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8. Oscillations Chapter 8 Oscillations Basic Requirements 1. Understand the concept of Simple Harmonic Motion (SHM); 2. Master the force law for Simple harmonic motion; 3. Understand the meaning of phase, in phase and out of phase of SHM; 4. Master the characteristic of composition of simple harmonic motion in the same direction with the same frequency; 5. Understand the superposition of two SHMs in the directions perpendicular to each other; 6. Understand damped simple harmonic motion; 7. Understand forced oscillations and resonance. Review and Summary Frequency The frequency f of periodic or oscillatory motion is the number of oscillations per second. In the SI system, it is measured in hertz: 1 1hertz= 1Hz = 1 oscillation per second = 1s 1 (8-1) Period The period T is the time required for one complete oscillation, or cycle. It is related to the frequency by 1 T (8-2) f Simple Harmonic Motion In simple harmonic motion (SHM), the displacement x(t ) of a particle from its equilibrium position is described by the equation x xm cos(t ) (displacement) (8-3) in which xm is the amplitude of the displacement, the quantity (t ) is the phase of the motion , and is the phase constant. The angular frequency is related to the motion by - 189 - 8. Oscillations 2 2 f (angular frequency) (8-5) T Differentiating Eq. 8-3 leads to equations for the particle’s velocity and acceleration during SHM as function of time: v xm sin(t ) (8-6) and a 2 xm cos(t ) (8-7) In Eq. 8-6, the positive quantity xm is the velocity amplitude vm of the motion. In Eq. 8-7, the positive quantity xm is the acceleration amplitude am of the 2 motion. The Linear Oscillation A particle with mass m that moves under the influence of a Hooker’s law restoring force given by F kx exhibits simple harmonic motion with k (angular frequency) (8-12) m m and T 2 (period) (8-13) k Such a system is called a linear simple harmonic motion oscillator. Energy A particle in simple harmonic motion has, at any time, kinetic energy 1 2 1 K mv and potential energy U kx 2 . If no friction is present, the 2 2 mechanical energy E K U remains constant even though K and U change. Pendulums Examples of devices that undergo simple harmonic motion are the torsion pendulum of Fig. 16-7, the simple pendulum of Fig. 8-9, and the physical pendulum of Fig. 8-10. Their periods of oscillation for small oscillations are, respectively, T 2 I (8-22) - 190 - 8. Oscillations T 2 I mgh (8-26) and T 2 L g (8-27) Simple Harmonic Motion and Uniform Circular Motion Simple harmonic motion is the projection of uniform circular motion onto the diameter of the circle in which the latter motion occurs. Figure 8-14 shows that all parameters of circular motion (position, velocity, and acceleration) project to the corresponding values for simple harmonic motion. Damped Harmonic Motion The mechanical energy E in a real oscillating system decreases during the oscillations because external forces, such as a drag force, inhibit the oscillations and transfer mechanical energy to thermal energy. The real oscillator and its motion are then said to be damped. If the damping force is given by Fd bv , where v is the velocity of the oscillator and b is a damping constant, then the displacement of the oscillator is given by x(t ) xm e bt 2 m cos( 't ) (8-40) where ' , the angular frequency of the damped oscillator, is given by k b2 ' 2 (8-41) m 4m If the damping constant is small ( b km ), then ' , where is the angular frequency of the undamped oscillator. For small b, the mechanical energy E of the oscillator is given by 1 2 bt m E (t ) kxme (8-42) 2 Force Oscillations and Resonance If an external driving force with angular frequency d acts on the oscillating system with natural angular frequency , the system oscillates with angular frequency d . The velocity amplitude vm of the system is greatest when - 191 - 8. Oscillations d (8-44) a condition called resonance. The amplitude xm of the system is (approximately) greatest under the same condition. Examples Example 1 Suppose that two springs