15 Oscillatory Motion CHAPTER OUTLINE 15.1 Motion of an Object Attached to a Spring 15.2 Analysis Model: Particle in Simple Harmonic Motion 15.3 Energy of the Simple Harmonic Oscillator 15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion 15.5 The Pendulum 15.6 Damped Oscillations 15.7 Forced Oscillations * An asterisk indicates a question or problem new to this edition ANSWERS TO OBJECTIVE QUESTIONS OQ15.1 Answer (d) The period of a simple pendulum is T = 2π g , and its frequency is f = T = ( 2π ) g Thus, if the length is doubled so ′ = 2, the new frequency is f′ = 2π g = ′ 2π g ⎛ = 2 ⎜⎝ 2π g⎞ f = ⎟ ⎠ OQ15.2 Answer (c) The equilibrium position is 15 cm below the starting point The motion is symmetric about the equilibrium position, so the two turning points are 30 cm apart OQ15.3 Answer (a) In this spring-mass system, the total energy equals the elastic potential energy at the moment the mass is temporarily at rest at x = A = cm (i.e., at the extreme ends of the simple harmonic motion) Thus, E = kA 2 and we see that as long as the spring constant k and the amplitude A remain unchanged, the total energy is unchanged 792 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 OQ15.4 793 Answer (c) The total energy of the object-spring system is 1 kA = mv + kx When the kinetic energy is twice the potential 2 ⎛1 ⎞ energy, mv = ⎜ kx ⎟ = kx , and the total energy is ⎝2 ⎠ 2 1 A kA = kx + kx → kA = kx → x = 2 2 OQ15.5 Answer (d) When the object is at its maximum displacement, the magnitude of the force exerted on it by the spring is Fs = k xmax = ( 8.0 N m )( 0.10 m ) = 0.80 N This force will give the mass an acceleration of a = Fs m = 0.80 N 0.40 kg = 2.0 m s OQ15.6 Answer (a) The car will continue to compress the spring until all of the car’s original kinetic energy has been converted into elastic 1 potential energy within the spring, i.e., until kx = mvi2 , or 2 x = vi 3.0 × 105 kg m = ( 2.0 m s ) = 0.77 m k 2.0 × 106 N m OQ15.7 Answer (c) When an object undergoes simple harmonic motion, the position as a function of time may be written as x = A cos ω t = A cos ( 2π ft ) Comparing this to the given relation, we see that the frequency of vibration is f = Hz, and the period is T = 1/f = 1/3 *OQ15.8 Answer (b) The frequency of vibration is f = ω = 2π 2π k m Thus, increasing the mass by a factor of will decrease the frequency to 1/3 of its original value OQ15.9 Answer (a) Higher frequency When it supports your weight, the center of the diving board flexes down less than the end does when it supports your weight—this is similar to a spring that stretches a smaller distance for the same force: its spring constant is greater because the displacement is smaller Therefore, the stiffness constant describing the center of the board is greater than the stiffness k constant describing the end And then f = ⎛⎜ ⎞⎟ is greater for ⎝ 2π ⎠ m you bouncing on the center of the board © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 794 Oscillatory Motion OQ15.10 (i) Answer (c) At 40 cm we have the midpoint between the turning points, so it is the equilibrium position and the point of maximum speed, and therefore, maximum momentum (ii) Answer (c) The position of maximum speed is also the position of maximum kinetic energy (iii) Answer (e) The total energy of the system is conserved, so it is the same at every position OQ15.11 The ranking is (c) > (e) > (a) = (b) > (d) The amplitude does not affect the period in simple harmonic motion; neither constant forces that offset the equilibrium position Thus (a) and (b) have equal periods The period is proportional to the square root of mass divided by spring constant So (c), with larger mass, has a larger period than (a) And (d) with greater stiffness has smaller period In situation (e) the motion is not quite simple harmonic, but has slightly smaller angular frequency and so a slightly longer period OQ15.12 (a) Yes In simple harmonic motion, one-half