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Version 001 – shmgravityII – holland – (1570) This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. AP B 1993 MC 43 001 10.0 points A particle oscillates up and down in simple harmonic motion. Its height y as a function of time t is shown in the diagram. 5 y (cm) t (s) 1 2 3 4 5 5 At what time t in the period shown does the particle achieve its maximum positive acceleration? 1 particle is slowing down. At t = 1 s, the particle is momentarily at rest and v = 0. Just after t = 1 s , the velocity is positive since the particle is speeding up. Remember that ∆v a= , acceleration is a positive maximum ∆t because the velocity is changing from a negative to a positive value. AP B 1993 MC 9 002 10.0 points When an object oscillating in simple harmonic motion is at its maximum displacement from the equilibrium position, which of the following is true of the values of its speed and the magnitude of the restoring force? Magnitude of Speed Restoring Force 1. Maximum Zero 1. t = 4 s 2. Zero Maximum correct 2. t = 1 s correct 3. Maximum 3. t = 3 s 4. 4. t = 2 s 5. Zero 5. None of these; the acceleration is constant. Explanation: This oscillation is described by πt y(t) = − sin , 2 dy π πt v(t) = = − cos dt 2 2 2 d y a(t) = 2 dt π 2 πt . sin = 2 2 Themaximum acceleration will occur when πt sin = 1, or at t = 1 s . 2 From a non-calculus perspective, the velocity is negative just before t = 1 s since the 1 maximum 2 1 maximum 2 1 maximum 2 Zero Explanation: The maximum displacement occurs at the turning points (where the velocity or speed is zero). The magnitude of restoring force is given by Hooke’s law F = −k x , where k is the spring constant and x is the displacement. Since x is a maximum, F is a maximum. From a different perspective, the displacement from the equilibrium position can be written as y = A sin θ , where θ is the phase of the oscillation. When the object is at its maximum displacement sin θ = 1 π θ= 2 Version 001 – shmgravityII – holland – (1570) so its speed is π v = ω A cos θ = ω A cos = 0 2 and the restoring force is π F = m A ω 2 sin θ = m A ω 2 = m A ω 2 , 2 at its maximum value. AP B 1998 MC 56 003 10.0 points An object moves up and down the y-axis with an acceleration given as a function of time t by the expression a = A sin ω t, where A and ω are constants. What is the period of this motion? 1. T = Which graph represents the kinetic energy K of the object as a function of displacement x? K Kmax 1. −xmax K x +xmax Kmax 2. 2π correct ω −xmax K x +xmax Kmax 2 2. T = ω A 3. T = 2 ω 2π 3. 4. T = 2 π ω correct −xmax K Kmax 5. T = ω Explanation: This motion is a simple harmonic motion. 1 ω = 2πf = 2π T 2π T = . ω AP B 1998 MC 8 004 10.0 points The graph below represents the potential energy U as a function of displacement x for an object on the end of a spring (F = −k x) oscillating in simple harmonic motion with amplitude xmax . U 4. −xmax K x +xmax Kmax 5. −xmax K Kmax x +xmax 6. Umax −xmax x −xmax x +xmax +xmax x +xmax Version 001 – shmgravityII – holland – (1570) 3 A) The block is hung from only one of the two springs: K Kmax 7. k x −xmax +xmax Explanation: At the equilibrium point (x = 0), the velocity is maximum and the kinetic energy is Umax due to conservation of energy. At the maximum displacement points +xmax and −xmax the velocity is zero and the kinetic energy is zero. From a different perspective, in simple harmonic motion of an object on the end of a spring, the total energy is conserved. At the maximum displacement xmax , the kinetic energy is 0, so E = U (xmax ) = Umax . m B) The block is hung from the same two springs, but the springs are connected in series rather than parallel: k U + K = E = Umax K(x) = Umax − U (x) . k m Thus, K(x) looks like an upsidedown