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by: Mayra Brakus

GeneralPhysicsII PHY106

Marketplace > La Salle University > Physics 2 > PHY106 > GeneralPhysicsII
Mayra Brakus
La Salle
GPA 3.59


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This 8 page Class Notes was uploaded by Mayra Brakus on Tuesday October 13, 2015. The Class Notes belongs to PHY106 at La Salle University taught by Staff in Fall. Since its upload, it has received 77 views. For similar materials see /class/222319/phy106-la-salle-university in Physics 2 at La Salle University.


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Date Created: 10/13/15
HRW 7e Chapter 15 Page 1 of 8 HallidayResnicldWalker 7e Chapter 15 Oscillations 1 a The amplitude is half the range of the displacement or xm 10 mm b The maximum speed vm is related to the amplitude xm by vm wxm where a is the angular frequency Since a 21Tf where f is the frequency vm 2am 27239120 Hz10gtlt10393 m075 ms c The maximum acceleration is am mix 27rf2 xm 27239120 17122 10gtlt10393 m 57x102 msz 5 a The motion repeats every 0500 s so the period must be T 0500 s b The frequency is the reciprocal ofthe periodf lT 10500 s 200 Hz c The angular frequency a is a 21tf 211200 Hz 126 rads d The angular frequency is related to the spring constant k and the mass m by a km We solve for k k mmz 0500 kg126 rads2 790 Nm e Let xm be the amplitude The maximum speed is vm wxm 126 rads0350 m 440 ms f The maximum force is exerted when the displacement is a maximum and its magnitude is given by Fm kxm 790 Nm0350 m 276 N 7 a During simple harmonic motion the speed is momentarily zero when the object is at a turning point that is when x xm or x ixm Consider that it starts at x xm and we are told that t 025 second elapses until the object reaches x ixm To execute a full cycle of the motion which takes a period T to complete the object which started at x xm must return to x xm which by symmetry will occur 025 second after it was at x 79cm Thus T 2t 050 s b Frequency is simply the reciprocal ofthe period f lT 20 Hz c The 36 cm distance between x xm and x ixm is 29cm Thus xm 362 18 cm 8 a Since the problem gives the frequency f 300 Hz we have a 21tf 611 rads understood to be valid to three signi cant figures Each spring is considered to support one fourth of the mass mcar so that Eq 1512 leads to HRW 7e Chapter 15 Page 2 of 8 1 a m 2 kZl450kg67239rads2l29gtlt105Nm b If the new mass being supported by the four springs is mtmal 1450 573 kg 1815 kg then Eq 1512 leads to 5 mm k 3 fnew 1 129gtlt10 Nm 2268HZ mum4 27239 18154 kg 13 The magnitude of the maximum acceleration is given by am mzxm where a is the angular frequency and xm is the amplitude a The angular frequency for which the maximum acceleration is g is given by a lg xm and the corresponding frequency is given by 2 f i L 9398 498 Hz 27239 27239 xm 27239 l0gtltlO m b For frequencies greater than 498 Hz the acceleration exceeds g for some part of the motion 14 From highest level to lowest level is twice the amplitude xm of the motion The period is related to the angular frequency by Eq 155 Thus xm d and a 0503 radh The phase constant in Eq 153 is zero since we start our clock when x0 xm at the highest point We solve for I when x is onefourth of the total distance from highest to lowest level or which is the same half the distance from highest level to middle level where we locate the origin of coordinates Thus we seek I when the ocean surface is at x iixm d x xm cosat id Gdjcosmjmt 0 4 l i cos0503t 2 which has I 208 h as the smallest positive root The calculator is in radians mode during this calculation 15 The maximum force that can be exerted by the surface must be less than usFN or else the block will not follow the surface in its motion Here it is the coefficient of static friction and F N is the normal force exerted by the surface on the block Since the block does not accelerate vertically we know that F N mg where m