of the time, the velocity is in the same direction as the displacement away from equilibrium (b) Yes Velocity and acceleration are in the same direction half the time (c) No The spring force and, therefore, the acceleration are always opposite to the position vector, and never in the same direction OQ15.13 Answer (d) We assume that the coils of the spring not hit one another When the spring with two blocks is set into oscillation in space, the coil in the center of the spring does not move We can imagine clamping the center coil in place without affecting the motion We can effectively duplicate the motion of each individual block in space by hanging a single block on a half-spring here on Earth The half-spring with its center coil clamped—or its other half cut off—has twice the spring constant as the original uncut spring because an applied force of the same size would produce only onehalf the extension distance Thus the oscillation frequency in space is 12 ⎛ ⎞ ⎛ 2k ⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ = f The absence of a force required to support the 2π m vibrating system in orbital free fall has no effect on the frequency of its vibration OQ15.14 Answer (d) is the only false statement At the equilibrium position, ⎛ ⎞ x = 0, the elastic potential energy of the system ⎜ PEs = kx ⎟ is a ⎝ ⎠ minimum and the kinetic energy is a maximum © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 OQ15.15 (i) Answer (e) We have Ti = 2π Lf 795 Li and g 4Li = 2Ti The period becomes larger by a g g factor of 2, to become s T f = 2π OQ15.16 = 2π (ii) Answer (c) Changing the mass has no effect on the period of a simple pendulum (i) Answer (b) The upward acceleration has the same effect as an increased gravitational acceleration (ii) Answer (a) The downward acceleration has the same effect as a decreased gravitational acceleration (iii) Answer (c) The absence of acceleration means that the effective gravitational field is the same as that for a stationary elevator OQ15.17 (i) Answer (c) At 120 cm we have the midpoint between the turning points, so it is the equilibrium position and the point of maximum speed (ii) Answer (a) In simple harmonic motion the acceleration is maximum when the displacement from equilibrium is maximum (iii) Answer (a), by the same logic as in part (ii) ANSWERS TO CONCEPTUAL QUESTIONS CQ15.1 An imperceptibly slight breeze blowing over the edge of a leaf can produce fluttering in the same way that a breeze can cause a flag to flap As a leaf twists in the wind, the fibers in its stem provide a restoring torque If the frequency of the breeze matches the natural frequency of vibration of one particular leaf as a torsional pendulum, that leaf can be driven into a large-amplitude resonance vibration Note that it is not the size of the driving force that sets the leaf into resonance, but the frequency of the driving force If the frequency changes, another leaf will be set into resonant oscillation CQ15.2 (a) No Since the acceleration is not constant in simple harmonic motion, none of the equations in Table 2.2 are valid (b) Equation Information given by equation x ( t ) = A cos (ω t + φ ) position as a function of time © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 796 Oscillatory Motion v ( t ) = −ω A sin (ω t + φ ) velocity as a function of time v ( x ) = ±ω ( A − x ) velocity as a function of position a ( t ) = −ω A cos (ω t + φ ) acceleration as a function of time 12 a ( t ) = −ω x ( t ) (c) CQ15.3 (a) (b) acceleration as a function of position The angular frequency ω appears in every equation The general equation of position is x ( t ) = A cos (ω t + φ ) If x = −A cos (ω t ) , then φ = π , or equally well, φ = −π At t = 0, the particle is at its turning point on the negative side of equilibrium, at x = –A CQ15.4 We assume the diameter of the bob is not very small compared to the length of the cord supporting it As the water leaks