U (x). AP M 1993 MC 24 005 10.0 points Two identical massless springs are hung from a horizontal support. A block of mass m is suspended from the pair of springs, as shown. C) An additional mass of m is attached to the block: k k k k 2m m When the block is in equilibrium, each spring is stretched an additional ∆x. Then the block is set into oscillation with amplitude A; when it passes through its equilibrium point it has a speed v. In which of the following cases will the block, when oscillating with amplitude A, also have speed v when it passes through its equilibrium point? The acceleration of gravity is g. 1. A and C only 2. A only 3. A, B, and C 4. B only 5. A and B only Version 001 – shmgravityII – holland – (1570) 4 For springs in series: k1 k2 6. C only m 7. None of these correct 8. B and C only Consider the forces from a spring’s point of view. The oscillating mass exerts the same force F (at some instant in time) on each spring, so Explanation: Let k1 = k , k2 = k . and Call the displacement of the mass x, and choose the positive direction to be to the right. For springs in parallel: k1 k2 m F = k 1 x1 ⇒ x1 = F k1 F = k 2 x2 ⇒ x2 = F . k2 Now consider the effective spring constant kseries , where x = x1 + x2 . F F = x x1 + x2 F k1 k2 k1 k2 = · = , F F k1 k2 k2 + k1 + k k2 r1 r 2 k k = . = 2km 2m kseries = Hooke’s law is F = −k x and the frequency of oscillation is ωseries ω f≡ . 2π Using Eq. 2, the velocity is The forces from the springs on the mass m are to the left: F1 = −k1 x , F2 = −k2 x, and F = F1 + F2 so that force equilibrium is v= dx = ω A cos(ω t + δ) . dt At the equilibrium point r d2 x −k1 x − k2 x = m a = m 2 . dt v = ωA = This is a differential equation for x(t) d2 x k1 + k2 x = 0, + dt2 m k A. m Therefore the angular velocity ω as presented in the question should be the same in cases A, B, and/or C. The question presents the springs in parallel (Eq. 3), so which has a sine solution of the form x(t) = A sin(ω t + δ) , where the angular frequency ω is r r k1 + k2 2k = . ω= m m ωparallel = Note: kparallel = k1 + k2 . so (1) r k1 + k2 = m r 2k . m Case A: Only one spring is present: r r k 2k ωsingle = 6= . m m Version 001 – shmgravityII – holland – (1570) Case B: Eq. 4, the springs are in series: r r k 2k ωseries = 6= . 2m m Case C: Eq. 1, but the mass is doubled: r r r 2k k k ω2 m = = 6= . 2m m 2m Consequently, none of the choices is the correct answer. AP M 1993 MC 07 08 006 (part 1 of 2) 10.0 points A block on a horizontal frictionless plane is attached to a spring, as shown below. The block oscillates along the x-axis with simple harmonic motion of amplitude A. k vx0 m −A 0 +A Which statement about the block is correct? 1. At x = A, its displacement is at a maximum. correct 2. At x = A, its velocity is at a maximum. 3. At x = 0, its velocity is zero. 4. At x = 0, its acceleration is at a maximum. 5. At x = A, its acceleration is zero. 5 3. The potential energy of the spring is at a minimum at x = A. 4. The kinetic energy of the block is always equal to the potential energy of the spring. 5. The kinetic energy of the block is at a maximum at x = A. Explanation: From conservation of energy, v = 0 at x = ±A, so the kinetic energy is zero and the spring potential energy is at its maximum. At x = 0, the spring potential energy is 0 and the kinetic energy is at its maximum. AP M 1993 MC 33 008 10.0 points A simple pendulum consists of a brass sphere of mass m = 1 kg suspended on a string of length L ≈ 1 meter. The pendulum oscillates with amplitude A = 1 cm and period T = 2 seconds. Which of the following alterations will change the period of the pendulum’s oscillations to T ′ = 1 second? 1. Using a longer string of length L′ ≈ 4 meters. 2. Using a longer string of length L′ ≈ 2 meters. 