Ondas-Período-Frequência-e-Movimento-Harmônico-Simples-Exercícios-Resolvidos

Copyright 2016 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 141 141 IDENTIFY We want to relate the characteristics of various waves such as the period frequency and angular frequency SET UP The frequency f in Hz is the number of cycles per second The angular frequency is 2 f and has units of radians per second The period T is the time for one cycle of the wave and has units of seconds The period and frequency are related by 1 T f EXECUTE a 3 1 1 215 10 s 466 Hz T f 3 2 2 466 Hz 293 10 rads f b 4 6 1 1 200 10 Hz 500 10 s f T 5 2 126 10 rads f c 2 f so f ranges from 15 14 27 10 rads 43 10 Hz 2 rad to 15 14 47 10 rads 75 10 Hz 2 rad 1 T f so T ranges from 15 14 1 13 10 s 75 10 Hz to 15 14 1 23 10 s 43 10 Hz d 7 6 1 1 20 10 s 50 10 Hz T f and 6 7 2 2 50 10 Hz 31 10 rads f EVALUATE Visible light has much higher frequency than either sounds we can hear or ultrasound Ultrasound is sound with frequencies higher than what the ear can hear Large f corresponds to small T 142 IDENTIFY and SET UP The amplitude is the maximum displacement from equilibrium In one period the object goes from x A to x A and returns EXECUTE a 0 120 m A b 0 800 s T 2 so the period is 160 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 143 IDENTIFY The period is the time for one vibration and 2 T SET UP The units of angular frequency are rads EXECUTE The period is 3 0 50 s 1 14 10 s 440 and the angular frequency is 3 2 5 53 10 rads T EVALUATE There are 880 vibrations in 10 s so 880 Hz f This is equal to 1 T PERIODIC MOTION 14 142 Chapter 14 Copyright 2016 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 144 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 100 cm The time for one cycle is 160 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 rads f EVALUATE After one cycle the motion repeats 145 IDENTIFY This displacement is 1 4 of a period SET UP 1 0 250 s T f EXECUTE t 0 0625 s EVALUATE The time is the same for x A to 0 x for 0 x to x A for x A to 0 x and for 0 x to x A 146 IDENTIFY and SET UP Note that 900 beats per minute is the frequency of the wing beat which we will convert into hertz We will use the relationship between frequency and period and we will also use the relationship between frequency and angular frequency 1 and 2 T f f respectively EXECUTE a 1 0 00111 min 0 067 s 900 beatsmin T b f 900 beats min 1 min 60 s 15 Hz c 2 f 94 rads EVALUATE We could have calculated the answer to part b first and then used 1 T f to find the period 147 IDENTIFY and SET UP The period is the time for one cycle A is the maximum value of x EXECUTE a From the figure with the problem T 0 800 s b 1 1 25 Hz f T c 2 7 85 rads f d From the figure with the problem 3 0 cm A e 2 m T k so 2 2 2 2 2 40 kg 148 Nm 0 800 s k m T EVALUATE The amplitude shown on the graph does not change with time so there must be little or no friction in this system 148 IDENTIFY Apply 1 2 2 m T f k 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 2 2 0 200 kg 0 292 N m 22 60 s k m m T EVALUATE 2 1 N 1 kg ms so 2 1 Nm 1 kgs 149 IDENTIFY and SET UP Use T 1f to calculate T 2 f to calculate and k m for m EXECUTE a 1 16 00 Hz 0 167 s T f b 2 2 6 00 Hz 37 7 rads f c ω km implies m kω2 120 Nm377 rads2 00844 kg EVALUATE We can verify that kω2 has units of mass 1410 IDENTIFY The mass and frequency are related by f 12π km SET UP