411_PartUniversity Physics Solution

411_PartUniversity Physics Solution - Periodic Motion 13-3...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Periodic Motion 13-3 (b) Eq.(13.18): 00 arctan( / ) x vx φω =− / (300 N/m)/2.00 kg 12.25 rad/s km ω == = (4 . 0 0 m / s ) arctan arctan( 1.633) 58.5 (12.25 rad/s)(0.200 m) φ ⎛⎞ = + = ° ⎜⎟ ⎝⎠ (or 1.02 rad) (c) cos( ) xA t ωφ =+ gives (0.383 m)cos([12.2rad/s] 1.02 rad) xt EVALUATE: 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 in the range 0 90 < gives 0 0. x v < 13.13. IDENTIFY: For SHM, 22 (2 ) x a ω x π fx . Apply Eqs.(13.13), (13.15) and (13.16), with A and from Eqs.(13.18) and (13.19). SET UP: 1.1 cm x = , 0 15 cm/s x v . 2 f π = , with 2.5 Hz f = . EXECUTE: (a) () 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) , ( 22.9 cm s) sin ((15.7 rad/s) 0.715 rad) x vt + and 2 ( 359 cm/s ) cos ((15.7 rad/s) 0.715 rad) . x at + EVALUATE: We can verify that our equations for x , x v and x a give the specified values at 0 t = . 13.14. IDENTIFY and SET UP: Calculate x using Eq.(13.13). Use T to calculate and 0 x to calculate . EXECUTE: 0 x = at 0 t = implies that /2 rad = ± Thus cos( /2). t 2/ T πω = so 2 / 2 /1.20 s 5.236 rad/s T ππ = (0.600 m)cos([5.236 rad/s][0.480 s] /2)= 0.353 m. x The distance of the object from the equilibrium position is 0.353 m. EVALUATE: The problem doesn’t specify whether the object is moving in the x + or x direction at 0. t = 13.15. IDENTIFY: Apply 2 m T k = . Use the information about the empty chair to calculate k . SET UP: When 42.5 kg m = , 1.30 s T = . EXECUTE: Empty chair: 2 m T π k = gives 4 4 (42.5 kg) 993 N/m (1.30 s) π m π k T = With person in chair: 2 m T k = gives (2.54 s) (993 N/m) 162 kg 44 Tk m = and person 162 kg 42.5 kg 120 kg m = . EVALUATE: For the same spring, when the mass increases, the period increases. 13.16. IDENTIFY and SET UP: Use Eq.(13.12) for T and Eq.(13.4) to relate x a and k . EXECUTE: , Tm k = 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 k EVALUATE: a x is negative when x is positive. / x ma x has units of N/m and / mk has units of s. 13.17. IDENTIFY: 2 m T k = . x k ax m so max k aA m = . Fk x = − . SET UP: x a is proportional to x so x 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 . EXECUTE: (a) 2 m T k = gives 0.20 s (250 N/m) 0.253 kg T = (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 A =
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
13-4 Chapter 13 EVALUATE: We can also calculate the maximum force from the maximum acceleration: 2 max max (0.253 kg)(12.0 m/s ) 3.04 N, Fm a == = which agrees with our previous results. 13.18. IDENTIFY: The general expression for ( ) x vt is ( ) sin( ) x A t ω φ = −+ . We can determine and A by comparing the equation in the problem to the general form. SET UP: 4.71 rad/s = . 3.60 cm/s 0.0360 m/s A . EXECUTE: (a) 22 r a d 1.33 s 4.71 rad/s T ππ = (b) 3 0.0360 m/s 0.0360 m/s 7.64 10 m 7.64 mm 4.71 rad/s A === × = (c) 3 2 max (4.71 rad/s) (7.64 10 m) 0.169 m/s aA × = (d) k m = so (0.500 kg)(4.71 rad/s) 11.1 N/m km = .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 5

411_PartUniversity Physics Solution - Periodic Motion 13-3...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online