14 1 Solve The frequency generated by a guitar string is 440 Hz The period is the inverse of the frequency hence T 1 f 1 440 Hz 2 27 10 s 3 2 27 ms 14 2 Model The air track glider oscillating on a spring is in simple harmonic motion Solve The glider completes 10 oscillations in 33 s and it oscillates between the 10 cm mark and the 60 cm mark a T b 3 3 s oscillation 33 s 10 oscillations 1 1 3 3 s T 2 2 d The oscillation from one side to the other is equal to 60 cm 10 cm 50 cm 0 50 m A 1 90 rad s 0 303 Hz 0 303 Hz 0 30 Hz 3 3 s c f f 1 2 0 50 m 0 25 m e The maximum speed is v max A A 1 90 rad s 0 25 m 0 48 m s 2 T Thus the amplitude is 14 3 Model The air track glider attached to a spring is in simple harmonic motion cos Visualize The position of the glider can be represented as Solve The glider is pulled to the right and released from rest at T 40 cm s 0 40 m s 0 s t t 2 0 s x t A It then oscillates with a period and a maximum speed max 2 T and A v 2 2 0 s t b The glider s position at 0 25 s is a v max rad s A 0 127 m 12 7 cm v max 0 40 m s rad s x 0 25 s 0 127 m cos rad s 0 25 s 0 090 m 9 0 cm 14 4 Model The oscillation is the result of simple harmonic motion Visualize Please refer to Figure EX14 4 Solve b The time to complete one cycle is the period hence a The amplitude 10 cm and 2 0 s T A f 0 50 Hz 1 T 1 2 0 s c The position of an object undergoing simple harmonic motion is x t A cos 0 t At t 0 s x 0 5 cm thus 5 cm 10 cm cos 0 s 0 cos 0 0 cos 1 rad or 120 5 cm 10 cm 1 2 2 3 1 2 Since the oscillation is originally moving to the left 120 0 14 5 Model The oscillation is the result of simple harmonic motion Visualize Please refer to Figure EX14 5 Solve b The period a The amplitude T A thus 20 cm 4 0 s f 0 25 Hz 1 T 1 4 0 s c The position of an object undergoing simple harmonic motion is x t A cos 0 t At t 0 s x 0 10 cm Thus 10 cm 20 cm cos 0 0 cos 1 10 cm 20 cm cos 1 rad 60 1 2 3 Because the object is moving to the right at t 0 s it is in the lower half of the circular motion diagram and thus must have a phase constant between and 2 radians Therefore rad 60 0 3 14 6 Visualize The phase constant 2 3 has a plus sign which implies that the object undergoing simple harmonic motion is in the second quadrant of the circular motion diagram That is the object is moving to the left Solve The position of the object is x t A cos t 0 A cos 2 ft 0 4 0 cm cos 4 rad s t 2 3 rad The amplitude is A 4 cm and the period is T 1 f 0 50 s A phase constant 2 3 rad 120 second 0 quadrant means that x starts at and is moving to the left getting more negative 1 2 A Assess We can see from the graph that the object starts out moving to the left 14 7 Visualize A phase constant of 2 in the lower half of the circular motion diagram That is the object is moving to the right Solve The position of the object is given by the equation implies that the object that undergoes simple harmonic motion is x t A cos t 0 A cos 2 ft 0 8 0 cm cos rad s t rad 2 2 0 A The amplitude is moving to the right getting more positive 8 0 cm and the period is T 1 f 4 0 s With 2 rad x starts at 0 cm and is Assess As we see from the graph the object starts out moving to the right 14 8 Solve The position of the object is given by the equation x t A cos t 0 A cos 2 ft 0 We can find the phase constant 0 from the initial condition 0 cm 4 0 cm cos 0 cos 0 0 0 cos 1 0 rad 1 2 Since the object is moving to the right the object is in the lower half of the circular motion diagram Hence 0 The final result with 4 0 Hz rad f is 1 2 x t 4 0 cm cos 8 0 rad s t 1 2 rad 14 9 Solve The position of the object is given by the equation x t A cos 0 t The amplitude A 8 0 cm The angular frequency 2 f 0 50 Hz rad s Since at t it has its 0 2 most negative velocity it must be at the equilibrium point x 0 cm and moving to the left so 0 Thus 2 x t 8 0 cm cos t rad s rad 2 14 10 Model The air track glider is in simple harmonic motion Solve a We can find the phase constant from the initial conditions for position and velocity Dividing the second by the first we see that x 0 A cos 0 v 0 x sin 0 A sin cos 0 0 tan 0 xv 0 x 0 The glider starts to the left x 0 2 s the angular frequency is 1 5 s 3 2 T rad s Thus 4 3 5 00 cm and is moving to the right xv 0 36 3 cm s With a period of 0 tan 1 36 3 cm s 4 3 rad s 5 00 cm 1 3 rad 60 or rad 120 2 3 The tangent function repeats every 180 so there are always two possible values when evaluating the arctan function We can distinguish between them because an object with a negative position but moving to the right is in the third quadrant of the corresponding circular motion Thus or 120 rad b At time t the phase is t 0 4 3 rad s t rad 2 3 2 3 rad 0 rad 2 3 rad and 2 0 3 This gives rad at respectively t 0 s 0 5 s 1 0 s and 1 5 s This is one period of the motion 4 3 14 11 Model The block attached to the …
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