in figure have different spring constants k1 and k2. Show that the frequency f of oscillation of the block is then given by f f12 f 22 where f1 and f2 are the frequencies at which the block would oscillate if connected only to spring 1or only to spring 2. Solution：The key idea is that the motion of block-spring system is SHM. Assuming that the spring 1 is stretched k1 k2 for a small displacement x and the spring 2 is compressed m by the same amount at the same time. From the Hook’s law we can write Fig.8-1 Example 1. Fx k1x k 2 x= (k1 k 2 )x where (k1 k 2 ) is the coefficient of a composite spring. Using the Newton’s Second Law, we can obtain d2x m (k1 k 2 )x dt 2 d2x m (k1 k 2 )x=0 dt 2 Thus the angular frequency is k1 k 2 m And the frequency f of oscillation of the oscillator is 1 k1 k2 1 f 12 2 2 f12 f12 (Answer) 2 2 m 2 - 192 - 8. Oscillations Example 2 A block-spring system of the constant k 9.8N/ m , and the block of mass m 20 g . Now the spring is stretched 2 2cm away from the equilibrium position, and the block has the initial speed the 7.0m / s . Find the kinematics equation of the oscillator. (SI) We Solution： choose the equilibrium position as the origin and the x axis horizontal to the right. So the equation of the spring oscillator is x A cos(0t ) (1) k 9.8( N / m), m 200 g k 9.8 (2) 0 7(rad / s) m 200 103 From the avaliable conditions given by the problem, we know t 0 x x0 2 2(cm), v v0 7.0(cm / s) so we can obtain 02 A x 2 3 102 (m) 2 (3) 0 0 x0 A cos 0.34(rad ) (4) v0 A0 sin Substituting the Eq.(2)-(4) into the Eq.(1), the kinematics equation of the oscillator can be written as: x 3 102 cos(7t 0.34) (Answer) Example 3 An object oscillates with simple harmonic motion along the x axis. Its displacement from the origin varies with time according to the equation x (4.00m) cos( t ) 4 where t is in seconds and the angles in the parentheses are in radians. (a) Determine the amplitude, frequency, and period of the motion. (b) Calculate the velocity and acceleration of the object at any time t. (c) Using the results of part (b), determine the - 193 - 8. Oscillations position, velocity, and acceleration of the object at t 1.00s . Solution： (a) By comparing tthe general equation for simple harmonic motion— x A cos(t ) —we see that A 4.00m , rad / s . Therefore, f 2 0.500 Hz , and T 1 f 2.00s . (Answer) (b) Calculate the velocity and acceleration of the object at any time t. d v dx (4.00m)sin( t ) ( t ) (4.00 m / s)sin( t ) dt 4 dt 4 d a dv (4.00 m / s) cos( t ) ( t ) (4.00 2 m / s 2 ) cos( t ) dt 4 dt 4 (Answer) (c) Noting that the angles in the trigonometric functions are in radians, we obtain, at t 1.00s , 5 x (4.00m) cos( ) (4.00m) cos( ) (4.00m)(0.707) 2.83m 4 4 5 v (4.00 m / s)sin( ) (4.00 m / s)(0.707) 8.89m / s 4 5 a (4.00 2 m / s 2 ) cos( ) (4.00 2 m / s 2 )(0.707) 27.9m / s 2 4 (Answer) Example 4 The thin circle with radius of R, resting on a knife edge O, is in its own plane for the small swing. Calculate (a) the period of the oscillation, (b) the length of the pendulum with the same period of oscillation, and (c) cutting off 2 3 of the circle and putting the center of the remaining arc on the knife. Find the ratio of periods between the remaining arc and the whole circle. Solution ： (a) From the period of the physical o pendulum, we know o I T 2 (1) Fig.8-2 Example 4. mgh where m is the mass of the thin circle.According to the Parallel axis theorem, we can get I 0 I 0 moo2 mR 2 mR 2 2mR 2 - 194 - 8. Oscillations Substituting the I 0 into the Eq.(1), yields 2mR 2 2R T 2 2 (2) (Answer) mgR g (b) From the period of the simple pendulum, we get T0 2 (3) g Because the period of the simple pendulum is the same as that of the circle, so T T0 . Combining the Eq.(2)-(3), we can obtain the length of the simple pendulum 2R (Answer) (c) The remain arc whose center is on the knife edge is still the physical pendulum, so its period is I T 2 (4) mgh From the symmetry property, we know the center of the mass is on the oo . Assuming m is the mass of the remaining arc. And h oc is the distance between the center of the remaining arc and its supporting point o, so the rotational inertia of the remain arc can be written as I 0 I C mh2 mR 2 m( R h) 2 mh2 2mRh (5) Substituting the Eq.