YF_ISM_13

# YF_ISM_13 - PERIODIC MOTION 13 13.1 IDENTIFY and SET UP The...

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13-1 P ERIODIC M OTION 13.1. IDENTIFY and SET UP: The target variables are the period T and angular frequency . ω We are given the frequency f , so we can find these using Eqs.(13.1) and (13.2) EXECUTE: (a) 220 Hz f = 3 1/ 1/220 Hz 4.54 10 s Tf == = × 2 2 (220 Hz) 1380 rad/s f ωπ π = (b) 2(220 Hz) 440 Hz f 3 1/ 1/440 Hz 2.27 10 s = × (smaller by a factor of 2) 2 2 (440 Hz) 2760 rad/s f = (factor of 2 larger) EVALUATE: The angular frequency is directly proportional to the frequency and the period is inversely proportional to the frequency. 13.2. IDENTIFY and SET UP: The amplitude is the maximum displacement from equilibrium. In one period the object goes from xA =+ to =− and returns. EXECUTE: (a) 0.120 m A = (b) 0.800 s /2 T = so the period is 1.60 s (c) 1 0.625 Hz f T EVALUATE: Whenever the object is released from rest, its initial displacement equals the amplitude of its SHM. 13.3. IDENTIFY: The period is the time for one vibration and 2 T π = . SET UP: The units of angular frequency are rad/s. EXECUTE: The period is 3 0.50 s 1.14 10 s 440 and the angular frequency is 3 2 5.53 10 rad s. π ω T × EVALUATE: There are 880 vibrations in 1.0 s, so 880 Hz f = . This is equal to 1/ T . 13.4. IDENTIFY: The period is the time for one cycle and the amplitude is the maximum displacement from equilibrium. Both these values can be read from the graph. SET UP: The maximum x is 10.0 cm. The time for one cycle is 16.0 s. EXECUTE: (a) 16.0 s T = so 1 0.0625 Hz f T . (b) 10.0 cm A = . (c) 16.0 s T = (d) 2 0.393 rad/s f ωπ EVALUATE: After one cycle the motion repeats. 13.5. IDENTIFY: This displacement is 1 4 of a period. SET UP: 1/ 0.200 s . EXECUTE: 0.0500 s t = EVALUATE: The time is the same for = to 0 x = , for 0 x = to , for to 0 x = and for 0 x = to = . 13.6. IDENTIFY: Apply Eq.(13.12). SET UP: The period will be twice the interval between the times at which the glider is at the equilibrium position. EXECUTE: 2 2 2 22 (0.200 kg) 0.292 N m. 2(2.60 s) ππ k ω mm T ⎛⎞ = = ⎜⎟ ⎝⎠ EVALUATE: 2 1 N 1 kg m/s =⋅ , so 2 1 N/m 1 kg/s = . 13

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13-2 Chapter 13 13.7. IDENTIFY and SET UP: Use Eq.(13.1) to calculate T , Eq.(13.2) to calculate , ω and Eq.(13.10) for m . EXECUTE: (a) 1/ 1/6.00 Hz 0.167 s Tf == = (b) 2 2 (6.00 Hz) 37.7 rad/s f ωπ π = (c) / km = implies 22 / (120 N/m)/(37.7 rad/s) 0.0844 kg mk = EVALUATE: We can verify that 2 / k has units of mass. 13.8. IDENTIFY: The mass and frequency are related by 1 2 k f m π = . SET UP: constant 2 k fm , so 11 2 2 fm fm = . EXECUTE: (a) 1 0.750 kg m = , 1 1.33 Hz f = and 2 0.750 kg 0.220 kg 0.970 kg m =+= . 1 21 2 0.750 kg (1.33 Hz) 1.17 Hz 0.970 kg m ff m = . (b) 2 0.750 kg 0.220 kg 0.530 kg m =−= . 2 0.750 kg (1.33 Hz) 1.58 Hz 0.530 kg f EVALUATE: When the mass increases the frequency decreases and when the mass decreases the frequency increases. 13.9. IDENTIFY: Apply Eqs.(13.11) and (13.12). SET UP: f = 1/ T EXECUTE: (a) 0.500 kg 2 0.375 s 140 N/m T π . (b) 1 2.66 Hz f T . (c) 2 16.7 rad/s. ωπ f EVALUATE: We can verify that 2 1 kg/(N/m) 1 s = . 13.10. IDENTIFY and SET UP: Use Eqs. (13.13), (13.15), and (13.16). EXECUTE: 440 Hz, f = 3.0 mm, A = 0 φ = (a) cos( ) xA t =+ 3 2 2 (440 hz) 2.76 10 rad/s f = × 33 (3.0 10 m)cos((2.76 10 rad/s) ) xt × (b) sin( ) x vA t ωω =− + max (2.76 10 rad/s)(3.0 10 m) 8.3 m/s × × = (maximum magnitude of velocity) 2 cos( ) x aA t + 23 2 34 2 max (2.76 10 rad/s) (3.0 10 m) 2.3 10 m/s × × = × (maximum magnitude of acceleration) (c) 2 cos x t 3 3 / sin [2 (440 Hz)] (3.0 10 m)sin([2.76 10 rad/s] ) x da dt A t t = × × 73 3 (6.3 10 m/s )sin([2.76 10 rad/s] ) t × Maximum magnitude of the jerk is 37 3 6.3 10 m/s A EVALUATE: The period of the motion is small, so the maximum acceleration and jerk are large.
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## This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.

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YF_ISM_13 - PERIODIC MOTION 13 13.1 IDENTIFY and SET UP The...

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