out, the center of mass of the bob moves down, increasing the effective length of the pendulum and slightly lowering its frequency As the last drops of water dribble out, the center of mass of the bob moves back up to the center of the sphere, and the pendulum frequency quickly increases to its original value CQ15.5 (a) No force is exerted on the particle The particle moves with constant velocity (b) The particle feels a constant force toward the left It moves with constant acceleration toward the left If its initial push is toward the right, it will slow down, turn around, and speed up in motion toward the left If its initial push is toward the left, it will just speed up (c) A constant force toward the right acts on the particle to produce constant acceleration toward the right (d) The particle moves in simple harmonic motion about the lowest point of the potential energy curve CQ15.6 Most everyday vibrations are damped, they eventually die down as their energy is transferred to their surroundings However, as you will learn later, atoms in the molecules have vibration modes that not damp out CQ15.7 The mechanical energy of a damped oscillator changes back and forth between kinetic and potential while it gradually and permanently decreases and transforms to internal energy CQ15.8 Yes An oscillator with damping can vibrate at resonance with © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 797 amplitude that remains constant in time Without damping, the amplitude would increase without limit at resonance CQ15.9 No If the resistive force is large compared to the restoring force of the spring (in particular, if b > 4mk ), the system will be overdamped and will not oscillate CQ15.10 The period of a pendulum depends on the acceleration of gravity: L T = 2π If the acceleration of gravity is different at the top of the g mountain, the period is different and the pendulum does not keep perfect time Two things can effect the acceleration of gravity, the top of the mountain is farther from the center of the Earth, and the nearby large mass of the mountain under the pendulum CQ15.11 Neither are examples of simple harmonic motion, although they are both periodic motion In neither case is the acceleration proportional to the displacement from an equilibrium position Neither motion is so smooth as SHM The ball’s acceleration is very large when it is in contact with the floor, and the student’s when the dismissal bell rings CQ15.12 The motion will be periodic—that is, it will repeat, though it is not harmonic at large angles The period is nearly constant as the angular amplitude increases through small values; then the period becomes noticeably larger as θ increases farther CQ15.13 The angle of the crank pin is θ = ω t Its x coordinate is x = A cos θ = A cos ω t, where A is the distance from the center of the wheel to the crank pin This is of the form x = A cos (θ t + φ ) , so the yoke and piston move with simple harmonic motion ANS FIG CQ15.13 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 798 Oscillatory Motion SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 15.1 P15.1 (a) Motion of an Object Attached to a Spring Taking to the right as positive, the spring force acting on the block at the instant of release is Fs = −kxi = − ( 130 N m ) ( +0.13 m ) = −17 N or 17 N to the left (b) At this instant, the acceleration is a= or P15.2 Fs −17 N = = −28 m s m 0.60 kg a = 28 m s to the left When the object comes to equilibrium (at distance y0 below the unstretched position of the end of the spring), ∑ Fy = −k ( −y ) − mg = and the force constant is ( ) mg ( 4.25 kg ) 9.80 m s k= = = 1.59 × 103 N = 1.59 kN/m −2 y0 2.62 × 10 m Section 12.2 P15.3 Analysis Model: Particle in Simple Harmonic Motion The spring constant is found from Fs mg ( 0.010 kg ) ( 9.80 m s ) k= = = = 2.5 N m x x 3.9 × 10−2 m When the object attached to the spring has mass m = 25 g, the period of oscillation is T = 2π P15.4 (a) 0.025 kg m = 2π = 0.63 s k 2.5 N m The equation for the piston’s position is given as π⎞ ⎛ x = ( 5.00 cm ) cos ⎜ 2t + ⎟ ⎝ 6⎠ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 799 At t = 0, ⎛π⎞ x = ( 5.00 cm ) cos ⎜ ⎟ = 4.33 cm ⎝ 6⎠ (b) Differentiating the equation for position with respect to time gives us the piston’s velocity: v= dx π⎞ ⎛ = − ( 10.0 cm/s ) sin ⎜ 2t + ⎟ ⎝ dt 6⎠ At t = 0, v = −5.00 cm s (c) Differentiating again gives its acceleration: a= dv π⎞ ⎛ = − ( 20.0 cm/s ) cos ⎜ 2t + ⎟ ⎝ dt 6⎠ At t = 0, a = −17.3 cm s (d) The period of motion is T= (e) 2π 2π = = 3.14 s ω We read the amplitude directly from the equation for x: A = 5.00 cm P15.5 P15.6 x = ( 4.00 m ) cos ( 3.00π t + π ) ; compare this with x = A cos (ω t + φ ) to find (a) ω = 2π f = 3.00π or (b) T= (c) A = 4.00 m (d) φ = π rad (e) x ( t = 0.250 s ) = ( 4.00 m ) cos ( 1.75π ) = 2.83 m f = 1.50 Hz = 0.667 s f From the information given, we write the equation for position as x = A cos ω t, with the amplitude given as A = 0.050 m Differentiating gives us the piston’s velocity, v = −Aω sin ω t © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 800 Oscillatory Motion and differentiating again gives its acceleration a = −Aω cos ω t Then, if f = 3600 rev/min = 60 Hz, then ω = 2π f = 120π s −1 15.7 (a) vmax = ω A = ( 120π )( 0.050 ) m s = 18.8 m s (b) amax = ω A = ( 120π ) ( 0.050 ) m s = 7.11 km s 2 The 0.500 s must elapse between one turning point and the other Thus the period is 1.00 s ω= 2π = 6.28 s −1 T and vmax = ω A = ( 6.28 s −1 )( 0.100 m ) = 0.628 m s P15.8 (a) From the information given, T= P15.9 12.0 s = 2.40 s 1 = = 0.417 Hz T 2.40 (b) f = (c) ω = 2π f = 2π ( 0.417 ) = 2.62 rad s An object hanging from a vertical spring moves with simple harmonic motion just like an object moving without friction attached to a horizontal spring We are given the period, which is related to the k , frequency of motion by T = 1/f Then, since ω = 2π f = m T= m = 2π f k Solving for k, k= *P15.10 4π m 4π ( 7.00 kg ) = = 40.9 N m T2 ( 2.60 s )2 For a simple harmonic oscillator, the maximum speed occurs at the equilibrium position and is given by Equation 15.17: vmax = A k m © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 801 Thus, kA ( 16.0 N/m )( 0.200 m ) m= = = 4.00 kg vmax ( 0.400 m/s )2 and Fg = mg = ( 4.00 kg ) ( 9.80 m/s ) = 39.2 N *P15.11 The mass of the cube is m = ρV = ( 2.70 × 103 kg/m ) ( 0.015 m )3 = 9.11 × 10−3 kg The spring constant of the strip of steel is P15.12 (a) k= F 1.43 N = = 52.0 N/m x 0.027 m f = ω = 2π 2π k = m 2π 52.0 N/m = 12.0 Hz 9.11 × 10−3 kg The spring constant of this spring is F mg k= = = x x ( 0.450 kg )( 9.80 m s2 ) 0.350 m = 12.6 N m We take the x axis pointing downward, so φ = ⎡ 12.6 N/m ⎤ x = A cos ω t = ( 18.0 cm ) cos ⎢ ( 84.4 s )⎥ ⎣ 0.450 kg ⎦ = ( 18.0 cm ) cos ( 446.6 rad ) = 15.8 cm (b) (c) Now 446.6 rad = 71 × 2π + 0.497 rad In each cycle the object moves 4(18) = 72 cm, so it has moved 71( 72 cm ) + ( 18 − 15.8 ) cm = 51.1 m By the same steps, k = x = A cos ( 0.440 kg )( 9.80 m/s2 ) 0.355 m = 12.1 N/m ⎡ 12.1 N/m ⎤ k t = (18.0 cm) cos ⎢ ( 84.4 s )⎥ m ⎣ 0.440 kg ⎦ = (18.0 cm)cos(443.4 rad) = −15.9 cm (d) 443.4 rad = 70.569 ( 2π ) Distance moved = 70.569 ( 0.72 m ) = 50.8 m © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 (h) 839 Utilizing the axis-crossing point, ⎛ 0.0589 ⎞ ms = ⎜ kg = grams ± 12% ⎝ 21.7 ⎟⎠ in agreement with 7.4 grams P15.77 The free-body diagram in ANS FIG P15.77 shows the forces acting on the balloon when it is displaced distance s = Lθ along the circular arc it follows The net force tangential to this path is Fnet = ∑ Fx = −Bsin θ + mg sin θ = − ( B − mg ) sin θ For small angles, sin θ ≈ θ = s / L Also, mg = ( ρHeV ) g and the buoyant force is B = ( ρairV ) g Thus, the net restoring force acting on the balloon is ⎡ ( ρ − ρHe ) Vg ⎤ Fnet ≈ − ⎢ air ⎥s L ⎣ ⎦ ANS FIG P15.77 Observe that this is in the form of Hooke’s law, F = −k