fm k2π constant so f1m1 f2m2 EXECUTE a m1 0750 kg f1 175 Hz and m2 0750 kg 0220 kg 0970 kg f2 f1 m1m2 175 Hz0750 kg0970 kg 154 Hz b m2 0750 kg 0220 kg 0530 kg f2 175 Hz0750 kg0530 kg 208 Hz EVALUATE When the mass increases the frequency decreases and when the mass decreases the frequency increases 1411 IDENTIFY For SHM the motion is sinusoidal SET UP xt Acosωt EXECUTE xt Acosωt where A 0320 m and ω 2πT 2π0900 s 6981 rads a x 0320 m at t1 0 Let t2 be the instant when x 0160 m Then we have 0160 m 0320 mcosωt2 cosωt2 0500 ωt2 1047 rad t2 1047 rad 6981 rads 0150 s It takes t2 t1 0150 s b Let t3 be when x 0 Then we have cosωt3 0 and ωt3 1571 rad t3 1571 rad 6981 rads 0225 s It takes t3 t2 0225 s 0150 s 00750 s EVALUATE Note that it takes twice as long to go from x 0320 m to x 0160 m than to go from x 0160 m to x 0 even though the two distances are the same because the speeds are different over the two distances 1412 IDENTIFY For SHM the restoring force is directly proportional to the displacement and the system obeys Newtons second law SET UP Fx max and f 12π km EXECUTE Fx max gives ax kxm so km axx 530 ms2 0280 m 1893 s2 f 12π km 12π 1893 s2 0692 Hz EVALUATE The period is around 15 s so this is a rather slow vibration 1413 IDENTIFY Use A x02 v0x2ω2 to calculate A The initial position and velocity of the block determine ϕ xt is given by x Acosωt ϕ SET UP cosθ is zero when θ π2 and sinπ2 1 EXECUTE a From A x02 v0x2ω2 A v0ω v0km 098 m 144 Chapter 14 Copyright 2016 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher b Since 0 0 x cos x A t requires 2 Since the block is initially moving to the left 0 0 v x and v0x Asin requires that sin 2 0 so c cos 2 sin so 0 98 m sin12 2 rads t t x t EVALUATE The x t result in part c does give 0 x at 0 t and 0 x for t slightly greater than zero 1414 IDENTIFY and SET UP We are given k m x0 and 0 v Use 2 2 2 2 2 0 0 0 0 x x A x v x mv k 0 0 arctan v x x and cos x A t EXECUTE a 2 2 2 2 2 0 0 0 0 x x A x v x mv k 2 2 0 200 m 2 00 kg 4 00 ms 300 Nm 0 383 m A b 0 0 arctan v x x 300 Nm2 00 kg 12 25 rads k m 400 ms arctan arctan 1633 585 or102 rad 1225 rads0200 m c cos x A t gives 0 383 mcos12 2rads 1 02 rad x t EVALUATE At 0 t the block is displaced 0200 m from equilibrium but is moving so 0 200 m A Since cos vx A t a phase angle in the range 0 90 gives 0 0 v x 1415 IDENTIFY For SHM 2 2 2 ax x f x Apply cos x A t cos vx A t and 2 cos ax A t with A and from 0 0 arctan x v x and 2 2 0 0 2 v x A x SET UP 1 1 cm x 0 15 cms v x 2 f with 2 5 Hz f EXECUTE a 2 2 2 2 2 5 Hz 1 1 10 m 2 71 ms ax b From 2 2 0 0 2 v x A x the amplitude is 146 cm and from 0 0 arctan x v x the phase angle is 0715 rad The angular frequency is 2 15 7 rads f so 1 46 cm cos 15 7 rads 0 715 rad x t 22 9 cms sin 15 7 rads 0 715 rad vx t and 2 359 cms cos 15 7 rads 0 715 rad ax t EVALUATE We can verify that our equations for x vx and x a give the specified values at 0 t 1416 IDENTIFY The motion is SHM and in each case the motion described is onehalf of a complete cycle SET UP For SHM cos x A t and 2 T EXECUTE a The time is half a period The period is independent of the amplitude so it still takes 270 s b 0090 m x at time 1 t 540 s T and 2 1164 rads T 1 1 cos x A t 1 cos 0500 t 1 1047 rad t and 1 08997 s t 0090 m x at time 2 t 2 cos 0500 m t 2 2094 rad t and 2 1800 s t The elapsed time is 2 1 1800 s 08997 s 0900 s t t EVALUATE It takes less time to travel from 0090 