(5) into the Eq.(4), the period of the remain arc is 2mRh 2R T 2 2 mgh g T 1 (Answer) T No matter how much the circle removed, its period of the oscillation doesn’t change as long as the knife edge higher than the center of the remaining arc. Example 5 A ball of mass m1 0.9kg is suspended on from the ceiling with a light cord of length 0.9m . Another ball of mass m1 0.1kg with velocity v 1.0m / s , moving horizontally strikes the ball in completely inelastic collision. Determine the - 195 - 8. Oscillations kinematics equation after collision of two balls. Solution ： Considering the balls m1 ,m 2 as a system, the linear momentum is conserved along the horizontal direction in the completely inelastic collision. m 2 v2x (m1 m 2 )vx m 2v2x vx 0.1(m / s) m1 m 2 After collision the m1 and the light line looked as the simple pendulum with the simple harmonic motion. Assuming its equation is A cos(0 t ) (1) At the initial time 0 , and then we can get d x 0.1 0 0, angular velocity dt 0.9 g 9.8 0 3.3 (2) 0.9 0 2 0 A 02 ( ) 0.0337 (3) 0 0 From the initial condition, we also know cos 0 , sin 0 A0 1 . Solving for , yields (4) 2 Combining the Eq.(1)-(4), the the kinematics equation can be written as 0.0337 cos(3.3t ) (Answer) 2 Example 6 A 0.500 kg cube connected to a light spring for which the force constant is 20.0N / m oscillates on a horizontal, frictionlesstrack. (a) Calculate the total energy of the system and the maximum speed of the cube if the amplitude of the motion is 3.00cm . (b) What is the velocity of the cube when the displacement is 2.00cm ? (c) Compute the kinetic and potential energies of the system when the displacement is 2.00cm . Solution：(a) Using Equation E K U , we obtain - 196 - 8. Oscillations 1 E K U (20.0 N / m)(3.00 102 )2 9.00 103 J 2 1 2 When the cube is at x 0 we know that U 0 , and E mvmax , therefore 2 1 2 18.0 103 mvmax 9.00 103 J vmax 0.190m / s (Answer) 2 0.500kg (b) We can calculate directly: k 2 20.0 v ( A x2 ) [(0.0030) 2 0.0020) 2 ] 0.141m / s (Answer) m 0.500 The positive and negative signs indicate that the cube could be moving to either the right or the left at this instant. (c) Using the result of (b), we find that 1 2 1 K mv (0.500kg )(0.141m / s) 2 5.00 103 J 2 2 1 2 1 U kx (20.0 N / m)(0.020m) 2 4.00 103 J (Answer) 2 2 Problem Solving 1 (2) Fig3 shows block 1 of mass 0.200kg sliding to the right over a frictionless elevated surface at a speed of 8.00m / s . The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1208.5N / m . (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of 0.140s , and block 1 slides off the opposite end of the elevated surface, landing a distance d from the base of that surface after falling height h 4.90m . What is the value of d? Solution：The statement that “the spring does not affect the collision” justifies the use of elastic collision formulas. We are told the period of SHM Fig.8-3 Problem 1. so that we can find the mass of block 2: m2 kT 2 T 2 m2 0.600kg k 4 2 - 197 - 8. Oscillations At this point, the rebound speed of block 1 can be found 0.200 0.600 v1 f (8.00m / s ) 4.00m / s 0.200 0.600 This becomes the initial speed v 0 of the projectile motion of block 1. A variety of choices for the positive axis directions are possible, and we choose left as the +x direction and down as the +y direction, in this instance. With the “launch” angle being zero, we can get 2h 2(4.90) d x x0 v0t v0 (4.00) 4.00m (Answer) g 9.8 2 (4) Two particles oscillate in simple harmonic motion along a common straight line segment of length A. Each particle has a period of 1.5s , but they differ in phase 6rad .