s, with k = ( ρair − ρHe ) Vg L Thus, the motion will be simple harmonic and the period is given by T= 2π m = = 2π = 2π f ω k ρHeV ( ρair − ρHe )Vg L ⎛ ρHe ⎞ L = 2π ⎜ ⎝ ρair − ρHe ⎟⎠ g © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 840 Oscillatory Motion This yields ⎛ ⎞ ( 3.00 m ) 0.179 kg/m T = 2π ⎜ = 1.46 s 3⎟ ⎝ 1.20 kg/m − 0.179 kg/m ⎠ ( 9.80 m s ) P15.78 (a) We require Ae −bt 2m = A or bt = ln 2m or 0.100 kg s t = 0.693 ( 0.375 kg ) → e +bt 2m = 2, which gives t = 5.20 s The spring constant is irrelevant (b) We can evaluate the energy at successive turning points, where cos (ω t + φ ) = ±1 and the energy is We require or 2 −bt m kx = kA e 2 −bt m ⎛ ⎞ kA e = ⎜ kA ⎟ ⎠ 2⎝2 e +bt m = which gives t= (c) m ( ln ) ( 0.375 kg ) ( 0.693 ) = = 2.60 s b 0.100 kg/s From E = kA , the fractional rate of change of energy over time is (d dt) ⎛ kA ⎞ k ( 2A )(dA dt) ⎝2 ⎠ dE dt dA dt = = =2 2 E A kA kA 2 which gives dA dt dE dt = A E which is twice as fast as the fractional rate of change in amplitude © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 P15.79 (a) 841 x = A cos (ω t + φ ) → v = −ω A sin (ω t + φ ) We have at, t = 0, v = −ω A sin φ = −vmax This requires φ = 90°, so x = A cos (ω t + 90° ) π⎞ ⎛ → x = A cos ⎜ ω t + ⎟ ⎝ 2⎠ ω= Numerically we have k 50.0 N m = = 10.0 s −1 m 0.500 kg and vmax = ω A → 20.0 m s = ( 10.0 s −1 ) A → A = 2.00 m π⎞ ⎛ So x = cos ⎜ 10t + ⎟ , where x is in meters and t in seconds ⎝ 2⎠ (b) Using 1 1 ⎛1 ⎞ mv + kx = kA , we require kx = ⎜ mv ⎟ ⎝2 ⎠ 2 2 which implies 1⎛ 2⎞ 2 2 ⎜⎝ kx ⎟⎠ + kx = kA → x = A 2 x=± which gives (c) A = ±0.866(2.00 m) = ±1.73 m π⎞ ⎛ The particle’s position is given by x = cos ⎜ 10t + ⎟ ⎝ 2⎠ The particle is at x = when 10t + π π 3π 5π = , , , → 10t = 0, π , 2π , 4 π 2 2 At t = 0, the particle is at the origin, but moving to the left The next time the particle is at the origin is when 10t = π when it is moving to the right The particle is first at x = 1.00 m when 10t + So then, 10t = π 3π π 11π = + = 2 4π The minimum time required for the particle to move from x = to x = 1.00 m is 10Δt = 4π π π − π = → Δt = = 0.105 s = 105 ms 3 30 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 842 P15.80 Oscillatory Motion g g 9.80 m/s →L= = = 0.098 m L ω (10 s−1 ) (d) ω= (a) The block moves with the board in what we take as the positive x direction, stretching the spring until the spring force –kx is equal in magnitude to the maximum force of static friction: kx = µ s n = µ s mg This occurs at x = (b) µ s mg k Since v is small, the block is nearly at the rest at this break point It starts almost immediately to move back to the left, the forces on it being –kx and + µ k mg While it is sliding the net force exerted on it can be written as Fnet = −kx + µ k mg = −kx + k µ k mg µ mg ⎞ ⎛ = −k ⎜ x − k ⎟ ⎝ k k ⎠ = −kxrel where xrel is the excursion of the block away from the point µ k mg k Conclusion: the block goes into simple harmonic motion centered about the equilibrium position where the spring is stretched by µ k mg k (c) The graph of the motion looks as shown in ANS FIG P15.80(c): ANS FIG P15.80(c) (d) The amplitude of its motion is its original displacement, µ mg µ k mg A= s − , because the block has been pulled out to k k µ mg , then it goes into simple harmonic motion centered x= s k µ mg about x = k k © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 It first comes to rest at spring extension 843 ( µk − µs ) mg µ k mg −A= k k Almost immediately at this point it latches onto the slowlymoving board to move with the board The board exerts a force of static friction on the block, and the cycle continues (e) The time during each cycle when the block is moving with the 2A ( µ s − µ k ) mg board is = The time for which the block is v kv springing back is one half a cycle of simple harmonic motion, 1⎛ m m⎞ (because the block slides from +A to –A 2π =π ⎜ 2⎝ k k ⎟⎠ during its SHM) We ignore the times at the end points of the motion