m in b than it originally did because the block has larger speed at 0090 m with the increased amplitude 1417 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 k gives 2 2 2 2 4 4 42 5 kg 993 Nm 1 30 s m k T Periodic Motion 145 Copyright 2016 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher With person in chair 2 m T k gives 2 2 2 2 2 54 s 993 Nm 162 kg 4 4 T k m and person 162 kg 42 5 kg 120 kg m EVALUATE For the same spring when the mass increases the period increases 1418 IDENTIFY and SET UP Use T 2 m k for T and x k a m x to relate x a and k EXECUTE 2 m T k 0 400 kg m Use ax 180 ms2 to calculate k x kx ma gives 2 0 400 kg 180 ms 240 Nm 0 300 m max k x so T 2 m k 2 57 s EVALUATE x a is negative when x is positive x ma x has units of Nm and m k has units of seconds 1419 IDENTIFY 2 m T k x k a m x so max k a m A F kx SET UP 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 x a extends from 0 10 s t to 0 30 s t so the period is 0 20 s T 2 50 Ncm 250 Nm k From the graph the maximum acceleration is 2 12 0 ms EXECUTE a 2 m T k gives 2 250 Nm 0 20 s 2 0 253 kg 2 2 T m k b 2 max 0 253 kg12 0 ms 0 0121 m 1 21 cm 250 Nm ma A k c max 250 Nm0 0121 m 3 03 N F kA EVALUATE We can also calculate the maximum force from the maximum acceleration 2 max max 0 253 kg12 0 ms 3 04 N F ma which agrees with our previous results 1420 IDENTIFY The general expression for vx t is sin x v t A t We can determine and A by comparing the equation in the problem to the general form SET UP 4 71 rads 3 60 cms 0 0360 ms A EXECUTE a 2 2 rad 1 33 s 4 71 rads T b 3 0 0360 ms 0 0360 ms 7 64 10 m 7 64 mm 4 71 rads A c 2 2 3 2 max 4 71 rads 7 64 10 m 0 169 ms a A d k m so 2 0 500 kg4 71 rads2 11 1 Nm k m EVALUATE The overall negative sign in the expression for x v t and the factor of 2 both are related to the phase factor in the general expression 1421 IDENTIFY Compare the specific x t given in the problem to the general form cos x A t SET UP 7 40 cm A 4 16 rads and 2 42 rad EXECUTE a 2 2 1 51 s 4 16 rads T b k m so 2 1 50 kg4 16 rads2 26 0 Nm k m c max 4 16 rads7 40 cm 30 8 cms v A d Fx kx so max 26 0 Nm0 0740 m 1 92 N F kA 146 Chapter 14 Copyright 2016 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher e x t evaluated at 1 00 s t gives 0 0125 m x sin 30 4 cms vx A t 2 2 0 216 ms ax kx m x f 260 Nm 00125 m 0325 N Fx kx EVALUATE The maximum speed occurs when 0 x and the maximum force is when x A 1422 IDENTIFY The frequency of vibration of a spring depends on the mass attached to the spring Differences in frequency are due to differences in mass so by measuring the frequencies we can determine the mass of the virus which is the target variable SET UP The frequency of vibration is 1 2 k f m EXECUTE a The frequency without the virus is s s 1 2 k f m and the frequency with the virus is s v s v 1 2 k f m m s v s s s s v s v v s 1 1 2 2 1 f k m m f m m k m m m m b 2 s v s v s 1 1 f f m m Solving for v m gives 2 2 15 16 15 s v s 14 s v 200 10 Hz 1 210 10 g 1 999 10 g 287 10 Hz f m m f or v 999 femtograms m EVALUATE When the mass increases the frequency of oscillation increases 1423 IDENTIFY and SET UP Use cos x A t sin vx A t and 2 cos ax A t EXECUTE 440 Hz f 3 0 mm A 0 a cos x A t 3 2 2 440 Hz 2 76 10 rads f 3 3 3 0 10 mcos2 76 10 rads x t b sin vx A t 3 3 max 2 76 10 rads3 0 10 m 8 3 ms v A maximum magnitude of velocity 2 cos ax A t 2 3 2 3 4 2 max 2 76 10 rads 3 0 10 m 2 3 10 