(a) How far apart are they (in terms of A) 0.50s after the lagging particle leaves one end of the path? (b) Are they then moving in the same direction, toward each other, or away from each other? Solution：(a) Let A 2 t x1 cos 2 T be the coordinate as a function of time for partical 1 and A 2 t x2 cos( ) 2 T 6 be the coordinate as a function of time for particle 2. Here T is the period. Note that since the range of the motion is A A, the amplitudes are both A 2 . The arguments of the cosine functions are in radians. Particle 1 is at one end of its path ( x1 A 2 ) when t 0 . Particle 2 is at A 2 when 2 t T 6 0 or t T 12 . That is, particle 1 lags particle 2 by one twelfth a period. We want the coordinates of the particles 0.50s later; that is, at t 0.50s , A (2 0.50s) x1 cos 0.25 A 2 1.5s and A 2 0.50s x2 cos( ) 0.43 A 2 1.5s 6 - 198 - 8. Oscillations Their separation at that time is x1 x2 0.25 A 0.43 A 0.18 A . (Answer) (b) The velocities of the particles are given by dx1 A 2 v1 sin( ) dt T T and dx2 A 2 v2 sin( ) (Answer) dt T T 6 We evaluate these expressions for t 0.50s and find they are both negative-valued, indicating that the particles are moving in the same direction. 3 (7) In Fig.3, two blocks ( m 1.0kg and M 10kg ) and a spring ( k 200N / m ) are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is 0.40 . What amplitude of simple harmonic motion of the springblocks system puts the smaller block on the verge of splipping over the large block? Solution： The key idea is that the smaller block gives a force which acted on the larger block, that is f mg 0.4 0.1kg 10m / s 2 0.4 N Fig.8-4 Problem 3. We know the sustem oscillate in SHM, and when the smaller block slipping over the large block, we have kA mg mg M m Solving for d and substituting known data, we can obtain the amplitude (m M ) g 0.4 11kg 10m / s 2 A 0.22m (Answer) k 200 N / m 4 (8) In Fig.5, two spring are joined and connected to a block of mass 0.245kg that is set oscillating over a frictionless floor. The spring each have spring constant k 6430N / m . What is the frequency of the oscillator? Solution：We wish to find the effective spring constant for the combination of springs shown in Fig. 5. We do this by finding the magnitude F of Fig.8-5 Problem 4. - 199 - 8. Oscillations the force exerted on the mass when the total elongation of the springs is Δx. Then keff F x . Suppose the left-hand spring is elongated by xl and the right-hand spring is elongated by xr The left-hand spring exerts a force of magnitude k xl on the right-hand spring and the right-hand spring exerts a force of magnitude k xr on the left-hand spring. By Newton’s third law these must be equal, so xl xr . The two elongations must be the same and the total elongation is twice the elongation of either spring: x 2xl . The left-hand spring exerts a force on the block and its magnitude is F k xl . Thus keff k xl 2xr k 2 . The block behaves as if it were subject to the force of a single spring, with spring constant k 2 . To find the frequency of its motion 1 keff replace keff in f with k 2 to obtain 2 m 1 k 1 6430 f 18.2 Hz (Answer) 2 2m 2 2 0.245 5(9) In Fig. 6, a block weight 14.0N , which can slide without friction on an incline at 40.00 , is connected to the top of the incline by a massless spring of unstreched length 0.450m and spring constant 120N / m . (a) How far from the top of the incline is the block’s equilibrium point? (b) If the block is pulled slightly down the incline and released, what is the period of the resulting oscillations? (a) Solution： The key idea is that the block stays in equilibrium is the block’s equilibrium point, so we have Fig.8-6 Problem 5. N mg sin Substitute the known data then yield: N 9.002N We have N kx, x 0.075m So the distance from the top of the incline to the block’s equilibrium point is - 200 - 8. Oscillations x x L 0.075m 0.45m 0.525m (b) If the block is pulled slightly down the incline and released, because the oscillation is SHM, so we have T 2 m k substituting the known data , yields 14 N T 2 m k 2 G kg 2 3.14 0.686 s (Answer) 120 N / m 9.8m / s 2 6 (10) In Fig.7, block 2 of mass 2.0kg oscillates on the end of a spring in SHM with a period of 20ms . The position of the block is given by x (1.0cm)cos(t 2). Block 1 of mass 4.0kg slides toward block 2 with a velocity of magnitude 6.0m / s , directed along the spring’s length. The two blocks udergo a completely inelastic collision at time t 5.0ms . (The duration of the collision is much less than the Fig.8-7 Problem 6. period of motion.) What is the amplitude of the SHM after the collision? Solution：We note that the spring constant is 4 2 m1 k 2 1.97 105 N / m T It is important to determine where in its simple harmonic motion (which “phase” of its motion) block 2 iswhen the impact occurs. Since 2 T and the given value of t (when the collision takes place) is one-fourth of T, then t 2 and the location then of block 2 is x (1.0cm)cos(t 2). where 2 which gives x 1.0m . This means block 2 is at a turning point in its motion (and thus has zero speed right before the impact occurs). Since the two blocks stick together in the process, we use momentum conservation and obtain m1v1 (m1 m2 )v - 201 - 8. Oscillations (4.0kg )(6.0m / s) Solving for v , yields v 4.0m / s 6.0kg And the system has kinetic energy and potential energy, respectively 1 1 K (m1 m2 )v 2 (6.0kg )(4.0m / s) 2 48 J 2 2 1 2 1 U kx (1.97 105 N / m)(0.010m) 2 10 J 2 2 meaning the total mechanical energy in the system at this stage is approximately 58J . When the system reaches its new turning point (at the new amplitude X ) then this amount must equal its (maximum) potential energy there: 1 (1.97 105 ) X 2 2 Therefore, we find X 2(58) /(1.97 105 ) 0.024m (Answer) 7 (11) A block of mass M 5.4kg , at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k=6000N/m. A bullet of mass m 9.5 g and velocity v of magnitude 630m / s strikes and is embedded in the block (Fig 8). Assuming the compression of the spring is negnigible untile the bullet is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion. Solution：(a) When the bullet strikes the block which is on the horizontal frictionless table, the momentum of the system1 composed with the bullet and the block is conservative. So we can obtain the speed of the block immediately after the collision mv (m M )v1 Fig.8-8 Problem 7. So the speed of the block immediately after the collision is mv 9.5 v1 1.11m / s (Answer) (m M ) 9.5 5400 (b) Then we can get the kinetic energy of the system1 - 202 - 8. Oscillations 1 1 ( mv) 2 m2v2 EK (m M )v1 ( m M ) 2 2 2 ( m M ) 2 2( m M ) Next, we consider the bullet、the block and the spring as the system2, the mechanic energy of the system2 is conserved. When the kinetic energy is zero, the spring potential energy is max and we can get the amplitude of the resulting SHM 1 2 m2v 2 UK kxm EK 2 2( m M ) Then we can get amplitude of the resulting simple harmonic motion 1 1 xm mv 0.0095 630 3.32 102 m 3.32cm (m M )k (0.0095 5.4)6000 (Answer) 8 (16) In Fig.9, a stick of length L =1.85m oscillates as a physical pendulum. (a)What value of distance x between the stick’s center of mass and its pivot point O gives the least period? (b)What is that least period? Solution ： The period of the physical pendulum I is T 2 , so the key idea here is what is I. And the mgx 1 rotational inertia of the pendulum is I mL2 mx 2 12 1 mL2 mx 2 L2 x So T 2 12 2 mgx 12 gx g Fig.8-9 Problem 8. L2 x When , we can get the minimum value of the time 12 gx g 2 2 So x L 1.85 0.534m (Answer) 12 12 (b) From (a) we know - 203 - 8. Oscillations 3L 3 1.85 T 2 2 2.07 s (Answer) 3g 3 9.8m / s 2 9 (18) The angle of the pendulum of Fig. 10 is given by m cos[(4.44rad / s )t ]. If at t 0 , 0.04rad and d dt 0.200 rad / s , what are (a) the phase constant and (b) the maximum angle m ? (Hint: Don't confuse the rate d dt at which changes with the w of the SHM.) Solution： As the angle of the pendulum is given by m cos[(4.44rad / s)t ] , we can differentiate with respect to t then we can get d 4.44 m sin[(4.44rad / s)t ] dt Substituting the data to the above equation, we can get Fig.8-10 Problem 9. m cos 0.04, 4.44 m sin 0.2 (a) tan 1.126, 49.24 0 (Answer) 0.04 (b) m 0.061rad (Answer) cos 10 (19) In Fig.11a, a metal plate is mounted on an axle through its center of mass. A spring with k 200N / m connects a wall with a point on the rim a distance r 2.5cm from the center of mass. Initially the spring is at its rest length. If the plate is rotated by 7 0 and released, it rotates about the axle in