when the speed of the block changes from v to and from 2A to v Since v is small compared to , these times are π m/ k negligible Then the period is T= P15.81 (a) ( µ s − µ k ) mg m +π kv k Let represent the length below water at equilibrium and M the tube’s mass: ∑ Fy = ⇒ −Mg + ρπ r g = Now with any excursion x from equilibrium −Mg + ρπ r ( − x ) g = Ma Subtracting the equilibrium equation gives ⎛ ρπ r g ⎞ − ρπ r gx = Ma → a = − ⎜ x ⎝ M ⎟⎠ The opposite direction and direct proportionality of a to x imply SHM (b) For SHM, F = −kx = ma → a = −(k/m)x = −ω x: the coefficient of x is the square of the angular frequency: ω= ρπ r g 2π πM →T = = M ω r ρg © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 844 P15.82 Oscillatory Motion From the oscillator information, find the natural frequency of the oscillator: ω = k 10.0 N/m = = 100 s −1 m 0.001 kg From the measurement information, find the value of b/2m: xmax ( 23.1 ms ) Ae − (b/2m)( 0.023 1 s) = 0.250 = = e − (b/2m)( 0.023 1 s) xmax ( ) A(e ) Solving, ln ( 0.250 ) b = − = 60.0 s −1 2m 0.0231 s If the damping constant is doubled, b/2m = 120 s −1 In this case, however, b/2m > ω and the system is overdamped Your design objective is not met because the system does not oscillate P15.83 The effective spring constant of a ball is k= F 1.60 × 103 N = = 80.0 MN/m x 0.200 × 10−3 m The half-cycle is from the equilibrium position of the model spring to maximum compression and back to equilibrium again The time is one-half the period: 1 m 0.067 kg T = ( 2π ) =π = 9.12 × 10−5 s 2 k 80.0 × 10 N/m © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 845 Challenge Problems P15.84 (a) ΔK + ΔU = Thus, K top + U top = K bot + U bot where K top = U bot = Therefore, mgh = Iω , but h = R − R cosθ = R ( − cosθ ) ω= v R ANS FIG P15.84 MR mr and I = + + mR 2 Substituting, we find mgR ( − cos θ ) = ⎛ MR mr 2⎞ v + + mR ⎟⎠ R 2 ⎜⎝ 2 ⎡ M mr m ⎤ mgR ( − cos θ ) = ⎢ + + ⎥v 2⎦ ⎣ 4R − cosθ ⎛ ⎞ and v = 4gR ⎜ , so 2 ⎝ M/m + r /R + ⎟⎠ ⎡ Rg ( − cos θ ) ⎤ v = 2⎢ ⎥ 2 ⎣ M/m + r /R + ⎦ (b) T = 2π 12 I mT gdCM Substituting mT = m + M and solving for dCM gives dCM = mR + M ( ) m+ M The period is then 1 MR + mr + mR ( MR2 + 2mR2 + mr ) 2 T = 2π = 2π mgR mgR ⎡ ( M + 2m) R + mr ⎤ = 2π ⎢ ⎥ 2mgR ⎣ ⎦ 12 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 846 P15.85 Oscillatory Motion (a) Total energy = 2 kA = ( 100 N m ) ( 0.200 m ) = 2.00 J 2 At equilibrium, the total energy is: 1 ( m1 + m2 ) v = (16.0 kg ) v = ( 8.00 kg ) v 2 Therefore, ( 8.00 kg ) v = 2.00 J , and v = 0.500 m s This is the speed of m1 and m2 at the equilibrium point Beyond this point, the mass m2 moves with the constant speed of 0.500 m/s while mass m1 starts to slow down due to the restoring force of the spring (b) The energy of the m1-spring system at equilibrium is: 1 m1 v = ( 9.00 kg ) ( 0.500 m s ) = 1.125 J 2 This is also equal to k ( A′ ) , where A’ is the amplitude of the m1-spring system Therefore, (100) ( A′ )2 = 1.125 or A′ = 0.150 m The period of the m1-spring system is T = 2π 9.00 kg m1 = 2π = 1.885 s k 100 N/m T = 0.471 s after it passes the equilibrium point for the spring to become fully stretched the first time The distance separating m1 and m2 at this time is and it takes ⎛T⎞ D = v ⎜ ⎟ − A′ = ( 0.500 m/s ) ( 0.471 s ) − 0.150 m ⎝ 4⎠ = 0.085 m = 8.56 cm © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 P15.86 847 The time interval for your competitor’s package to arrive is half of the orbital period found from Kepler’s third law, Equation 13.11: 1 4π RE3 Δt = T = RE ) = π ( 2 GME GME Now, consider your proposal The force on the package at an arbitrary position r is M m m (3 πr ) Mm Fg = −G closer than r = −G ME = −G E3 r 2 r r ( π RE ) RE This force is of the form of Hooke’s law! The “spring constant” for this motion is k = G ME m RE Because the force on the package is a Hooke’s-law force, the package will oscillate between opposite points on the Earth in simple harmonic motion