ms a A maximum magnitude of acceleration c 2 cos ax A t 3 3 3 3 sin 2 440 Hz 3 0 10 msin2 76 10 rads x da dt A t t 7 3 3 6 3 10 ms sin 2 76 10 rads t Maximum magnitude of the jerk is 3 7 6 3 10 ms3 A EVALUATE The period of the motion is small so the maximum acceleration and jerk are large 1424 IDENTIFY The mechanical energy of the system is conserved The maximum acceleration occurs at the maximum displacement and the motion is SHM SET UP Energy conservation gives 2 2 max 1 1 2 2 mv kA 2 m T k and max kA a m EXECUTE a From the graph we read off T 160 s and A 100 cm 0100 m 2 2 max 1 1 2 2 mv kA gives max k v A m 2 m T k so 2 k m T Therefore max 2 2 0100 m 00393 ms 160 s v A T b amax kAm 2πT2 A 2π160 s2 0100 m 00154 ms2 EVALUATE The acceleration is much less than g 1425 IDENTIFY The mechanical energy of the system is conserved The maximum acceleration occurs at the maximum displacement and the motion is SHM SET UP Energy conservation gives 12 mvmax2 12 kA2 and amax kAm EXECUTE A 0165 m 12 mvmax2 12 kA2 gives km vmaxA2 390 ms 0165 m2 5587 s2 amax kAm 5587 s20165 m 922 ms2 EVALUATE The acceleration is much greater than g 1426 IDENTIFY The mechanical energy of the system is conserved Newtons second law applies and the motion is SHM SET UP Energy conservation gives 12 mvx2 12 kx2 12 kA2 Fx max Fx kx and the period is T 2πmk EXECUTE Solving 12 mvx2 12 kx2 12 kA2 for vx gives vx kmA2 x2 T 2πmk so km 2πT 2π320 s 1963 s1 vx 1963 s10250 m2 0160 m2 0377 ms ax kxm 1963 s12 0160 m 0617 ms2 EVALUATE The block is on the positive side of the equilibrium position x 0 If vx 0377 ms the block is moving in the positive direction and slowing down since the acceleration is in the negative direction If vx 0377 ms the block is moving in the negative direction and speeding up 1427 IDENTIFY and SET UP Use E 12 mv2 12 kx2 12 kA2 x A when vx 0 and vx vmax when x 0 EXECUTE a E 12 mv2 12 kx2 E 12 0150 kg0400 ms2 12 300 Nm0012 m2 00336 J b E 12 kA2 so A 2Ek 200336 J300 Nm 00150 m c E 12 mvmax2 so vmax 2Em 200336 J0150 kg 0669 ms EVALUATE The total energy E is constant but is transferred between kinetic and potential energy during the motion 1428 IDENTIFY and SET UP Use E 12 mv2 12 kx2 12 kA2 to relate K and U U depends on x and K depends on vx EXECUTE a U K E so U K says that 2U E 212 kx2 12 kA2 and x A2 magnitude is A2 But U K also implies that 2K E 212 mvx2 12 kA2 and vx km A2 ωA2 magnitude is ωA2 148 Chapter 14 Copyright 2016 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher b In one cycle x goes from A to 0 to A to 0 to A Thus 2 x A twice and 2 x A twice in each cycle Therefore U K four times each cycle The time between U K occurrences is the time at for 1 2 x A to 2 2 x A time bt for 1 2 x A to 2 2 x A time ct for 1 2 x A to 2 2 x A or the time dt for 1 2 x A to 2 2 x A as shown in Figure 1428 a b t t c d t t Figure 1428 Calculation of at Specify x in cos x A t choose 0 so x A at 0 t and solve for t 1 2 x A implies 1 2 cos A A t 1 cos 1 2 t so 1 arccos1 2 4 rad t 1 4 t 2 2 x A implies 2 2 cos A A t 2 cos 1 2 t so 1 3 4 rad t 2 3 4 t 2 1 3 4 4 2 at t t Note that this is 4 T onefourth period Calculation of dt 1 2 x A implies 1 3 4 t 2 2 x A 2t is the next time after 1t that gives 2 cos 1 2 t Thus 2 1 2 5 4 t t and 2 5 4 t 2 1 5 4 3 4 2 dt t t so is the same as at Therfore the occurrences of K U are equally spaced in time with a time interval between them of 2 EVALUATE This