is the mass of the block Ifthe block follows the table and moves in simple harmonic motion the magnitude of the maximum force exerted on it is HRW 7e Chapter 15 Page 3 of 8 given byF mam mwzxm m21tf2xm where am is the magnitude of the maximum acceleration a is the angular frequency and f is the frequency The relationship a 21Tf was used to obtain the last form We substitute F m211fzxm and F N mg into F lt qu to obtain m21tfzxm lt uxmg The largest amplitude for which the block does not slip is y g 05098 msz 0031 S m 2nf2 21 x 20 Hz2 xm A larger amplitude requires a larger force at the end points of the motion The surface cannot supply the larger force and the block slips 17 a Eq 158 leads to a m2x3m1 a1 x 0100 which yields a 3507 rads Thereforef m211 558 Hz b Eq 1512 provides a relation between a found in the previous part and the mass aE 2 MM0325kg m 3507 rads c By energy conservation the energy of the system at a turning point is equal to the sum of kinetic and potential energies at the time I described in the problem lkxfn lmv2 Jrlkx2 3 xm v2 x2 2 2 2 k Consequently xm 0325 400l362 01002 0400m 25 To be on the verge of slipping means that the force exerted on the smaller block at the point of maximum acceleration is fmax u mg The textbook notes in the discussion immediately after Eq 157 that the acceleration amplitude is am mzxm where a Jk ml is the angular frequency from Eq 1512 Therefore using Newton s second law we have k ma m 37x m I g mM m g which leads to xm 022 m HRW 7e Chapter 15 Page 4 of 8 3 l The total energy is given by E H1392 where k is the spring constant and xm is the amplitude We use the answer from part b to do part a so it is best to look at the solution for part b rst a The fraction of the energy that is kinetic is 1g1l2075 E E E 4 4 where the result from part b has been used b When x xm the potential energy is U kx2 The ratio is i 2 kam 025 l kx24 L 2 m mlq c Since Ekxfn and Ukx2 UE xzxfn We solve xzx 12 for x We should get x xm 2 32 We infer from the graph since mechanical energy is conserved that the total energy in the system is 60 J we also note that the amplitude is apparently xm 12 cm 012 m Therefore we can set the maximum potential energy equal to 60 J and solve for the spring constant k kxmz 60 J 2 k 83 x102 Nm 33 a Eq 1512 divided by 211 yields f 225112 b With x0 0500 m we have U0 125 J c With v0 100 ms the initial kinetic energy is K0 mvg 250 J d Since the total energy E K0 U0 375 J is conserved then consideration of the energy at the turning point leads to Eikxfnxm 2 0866 m 2 k 36 We note that the spring constant is k 4112m1T 2 197 X 105 Nm It is important to determine where in its simple harmonic motion which phase of its motion block 2 is when HRW 7e Chapter 15 Page 5 of 8 the impact occurs Since 03 21tT and the given value of I when the collision takes place is onefourth of T then oat n2 and the location then of block 2 is x xmcosnt 1 where I n2 which gives x xmcos112 112 ixm This means block 2 is at a turning point in its motion and thus has zero speed right before the impact occurs this means too that the spring is stretched an amount of 1 cm 001 m at this moment To calculate its aftercollision speed which will be the same as that of block 1 right after the impact since they stick together in the process we use momentum conservation and obtain 40 kg60 ms60 kg 40 ms Thus at the end of the impact itself while block 1 is still at the same position as before the impact the system consisting now of a total mass M 60 kg has kinetic energy 60 kg40 ms2 48 J and potential energy l97 X 105 Nm0010 m2 m 10 J meaning the total mechanical energy in the system at this stage is approximately 58 J When the system reaches its new turning point at the new amplitude X then this amount must equal its maximum potential energy there 197 X 105X2 Therefore we find X 258197 x105 0024 m 42 a Comparing the given expression to Eq 153 after changing notation x a 9 we see that a 