To deliver the package to the other side of the Earth, someone must grab the package before it begins its return journey The time interval for the package to travel to the other side of the Earth is half of a period of oscillation: ⎡ ⎛ M m⎞ 1⎡ m⎤ 1⎢ Δt = T = ⎢ 2π = 2π m ⎜ G E3 ⎟ ⎥ 2⎣ k ⎦ 2⎢ RE ⎠ ⎝ ⎣ −1 ⎤ ⎥ = π RE ⎥ GME ⎦ This is exactly the same time interval as for your competitor, so you have no advantage! In fact, you have the disadvantage of the initial capital outlay to bore through the entire Earth! P15.87 (a) For each segment of the spring: dK = ( dm) vx2 Also, ANS FIG P15.87 vx = x m v and dm = dx Therefore, the total kinetic energy of the block-spring system is 1 ⎛ x2 v2 ⎞ m 1⎛ m⎞ 2 K = Mv + ∫ ⎜ ⎟ dx = ⎜⎝ M + ⎟⎠ v 2 0⎝ ⎠ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 848 Oscillatory Motion (b) ω= k meff 1⎛ m⎞ meff v = ⎜ M + ⎟ v 2 2⎝ 3⎠ and Therefore, T= P15.88 (a) M+m 2π = 2π ω k Note that as the spring passes through the vertical position, the object is moving in a circular arc of radius L − yf , where the y coordinate of the object at this point must be negative (yf < 0) When the object is at yf , the spring is stretched x = yf – L At position yf , the spring is stretched and exerting an upward tension force of magnitude greater than the object’s weight This is necessary so the object experiences a net force toward the pivot to supply the needed centripetal acceleration in this position This is summarized by Newton’s second law applied to the object at this point, stating (remember, yf is negative) mv ∑ Fy = ma → −ky f − mg = L− yf [1] The system is isolated, so conservation of energy requires that E = KEi + PEg , i + PEs, i = KE f + PEg , f + PEs, f or E = + mgL + = 1 mv + mgy f + ky 2f 2 reducing to ( ) 2mg L − y f = mv + ky 2f [2] From equation [1], observe that mv = −(L − y f )(ky f + mg) Substituting this into equation [2] gives 2mg(L − y f ) = −(L − y f )(ky f + mg) + ky f After expanding and regrouping terms, this becomes (2k)y 2f + (3mg − kL)y f + (−3mgL) = which is a quadratic equation ay 2f + by f + c = , with a = 2k = ( 1250 N m ) = 2.50 × 103 N m © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 849 b = 3mg − kL = ( 5.00 kg ) ( 9.80 m s ) − ( 250 N m )( 1.50 m ) = −1.73 × 103 N and c = −3mgL = −3 ( 5.00 kg ) ( 9.80 m s ) ( 1.50 m ) = −221 N ⋅ m Applying the quadratic formula, keeping only the negative solution [see the discussion in part (a)], and suppressing units, gives −b − b − 4ac yf = 2a = or P15.89 − ( −1.73 × 103 ) − ( −1.73 × 10 ) − ( 2.50 × 10 )( −221) ( 2.50 × 10 ) 3 y f = −0.110 m (b) Because the length of this pendulum varies and is longer throughout its motion than a simple pendulum of length L, its period will be longer than that of a simple pendulum (a) The period of the pendulum is given by T = 2π L g and changes as ANS FIG P15.89 dT π dL = dt g L dt [1] We need to find L (t) and dL From the diagram in ANS FIG dt P15.89(a), L = Li + But a h − 2 and dL ⎛ ⎞ dh = −⎜ ⎟ ⎝ ⎠ dt dt dM dV dh =ρ = − ρ A Therefore, dt dt dt dh dM =− dt ρ A dt → dL ⎛ ⎞ dM = dt ⎜⎝ ρ A ⎟⎠ dt [2] © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 850 Oscillatory Motion Also, L ⎛ ⎞ ⎛ dM ⎞ ⎟ t = L − Li dt ⎠ ∫ dL = ⎜⎝ ρ A ⎟⎠ ⎜⎝ Li [3] Substituting equations [2] and [3] into [1] gives: dT π ⎛ ⎞ ⎛ dM ⎞ = ⎟⎠ 2⎟⎜ ⎜ ⎝ dt g ⎝ ρ a ⎠ dt Li + ( t / ρ a ) ( dM / dt ) Integrating, we get T= t π ⎛ ⎞ ⎛ dM ⎞ dt ⎟ 2⎟⎜ ∫ ⎜ g ⎝ ρ a ⎠ ⎝ dt ⎠ Li + ( t / ρ a ) ( dM / dt ) π ⎛ ⎞ ⎛ dM ⎞ Li + ( t / ρ a ) ( dM / dt ) T= ⎜ ⎟ g ⎜⎝ ρ a ⎟⎠ ⎝ dt ⎠ (1/ ρa2 )( dM / dt ) T= (b) 2π ⎛ dM ⎞ Li + ⎜ ⎟t ρ a ⎝ dt ⎠ g When the liquid is gone, the CM of the bob is suddenly again at the center of the cube We had ignored the mass of the cube up until now since it was small compared to the mass of the liquid Thus, once the liquid is gone, L = Li T = 2π Li g © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 851 ANSWERS TO EVEN-NUMBERED PROBLEMS P15.2 1.59 k N/m P15.4 (a) 4.33 cm; (b) −5.00 