is onefourth T as it must be if there are 4 equally spaced occurrences each period c EXECUTE 2 x A and U K E 2 2 2 2 2 2 2 1 1 1 1 1 1 2 2 2 2 2 8 2 3 8 K E U kA kx kA k A kA kA kA Then 2 2 1 2 3 8 3 4 K kA E kA and 2 1 8 2 1 2 1 4 kA U E kA EVALUATE At 0 x all the energy is kinetic and at x A all the energy is potiential But K U does not occur at 2 x A since U is not linear in x 1429 IDENTIFY Velocity and position are related by 2 2 2 1 1 1 2 2 2 x E kA mv kx Acceleration and position are related by x kx ma SET UP The maximum speed is at 0 x and the maximum magnitude of acceleration is at x A EXECUTE a For 0 x 2 2 1 1 max 2 2 mv kA and max 0 040 m 450 Nm 1 20 ms 0 500 kg k v A m b 2 2 2 2 450 Nm 0 040 m 0 015 m 1 11 ms 0 500 kg x k v A x m The speed is 1 11 ms v Periodic Motion 149 Copyright 2016 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher c For x A 2 max 450 Nm 0 040 m 36 ms 0 500 kg k a m A d 2 450 Nm 0 015 m 13 5 ms 0 500 kg x kx a m e 2 2 1 1 2 2 450 Nm0 040 m 0 360 J E kA EVALUATE The speed and acceleration at 0 015 m x are less than their maximum values 1430 IDENTIFY and SET UP x a is related to x by x k a m x and x v is related to x by 2 2 2 1 1 1 2 2 2 E mv kx kA x a is a maximum when x A and x v is a maximum when 0 x t is related to x by cos x A t EXECUTE a x kx ma so ax k m x But the maximum x is A so 2 max a k m A A 0 850 Hz f implies 2 2 0 850 Hz 5 34 rads k m f 2 2 2 max 5 34 rads 0 180 m 5 13 ms a A 2 2 2 1 1 1 2 2 2 mvx kx kA max vx v when 0 x so 2 2 1 1 max 2 2 mv kA max 5 34 rads0 180 m 0 961 ms v k mA A b 2 2 2 5 34 rads 0 090 m 2 57 ms ax k m x x 2 2 2 1 1 1 2 2 2 mvx kx kA says that 2 2 2 2 vx k m A x A x 2 2 5 34 rads 0 180 m 0 090 m 0 832 ms vx The speed is 0832 ms c cos x A t Let 2 so that 0 x at 0 t Then cos 2 sin x A t A t Using the trig identity cos 2 sin a a Find the time t that gives 0 120 m x 0 120 m 0 180 msin t sin 0 6667 t arcsin06667 07297 rad534 rads 0137 s t EVALUATE It takes onefourth of a period for the object to go from 0 x to 0 180 m x A So the time we have calculated should be less than T 4 1 10 850 Hz 1 18 s T f 4 0 295 s T and the time we calculated is less than this Note that the x a and x v we calculated in part b are smaller in magnitude than the maximum values we calculated in part a d The conservation of energy equation relates v and x and F ma relates a and x So the speed and acceleration can be found by energy methods but the time cannot Specifying x uniquely determines x a but determines only the magnitude of vx at a given x the object could be moving either in the x or x direction 1431 IDENTIFY and SET UP Use the results of Example 145 and also that 2 1 2 E kA EXECUTE In the example 2 1 M A A M m and now we want 1 2 2 1 A A Therefore 1 2 M M m or 3 m M For the energy 2 1 2 2 2 E kA but since 1 2 2 1 A A 1 2 4 1 E E and 3 1 4 E is lost to heat EVALUATE The putty and the moving block undergo a totally inelastic collision and the mechanical energy of the system decreases 1432 IDENTIFY Newtons second law applies to the system and mechanical energy is conserved SET UP ΣFx max K1 U1 K2 U2 U 12 kx2 Fspring kx EXECUTE a ΣFx max gives max kx which gives 0300 kg120 ms2 k0240 m Solving for k gives k 150 Nm b Applying K1 U1 K2 U2 gives 12 mv2 12 kx2 12 kA2 Putting in the numbers gives 0300 kg400 ms2 150 Nm0240 m2 150 NmA2 so A 061449 m which rounds to 0614 m c The kinetic energy is maximum when the