443 rads Since a gL then we can solve for the length L 0499 m b Since vm wxm mLGm 443 rads0499 m00800 rad and m 00600 kg then we can find the maximum kinetic energy mvm2 940 X 10 4 J 53 Replacing x and v in Eq 153 and Eq 156 with 9 and dBdt respectively we identify 444 rads as the angular frequency on Then we evaluate the expressions at t 0 and divide the second by the first dBalt 7 0 att0 7 Dtan 39 a The value of 9 at t 0 is 00400 rad and the value of dGdt then is 70200 rads so we are able to solve for the phase constant tan 1020000400 x 444 0845 rad b Once is determined we can plug back in to 90 Bmcoscl to solve for the angular amplitude We find 0m 00602 rad 58 Since the energy is proportional to the amplitude squared see Eq 1521 we find the fractional change assumed small is E39 Ed dx 2xmdxM2dx7m E E x x2 x I m m HRW 7e Chapter 15 Page 6 of 8 Thus if we approximate the fractional change in xm as dxmxm then the above calculation shows that multiplying this by 2 should give the fractional energy change Therefore if xm decreases by 3 then E must decrease by 60 65 a From the graph we find xm 70 cm 0070 m and T 40 ms 0040 s Thus the angular frequency is n 21tT 157 rads Using m 0020 kg the maximum kinetic energy is 1 1 thenamv2 5m 032x 2 12 J b Using Eq 155 we have f 03211 50 oscillations per second Of course Eq 152 can also be used for this 75 a Hooke s law readily yields k 15 kg98 ms20l2 m 1225 Nm Rounding to three significant figures the spring constant is therefore 123 kNm b We are toldf 200 Hz 200 cyclessec Since a cycle is equivalent to 211 radians we have a 211200 411 rads understood to be valid to three significant figures Using Eq 1512 we find HE 3 mzLNrgz776kg 7 47rrads Consequently the weight of the package is mg 760 N 78 a The textbook notes in the discussion immediately after Eq 157 that the acceleration amplitude is am mzxm where a is the angular frequency a 211 f since there are 211 radians in one cycle Therefore in this circumstance we obtain am 2111000 Hz2000040 m 16 gtlt104 msz b Similarly in the discussion after Eq 156 we find vm wxm so that vm 2111000 Hz000040 m 25 m s c From Eq 158 we have in absolute value lal 2111000 Hz2000020 m 79 gtlt103 msz d This can be approached with the energy methods of 154 but here we will use trigonometric relations along with Eq 153 and Eq 156 Thus allowing for both roots stemming from the square root HRW 7e Chapter 15 Page 7 of 8 v x2 sinat till cos2 wt 3 i l 2 wxm xm Taking absolute values and simplifying we obtain M 27mm x2 272391000 0000402 0000202 22 ms 86 Since the centripetal acceleration is horizontal and Earth s gravitational g is downward we can de ne the magnitude of an effective gravitational acceleration using the Pythagorean theorem Then since frequency is the reciprocal of the period Eq 1528 leads to f Bin Ng2v RZ 211 L 211 L 39 With v 70 ms R 50m andL 020 m we have f353 s391 353 Hz 89 a Hooke s law readily yields 0300 kg98 ms200200 m 147 Nm T 2713 0733 S k Elkx3 gt xm JE l ZM39O 020 m 2 k 200 b Since T Zn lmk 2n 080200 m 04 s then the block completes 1004 25 cycles during the speci ed interval b With m 200 kg the period is 100 a Eq 1521 leads to c The maximum kinetic energy is the total energy 40 J d This can be approached more than one way we choose to use energy conservation HRW 7e Chapter 15 Page 8 of 8 l l EKU340Emv2 Ekx2 Therefore when x 015 m we nd v 21 ms 102 The period formula Eq 1529 requires knowing the distance h from the axis of rotation and the center of mass of the system We also need the rotational inertia I about the axis of rotation From Figure 1559 we see h L R where R 015 m Using the parallelaxis theorem we nd 11ij2 MLR2 2 where M 10 kg Thus Eq 1529 with T 20 s leads to MR2 MLR2 20 211 MgLR which leads to L 08315 m 0108 kg 106 m 602 gtlt1023 18 X 103925 kg Using Eq 1512 and the fact thatf w21t we have 1 k 1x10 Hz2 3k27rgtlt1013218X10 25m7gtlt102Nm 7239 m


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