cm/s; (c) −17.3 cm/s2; (d) 3.14 s; (e) 5.00 cm P15.6 (a) 18.8 m/s; (b) 7.11 km/s2 P15.8 (a) 2.40 s; (b) 0.417 Hz; (c) 2.62 rad/s P15.10 39.2 N P15.12 (a) 15.8 cm; (b) 51.1 m; (c) −15.9 cm; (d) 50.8 m; (e) The patterns of oscillation diverge from each other, starting out in phase but becoming completely out of phase To calculate the future, we would need exact knowledge of the present; an impossibility P15.14 (a) motion is periodic; (b) 1.81 s; (c) The motion is not simple harmonic The net force acting on the ball is a constant given by F = −mg (except when it is in contact with the ground), which is not in the form of Hooke’s law P15.16 (a) See P15.16(a) for complete solution; (b) See P15.16(b) for complete solution P15.18 (a) 1.26 s; (b) 0.150 m/s, 0.750 m/s2; (c) x = 3.00 cos ( 5.00t + π ) , −15.0 sin ( 5.00t + π ) , and −75.0 cos ( 5.00t + π ) P15.20 (a) yes; (b) We see that finding the period does not depend on knowing the mass: T = 0.859 s P15.22 (a) 126 N/m; (b) 0.178 m P15.24 (a) 0.153 J; (b) 0.784 m/s; (c) 17.5 m/s2 P15.26 (a) E increases by a factor of 4; (b) vmax is doubled; (c) amax also doubles; (d) the period is unchanged P15.28 (a) 100 N/m; (b) 1.13 Hz; (c) 1.41 m/s; (d) x = 0; (e) 10.0 m/s2; (f) ±0.200 m; (g) 2.00 J; (h) 1.33 m/s; (i) 3.33 m/s2 P15.30 (a) Particle under constant acceleration; (b) 1.50 s; (c) isolated; (d) 73.4 N/m; (e) 19.7 m below the bridge; (f) 1.06 rad/s; (g) +2.01 s; (h) 3.50 s P15.32 (a) 5.98 m/s; (b) 206 N/m; (c) 0.238 m P15.34 1.001 P15.36 ω= k = m g R © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 852 P15.38 Oscillatory Motion I= mgd 4π f (I CM + md ) ; (b) I CM = md P15.40 (a) 2π P15.42 (a) 2.09 s; (b) 4.08% P15.44 For Length, L (m): 1.000, 0.750, 0.500 and Period, T (s): 2.00, 1.73, 1.42; (b) For Period T(s): 2.00, 1.73, 1.42 and g (m/s2): 9.87, 9.89, 9.79 This agrees with the accepted value of g = 9.80 m/s2 within 0.5%; (c) 9.94 m/s2 P15.46 1.00 × 10–3 s–1 P15.48 dE = −bv < dt P15.50 (a) 1.19 Hz; (b) 17.5 cm P15.52 318 N P15.54 See P15.54 for complete solution P15.56 0.919 × 1014 Hz P15.58 (a) 0.368 m/s; (b) 3.51 cm; (c) 40.6 mJ; (d) 27.7 mJ P15.60 (a) 4.31 cm; (b) When the rock is on the point of lifting off, the surrounding water is also barely in free fall No pressure gradient exists in the water, so no buoyant force acts on the rock The effect of the surrounding water disappears at that instant P15.62 (a) See P15.62(a) for complete solution; (b) 1.04 m/s; (c) 3.40 m P15.64 (a) A = 2.00 cm; (b) T = 4.00 s; (c) mgd π rad/s ; (d) π cm/s; ⎛π ⎞ (e) 4.93 cm/s ; (f) x = 2.00 sin ⎜ t ⎟ , where x is in centimeters and t is ⎝2 ⎠ in seconds P15.66 µs g 4π f P15.68 (a) 2π P15.70 ω= m ( k1 + k ) ; (b) 2π k1 k m ( k1 + k ) 3k m © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 15 P15.72 −2Ty ˆj; (b) ω = (a) ∑ F = L 853 2T mL P15.74 If he encounters washboard bumps at the same frequency as the free vibration, resonance will make the motorcycle bounce a lot It may bounce so much as to interfere with the rider’s control of the machine; ~101 m P15.76 (a) See ANS FIG P15.76(a); (b) 1.74 N/m ± 6%; (c) See table in P15.76(c); (d) See table in P15.76(d); (e) See ANS FIG P15.64(e); (f) 1.82 N/m ± 3%; (g) they agree; (h) grams ± 12% in agreement P15.78 (a) 5.20 s; (b) 2.60 s; (c) P15.80 See P15.80 for complete solution P15.82 If the damping constant is doubled, b/2m = 120 s−1 In this case, however, b/2m > ω and the system is overdamped Your design objective is not met because the system does not oscillate P15.84 dA/dt dE/dt = A E ⎡ Rg ( − cos θ ) ⎤ (a) v = ⎢ ⎥ 2 ⎣ M/m + r /R + ⎦ 1/2 ⎡ ( M + 2m) R + mr ⎤ ; (b) 2π ⎢ ⎥ 2mgR ⎣ ⎦ 1/2 P15.86 This is exactly the same time interval as for your competitor, so you have no advantage! In fact, you have the disadvantage of the initial capital outlay to bore through the entire Earth! P15.88 (a) y f = −0.110 m ; (b) its period will be longer © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part