potential energy is zero which is when x 0 Therefore 12 kA2 12 mv2 which gives 150 Nm061449 m2 0300 kgv2 v 4345 ms which rounds to 435 ms d The maximum force occurs when x A so Newtons second law gives Fmax max kA 150 Nm061449 m 0300 kgamax which gives amax 307 ms2 EVALUATE It is frequently necessary to use a combination of energy conservation and Newtons laws 1433 IDENTIFY Conservation of energy says 12 mv2 12 kx2 12 kA2 and Newtons second law says kx max SET UP Let x be to the right Let the mass of the object be m EXECUTE k maxx m 840 ms2 0600 m 140 s2m A x2 mkv2 0600 m2 m140 s2m220 ms2 0840 m The object will therefore travel 0840 m 0600 m 0240 m to the right before stopping at its maximum amplitude EVALUATE The acceleration is not constant and we cannot use the constant acceleration kinematic equations 1434 IDENTIFY and SET UP Energy conservation gives 12 mvx2 12 kx2 12 kA2 At t 0 x 0 and vx 120 ms Newtons second law also applies EXECUTE a x 0 gives 12 mvx2 12 kA2 and A vx mk 120 ms200 kg 315 Nm 0956 m b ax km x so amax km A 315 Nm 200 kg0956 m 151 ms2 c Fx kx so Fmax kA 315 Nm0956 m 301 N EVALUATE We can use ΣFx max to get Fmax m amax 200 kg151 ms2 302 N the same result found in c apart from a small roundoff effect 1435 IDENTIFY and SET UP Velocity position and total energy are related by E 12 kA2 12 mvx2 12 kx2 Acceleration and position are related by kx max The maximum magnitude of acceleration is at x A EXECUTE a E 12 mvx2 12 kx2 12 200 kg400 ms2 12 315 Nm0200 m2 E 160 J 63 J 223 J E 12 kA2 and A 2Ek 2223 J315 Nm 0376 m Periodic Motion 1411 Copyright 2016 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher b 2 max 315 Nm 0 376 m 59 2 ms 2 00 kg k a m A c 2 max max 2 00 kg59 2 ms 118 N F ma Or Fx kx gives max 315 Nm0 376 m 118 N F kA which checks EVALUATE The maximum force and maximum acceleration occur when the displacement is maximum and the velocity is zero 1436 IDENTIFY The mechanical energy the sum of the kinetic energy and potential energy is conserved SET UP K U E with 2 1 2 E kA and 2 1 2 U kx EXECUTE U K says 2 U E This gives 2 2 1 1 2 2 2 kx kA so 2 x A EVALUATE When 2 x A the kinetic energy is three times the elastic potential energy 1437 IDENTIFY Initially part of the energy is kinetic energy and part is potential energy in the stretched spring When x A all the energy is potential energy and when the glider has its maximum speed all the energy is kinetic energy The total energy of the system remains constant during the motion SET UP Initially 0 815 ms vx and 0 0300 m x EXECUTE a Initially the energy of the system is 2 2 2 2 1 1 1 1 2 2 2 2 0 175 kg0 815 ms 155 Nm0 0300 m 0 128 J E mv kx 2 1 2 kA E and 2 20 128 J 0 0406 m 4 06 cm 155 Nm E A k b 2 1 2 mvmax E and max 2 20 128 J 1 21 ms 0 175 kg E v m c 155 Nm 29 8 rads 0 175 kg k m EVALUATE The amplitude and the maximum speed depend on the total energy of the system but the angular frequency is independent of the amount of energy in the system and just depends on the force constant of the spring and the mass of the object 1438 IDENTIFY Use the amount the spring is stretched by the weight of the fish to calculate the force constant k of the spring 2 T m k max 2 v A fA SET UP When the fish hangs at rest the upward spring force Fx kx equals the weight mg of the fish 1 f T The amplitude of the SHM is 00500 m EXECUTE a mg kx so k mg x 65 0 kg9 80 ms2 0 180 m 354 103 Nm b T 2 m k 2 65 0 kg 