3. Reducing the oscillation amplitude to A′ = 0.25 cm. Explanation: The block oscillates from maximum displacements at x = A and x = −A. At those points the velocity is momentarily zero. 4. Using a lighter brass sphere of mass m′ ≈ 0.25 kg. 007 (part 2 of 2) 10.0 points Which statement about energy is correct? 5. Using a shorter string of length L′ ≈ 0.25 meters. correct 1. The kinetic energy of the block is at a minimum at x = 0. 6. Using a shorter string of length L′ ≈ 0.5 meters. 2. The potential energy of the spring is at a minimum at x = 0. correct 7. Increasing the oscillation amplitude to A′ = 4 cm. Version 001 – shmgravityII – holland – (1570) ′ 8. Using a heavier brass sphere of mass m ≈ 4 kg. Explanation: As long as the oscillation amplitude is small compared to the string length, the pendulum’s period is given by a simple formula s L T = 2π . g Note that this period does not depend on the mass of the pendulum’s bob (the brass sphere) or the oscillation amplitude, but only on the pendulum’s length L and the local gravitational field g. And since we cannot change g, the only way to change the period is to change the √ length L. Since T ∝ L and we want to halve the period, T′ 1s 1 = = , T 2s 2 we need √ 2 1 L′ L′ 1 1 √ = ⇒ = = 2 L 2 4 L 1 and hence L′ = L ≈ 0.25 meters. 4 AP M 1998 MC 10 009 10.0 points A pendulum with a period of 1 s on Earth, where the acceleration due to gravity is g, is taken to another planet, where its period is 2 s. The acceleration due to gravity on the other planet is most nearly 1. ag = g 2. ag = 2 g 3. ag = 4 g g 2 g 5. ag = correct 4 Explanation: 4. ag = 6 For a pendulum, the relationship between s ℓ the period and the acceleration is T = 2 π . g 1 T ∝ √ , so g r g2 T1 = T2 g1 r 1 g2 = 2 g1 1 g2 = g1 4 1 g2 = g1 , 4 which gives the acceleration on the other g planet as . 4 AP M 1998 MC 35 010 10.0 points An ideal massless spring is fixed to the wall at one end, as shown below. A block of mass M attached to the other end of the spring oscillates with amplitude A on a frictionless, horizontal surface. The maximum speed of the block is vm . k vm m −A 0 +A What is the force constant k of the spring? 1. k = 2. k = 3. k = 4. k = 5. k = 2 m vm 2A 2 m vm correct A2 mg A m g vm 2A 2 m vm 2 A2 Explanation: For the ideal harmonic oscillation of the spring system, the kinetic energy maximum is Version 001 – shmgravityII – holland – (1570) equal to the potential energy maximum which is also the total energy of the system, so we obtain 1 1 2 k A2 = m vm 2 2 2 m vm . k= A2 AP M 1998 MC 9 011 10.0 points The equation of motion of a simple harmonic oscillator is d2 x = −9 x , dt2 where x is displacement and t is time. What is the period of oscillation? 3 2π 2π 2. T = 9 9 3. T = 2π 2π 4. T = correct 3 1. T = 5. T = 6 π Explanation: d2 x = −ω 2 x , dt2 where ω is the angular frequency, so the period of oscillation is T = 2π 2π 2π =√ = . ω 3 9 AP B 1993 MC 6 012 10.0 points If Spacecraft X has twice the mass of Spacecraft Y , then what is true about X and Y ? I) On Earth, X experiences twice the gravitational force that Y experiences; II) On the Moon, X has twice the weight of Y; III) When both are in the same circular orbit, X has twice the centripetal acceleration of Y . 7 1. I, II, and III 2. III only 3. I only 4. I and II only correct 5. II and III only Explanation: I) gravitational force ∝ mass. II) weight ∝ mass. III) The centripetal acceleration is determined by ac = v2 , r so X and Y should have the same centripetal acceleration when they are in the same circular orbit. AP B 1993 MC 6 02 013 10.0 points If Spacecraft X has twice the mass of Spacecraft Y , then true statements about X and Y include which of the following? I) On Earth, X experiences twice the gravitational force that Y experiences. II) On the Moon, X has twice the weight of Y. III) When both are in the same circular orbit, X has twice the centripetal acceleration of Y . 