SHM, with its angular position given by Fig.11b.What is the rotational inertia of the plate about its center of mass? Fig.8-11 Problem 10. Solution：We note that the initial angle is 0 7 0.122rad (though it turns out 0 this value will cancel in later calculations). If we approximate the initial stretch of the - 204 - 8. Oscillations spring as the arclength that the corresponding point on the plate has moved through ( x r 0 , where r 0.025m ) then the initial potential energy is approximately, 1 2 kx 2000 (0.025 0.122)2 0.0093J 2 1 This should equal to the kinetic energy of the plate ( I 2 where this m is the 12 maximum angular speed of the plate, not the angular frequency ). Noting that the maximum angular speed of the plate is m 0 , where 2 T with T 20ms 0.02m . as determined from the graph, then we can find the rotational 1 inertial from I 2 0.0093J , thus 12 I 1.3 105 kg m 2 . (Answer) 11 (20) A pendulum is formed by pivoting a long thin rod about a point on the rod. In a series of experiments, the period is measured as a function of the distance d between the pivot point and the rod’s center. (a) If the rod’s length is L 2.20m and its mass is m 22.1g , what is the minimum period? (b) If d is chosen to minimize the period and then L is increased, does the period increase, decrease, or remain the same? (c) If, instead, m is increased without L increasing, does the period increase, decrease, or remain the same? Solution：A key idea here is that the system is a physical pendulum, the equation of period for physical pendulum can be used. (a) As we know the thin rod about axis through center perpendicular to the length is 1 12 ML2 . And according to the parallel-axis theroem, the rotational inetia of the thin rod about pivot point is 1 2 I I com md 2 ml md 2 12 And the period of the physical pendulum is I I com md 2 l2 d T 2 2 2 mgd mgd 12 gd g - 205 - 8. Oscillations l2 d l When , that is d , the period has its minimum value 12 gd g 2 3 3l 3 2.20 Tmin 2 2 3.14 2.26s (1) (Answer) 3g 3 9.80 (b) If L is increased, we can esily draw a conclusion based on Eq.(1) that the period will increase. (Answer) (c) Because the expression of Tmin is absent from the mass of the thin rod, the period would remain the same value when m is increased. (Answer) 12 (21) In Fig. 12, a 2.50kg disk of diameter D 42.0cm is supported by a rod of length L 76.0cm and negligible mass that is pivoted at its end. (a)With the massless torsion spring unconnected, what is the period of oscillation? (b)With the torsion spring connected, the rod is vertical at equilibrium. What is the torsion constant of the spring if the period of oscillation has been decreased by 0.500s ? Solution： The key idea is that the system can be seen as a simple pendulum and the torsion spring would decrease the period when connected. Fig.8-12 Problem 12. (a) With the massless torsion spring unconnected, the torque arm of the pendulum is D 42.0 L0 L 76.0cm cm 97.0cm 2 2 So the oscillation’s period is L0 0.97 T 2 2 2.00s (Answer) g 9.80m / s 2 (b) Now we have two “restoring torques” acting in tandem to pull the pendulum back to the vertical position when it is displaced. The magnitude of the net torque is ( Mgh ) , where the small angle approximation ( sin in radians). From the equation of the period of torsion pendulum, making the appropriate adjustment to the period formula, we have - 206 - 8. Oscillations IP T 2 Mgh where Mgh is the composite torsion constant. The problem statement requires T T 0.05s . Thus T (2.00 0.50) s 1.50s . Solving for , yields 4 2 I P Mgh 18.5 N .m / rad (Answer) T 2 “ 13 (23) A 1000kg car carrying four 82kg people travels over a washboard” dirt road with corrugations 4.0m apart. The car bounces with maximum amplitude when its speed is 16km / h . When the car stops, and the people get out, by how much does the car body rise on its suspension? Solution： With M 1000kg and m 82kg , we can write the period of the system M 4m T 2 k If d 4.0m is the distance traveled (at constant car speed v ) between impulses, then we may write T d v , in which case the above equation may be