354 103 Nm 0 8514 s which rounds to 0851 s c max 2 2 0 0500 m 2 0 369 ms 0 8514 s A v fA T EVALUATE Note that T depends only on m and k and is independent of the distance the fish is pulled down But vmax does depend on this distance 1439 IDENTIFY 2 1 2 K mv Ugrav mgy and 2 1 el 2 U kx SET UP At the lowest point of the motion the spring is stretched an amount 2A EXECUTE a At the top of the motion the spring is unstretched and so has no potential energy the cat is not moving and so has no kinetic energy and the gravitational potential energy relative to the bottom is 2 2 24 00 kg9 80 ms 0 050 m 3 92 J mgA This is the total energy and is the same total for each part b grav spring 0 0 so 3 92 J U K U 1412 Chapter 14 Copyright 2016 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher c At equilibrium the spring is stretched half as much as it was for part a and so 1 spring 4 3 92 J 0 98 J U 1 grav 2 3 92 J 1 96 J U and so 0 98 J K EVALUATE During the motion work done by the forces transfers energy among the forms kinetic energy gravitational potential energy and elastic potential energy 1440 IDENTIFY The torsion constant is defined by z 1 2 f I and 1 T f cos t t SET UP For the disk 2 1 2 I MR z FR At 0 t 3 34 0 0583 rad so 0 EXECUTE a 4 23 N0 120 m 8 71 N mrad 0 0583 rad 0 0583 rad z FR b 2 2 1 1 2 1 28 71 N mrad 2 17 Hz 2 2 2 6 50 kg0 120 m f I MR 1 0 461 s T f c 2 13 6 rads f 3 34 cos13 6 rads t t EVALUATE The frequency and period are independent of the initial angular displacement so long as this displacement is small 1441 IDENTIFY and SET UP The number of ticks per second tells us the period and therefore the frequency We can use a formula from Table 92 to calculate I Then 1 2 f I allows us to calculate the torsion constant EXECUTE Ticks four times each second implies 025 s per tick Each tick is half a period so 0 50 s T and 1 10 50 s 2 00 Hz f T a Thin rim implies 2 I MR from Table 92 3 2 2 8 2 0 900 10 kg0 55 10 m 2 7 10 kg m I b 2 T I so 2 8 2 2 6 2 2 7 10 kg m 2 0 50 s 4 3 10 N mrad I T EVALUATE Both I and are small numbers 1442 IDENTIFY 1 2 f I and 1 T f says 2 I T SET UP 2 1 2 I mR EXECUTE Solving 1 2 f I for in terms of the period 2 2 3 2 2 5 2 2 1 2 00 10 kg2 20 10 m 1 91 10 N mrad 1 00 s 2 I T EVALUATE The longer the period the smaller the torsion constant 1443 IDENTIFY 1 2 f I SET UP f 165265 s the number of oscillations per second EXECUTE 2 2 2 0 450 N mrad 0 0294 kg m 2 2 165265 s I f EVALUATE For a larger I f is smaller 1444 IDENTIFY t is given by cos t t Evaluate the derivatives specified in the problem SET UP cos sin d t dt t sin cos d t dt t 2 2 sin cos 1 In this problem 0 EXECUTE a 2 2 sin and cos 2 d d t t dt dt b When the angular displacement is Θ Θ Θ coswt This occurs at t 0 so w 0 α w²Θ When the angular displacement is Θ2 Θ2 Θ coswt or 12 coswt dθdt wΘ32 since sinwt 32 α w²Θ2 since coswt 12 EVALUATE coswt 12 when wt π3 rad 60 At this t coswt is decreasing and θ is decreasing as required There are other larger values of wt for which θ Θ2 but θ is increasing 1445 IDENTIFY T 2πLg is the time for one complete swing SET UP The motion from the maximum displacement on either side of the vertical to the vertical position is onefourth of a complete swing EXECUTE a To the given precision the smallangle approximation is valid The highest speed is at the bottom of the arc which occurs after a quarter period T4 π2 Lg 025 s b The same as