solved for the spring constant d M 4m 2 2 k ( M 4m)( ) 2 v k d Before the people got out, the equilibrium compression is xi ( M 4 m) g k , and afterward it is x f Mg v 16000 4.44m / s , we find k . Therefore, with 3600 the rise of the car body on its suspension is 4mg 4mg d xi x f ( )2 0.050m (Answer) k M 4m 2 14 (24) Figure13 five the position of - 207 - 8. Oscillations a 20g block oscillating in SHM on the end of a spring. What are (a) the maximum kinetic energy of the block and (b) the number of times per second that maximum is reached? (Hint: Measuring a slope will probably not be very accurate. Can you think of another approach?) Solution： The key idea is that the motion of the block is SHM. (a) From the figure, we can get the period T and amplitude of the system, they are Fig.8-13 Problem 14. T 40ms, xm 7cm so the angular frequency is 2 50 rad / s T The total mechanical energy of the SHM is 1 1 E m 2 xm 0.02kg (50 rad / s) 2 (0.07m) 2 1.21J 2 2 2 When the block is at its equilibrium point, it has a maximum kinetic energy, and it equals the total mechanical energy, that is Em E 1.21J (Answer) (b) Because the frquency is f 25Hz , in one period, the block passes 2 through the equilibrium point twice. So the number of times per second that maximum is 50 . (Answer) 15 (29) In Fig. 14, three 10000kg ore cars are held at rest on a mine railway using a cable that is parallel to the rails, which are inclined at angle 300 . The cable stretches 15cm just before the coupling between the two lower cars breaks, detaching the lowest car. Assuming that the cable obeys Hooker’s law, find the (a) frequency and (b) amplitude of the resulting oscillations of the remaining two cars. Solution： The magnitude of the downhill component of the gravitational force acting on Fig.14 Problem 15. each ore car is Fx mg sin (10000kg )(9.8m / s 2 ) sin - 208 - 8. Oscillations where 30 0 (and it is important to have the calculator in degrees mode during this problem). We are told that a downhill pull of 3 x causes the cable to stretch x 0.15m . Since the cable is expected to obey Hook’s law, its spring constant is 3F k x 9.8 105 N / m . x (a) Noting that the oscillating mass is that of two of the cars, we can get the frequency 1 9.8 105 f 1.1Hz (Answer) 2 20000kg (b) The difference between the equilibrium positions of the end of the cable when supporting two as opposed to three cars is 3Fx 2 Fx x 0.050m (Answer) k 16 (32) A 0.10 kg block oscillates back and forth along a straight line on a frictionless horizontal surface, Its displacement from the origin is given by x (10cm) cos[(10rad / s)t 2rad ]. (a)What is the oscillation frequency? (b) What is the maximum speed acquired by the block (c) At what value of x does this occur? (d) What is the magnitude of the maximum acceleration of the block? (e)At what value of x does this occur? (f)What force, applied to the block by the spring, results in the given oscillation? Solution： (a) A key idea here is that the motion of the block is SHM. From equation given by the title, we know that 10 f 1.6 Hz 2 2 dx v (10cm) sin[(10rad / s )t 2rad ] (10rad / s ) dt (1) (100cm / s ) sin[(10rad / s )t 2rad ] dv a (100cm / s ) cos[(10rad / s )t 2rad ](10rad / s ) dt (2) (Answer) (1000cm / s ) cos[(10rad / s )t 2rad ] 2 - 209 - 8. Oscillations (b) From the Eq.(1), the maximum speed is vmax 100cm / s (Answer) (c) When the speed is max, the value of x is zero. (Answer) 2 (d) The maximum acceleration of the block is 1000cm / s , given by the Eq.(2). (Answer) (e) When the acceleration is max, the value of x is 10cm . (Answer) (f) According to the Newton’s second law, we can get the force applied to the block F ma 0.1kg (10m / s 2 ) cos[(10rad / s)t 2rad ] (Answer) cos[(10rad / s)t 2rad ] 17 (33) In Fig.15, a solid cylinder attached to a horizontal spring ( k 3.00N / m ) rolls without slipping along a horizontal surface. If the system is released from rest when the spring is stretched by 0.250m , find (a) the translational kinetic energy and (b) the rotational kinetic energy of the cylinder as it passes through the equilibrium position. (c) Show that under these conditions the cylinder’s center lf mass executes simple harmonic motion with period 3M T 2 2k Where M is the cylinder mass. (Hint: Find the time derivative of the total mechanical energy.) Solution：(a) The potential energy at the turning point is equal (in the absence of friction) to the total kinetic energy (translational plus rotational) as it passes through the equilibrium position: Fig.15 Problem 17. 1 2 1 1 2 1 1 1 v kxm Mvcom I com 2 Mvcom ( MR 2 )( com ) 2 2 2 2 2 2 2 2 2 R 1 1 3 Mvcom Mvcom Mvcm 2 2 2 2 4 4 which leads to Mvcom 2kxm 3 0.125 J . The translational kinetic energy is 2 2 - 210 - 8. Oscillations 1 Mvcom kxm 3 0.0625 J 2 2 (Answer) 2 (b) And the rotational kinetic energy is 1 Mvcom kxm 6 3.13 102 J 2 2 (Answer) 4 (c) In this part, we use vcom to denote the speed at any instant (and not just the maximum speed as we had done in the previous parts). Since the energy is constant, then dE d 3 d 1 3 ( Mvcom ) ( kx 2 ) Mvcom acom kxvcom 0 2 dt dt 4 dt 2 2 which leads to 2k acom ( )x 3M We see that 2k 3M for this system. Since 2 T , we obtain the desired result: 3M T 2 (Answer) 2k 18 (34) An engineer has an odd-shaped 10kg object and needs to find its rotational inertia about an axis through its center of mass. The object is supported on a wire stretched along the desired axis. The wire has a torsion constant 0.50N m If this torsion pendulum oscillates through 20 cycles in 50s , what is the rotational inertia of the object? Solution： The key idea is that T 2 I . So we substitute the data into the equation and then get the rotational inertia of the object 50 s T 2 I 20 (Answer) 2.5 I ( ) 2 0.5 0.079kg m 2 2 19 (35) A block weighing 10.0N is attached to the lower end of a vertical spring - 211 - 8. Oscillations ( K 200.0N / m ), the other end of which is attached to a ceiling. The block oscillates vertically and has a kinetic energy of 2.00J as it passes through the point at which the spring is unstretched. (a)What is the period of the oscillation? (b) Use the law of conservation of energy to determine the maximum distance the block moves both above and below the point at which the spring is unstretched. (These are not necessarily the same.) (c) What is the amplitude of the oscillation? (d) What is the maximum kinetic energy of the block as it oscillates? Solution： The key idea is that the motion of the block is SHM, so we can get (a) T 2 m k 2 10 9.8 200.0 0.45s (Answer) (b) For a vertical spring, the distance between the unstretched length and the equilibrium length (with a mass m attached) is mg 10 N x 0.05m k 200.0 N / m During simple harmonic motion, the convention is to establish x 0 at the equilibrium position (the middle level for the oscillation) and to write the total energy 1 E K U K kx 2 2 Thus, as the block passes through the unstretched position, the energy is 1 1 E K kx 2 2.0 J (200 N / m) 2 (0.05m) 2 2.25 J 2 2 At its topmost and bottommost points of oscillation, the energy (using this convention) 1 kxm 2 . is all elastic potential 2 1 kxm 2 2.25 J xm 0.15m (Answer) 2 This gives the amplitude of oscillation as 0.15m , We add (or subtract) the 0.05m value found above and obtain 0.10m for the top-most position and 0.20m for the bottom-most position. (c) As noted in part (b), xm 0.15m (Answer) (d) The maximum kinetic energy equals the maximum potential energy (found in part (b))and is equal to 2.25J . (Answer) - 212 - 8. Oscillations 20 (38) In Fig. 16, a 2500kg demolition ball swings from the end of a crane. The length of the swinging segment of cable is 17m . (a) Find the period of the swinginng, assuming that the system can be treated as a simple pendulum. (b) Does the period depend on the ball’s mass? Solution：(a) Because the system can be treated as a simple pendulum, the period is Fig.16 Problem 20. L 17m T 2 2 8.3s (Answer) g 9.80m / s 2 I (b) Plugging I mL into T 2 2 , we see that the mass m cancels out. mgL Thus, the characteristics (such as the period) of the periodic motion do not depend on the mass. (Answer) - 213 -

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