calculated in a 025 s The period is independent of amplitude EVALUATE For small amplitudes of swing the period depends on L and g 1446 IDENTIFY Since the rope is long compared to the height of a person the system can be modeled as a simple pendulum Since the amplitude is small the period of the motion is T 2πLg SET UP From his initial position to his lowest point is onefourth of a cycle He returns to this lowest point in time T2 from when he was previously there EXECUTE a T 2π650 m 980 ms² 512 s t T4 128 s b t 3T4 384 s EVALUATE The period is independent of his mass 1447 IDENTIFY Since the cord is much longer than the height of the object the system can be modeled as a simple pendulum We will assume the amplitude of swing is small so that T 2πLg SET UP The number of swings per second is the frequency f 1T 12π gL EXECUTE f 12π 980 ms² 150 m 0407 swings per second EVALUATE The period and frequency are both independent of the mass of the object 1448 IDENTIFY Use Eq 1434 to relate the period to g SET UP Let the period on earth be TE 2πLgE where gE 980 ms² the value on earth Let the period on Mars be TM 2πLgM where gM 371 ms² the value on Mars We can eliminate L which we dont know by taking a ratio EXECUTE TMTE 2πLgM 2πLgE gEgM TM TE gEgM 160 s 980 ms² 371 ms² 260 s EVALUATE Gravity is weaker on Mars so the period of the pendulum is longer there 1449 IDENTIFY Apply T 2πLg SET UP The period of the pendulum is T 136 s100 136 s 1414 Chapter 14 Copyright 2016 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher EXECUTE 2 2 2 2 2 4 4 0 500 m 10 7 ms 1 36 s L g T EVALUATE The same pendulum on earth where g is smaller would have a larger period 1450 IDENTIFY and SET UP The period is for the time for one cycle The angular amplitude is the maximum value of EXECUTE a From the graph with the problem 1 60 s T 1 0 625 Hz f T 2 3 93 rads f From the graph we also determine that the amplitude is 6 degrees b 2 L T g so 2 2 2 9 80 ms 1 60 s 0 635 m 2 2 T L g c No The graph is unchanged if the mass of the bob is changed while the length of the pendulum and amplitude of swing are kept constant The period is independent of the mass of the bob EVALUATE The amplitude of the graph in the problem does not decrease over the time shown so there must be little or no friction in this pendulum 1451 IDENTIFY If a small amplitude is assumed 2 L T g SET UP The fourth term in Eq 1435 would be 2 2 2 6 2 2 2 1 3 5 sin 2 2 4 6 EXECUTE a 2 2 00 m 2 2 84 s 9 80 ms T b 2 4 6 1 9 225 2 84 s 1 sin 15 0 sin 15 0 sin 15 0 2 89 s 4 64 2304 T c Eq 1435 is more accurate 2 T L g is in error by 2 84 s 2 89 s 2 2 89 s EVALUATE As Figure 1422 in Section 145 shows the approximation F mg is larger in magnitude than the true value as increases The equation 2 T L g therefore overestimates the restoring force and this results in a value of T that is smaller than the actual value 1452 IDENTIFY tan a L 2 arad L and 2 2 tan rad a a a Apply conservation of energy to calculate the speed in part c SET UP Just after the sphere is released 0 and rad 0 a When the rod is vertical tan 0 a EXECUTE a The forces and acceleration are shown in Figure 1452a rad 0 a and tan sin a a g b The forces and acceleration are shown in Figure 1452b c The forces and acceleration are shown in Figure 1452c i f U K gives mgL1 cos 1 2 mv2 and v 2gL1 cos EVALUATE As the rod moves toward the vertical v increases arad increases and atan decreases