1. II and III only 2. I only 3. I and II only correct 4. III only 5. I, II, and III Explanation: I) gravitational force ∝ mass II) weight ∝ mass III) The centripetal acceleration is deter- Version 001 – shmgravityII – holland – (1570) 8 mined by M , r2 where G is the gravitational constant, M is the planet’s mass and r is the radius of the spacecraft’s orbit. Thus X and Y should have the same centripetal acceleration when they are in the same circular orbit. ac = G AP B 1998 MC 39 014 10.0 points An object has a weight W when it is on the surface of a planet of radius R. What will be the gravitational force on the object after it has been moved to a distance of 4 R from the center of the planet? 1. F = W 2. F = 1 W correct 16 3. F = 16 W 4. F = 4 W 1 W 4 Explanation: On the surface of the planet, 5. F = GM m . R2 When the object is moved to a distance 4 R from the center of the planet, the gravitational force on it will be GM m F = (4 R)2 GM m = 16 R2 1 GM m = 16 R2 1 W . = 16 W= AP B 1993 MC 8 015 10.0 points Two spheres have equal densities and are subject only to their mutual gravitational attraction. Which quantity must have the same magnitude for both spheres? 1. kinetic energy 2. displacement from the center of mass 3. velocity 4. acceleration 5. gravitational force correct Explanation: Two spheres with the same density have different masses due to their relative sizes. Using Newton’s third law, F~1 = −F~2 . All of the other quantities (acceleration, velocity, kinetic energy, and displacement from the center of mass) have different magnitudes because the two spheres have different masses. AP B 1998 MC 39 01 016 10.0 points An object has a weight W when it is on the surface of a planet of radius R. What will be the gravitational force on the object after it has been moved to a distance of 4 R from the center of the planet? 1 W 2 1 2. F = W 64 1 3. F = W 4 1. F = 4. F = 16 W 5. F = 2 W 6. F = W 7. F = 1 W 8 Version 001 – shmgravityII – holland – (1570) 8. F = 4 W 1 W correct 16 Explanation: On the surface of the planet, 9. F = GM m . R2 When the object is moved to a distance 4 R from the center of the planet, the gravitational force on it will be GM m F = (4 R)2 GM m = 16 R2 1 GM m = 16 R2 1 = W . 16 W= AP B 1998 MC 40 017 10.0 points What is the kinetic energy of a satellite of mass m that orbits the Earth of mass M in a circular orbit of radius R? 1 GM m correct 2 R 1 GM m 2. K = 2 R2 GM m 3. K = R2 1 GM m 4. K = 4 R 1. K = 5. K = 0 Explanation: The gravitational force on the satellite provides the centripetal force needed to keep it in circular orbit: GM m v2 = F = F = m c G R2 R G M m , so m v2 = R K= 1 GM m 1 m v2 = . 2 2 R 9 AP M 1993 MC 22 018 10.0 points A newly discovered planet has twice the mass of the Earth, but the acceleration due to gravity on the new planet’s surface is exactly the same as the acceleration due to gravity on the Earth’s surface. What is the radius Rp of the new planet in terms of the radius R of Earth? 1. Rp = 4 R 2. Rp = 2 R √ 2 3. Rp = R 2 1 4. Rp = R 2 √ 5. Rp = 2 R correct Explanation: From Newton’s second law and the law of universal gravitation, the gravitational force near the surface is Mm r2 GM g= 2 . r Fg = m g = G Mp = 2 Me and gp = ge , so G Mp G Me 2 G Me = = 2 2 R Rp Rp2 2 1 = 2 2 R Rp √ Rp = 2 R . AP M 1993 MC 32 019 10.0 points A satellite is in an elliptical orbit around a planet as shown, with r1 and r2 being its closest and farthest distances, respectively, from the center of the planet. Version 001 – shmgravityII – holland – (1570) 1 r1 Planet r2 ~v2 Satellite 2 10 v M v M ~v1 D If the satellite has a speed v1 ≡ k~v1 k at its closest distance, what is its speed at its farthest distance? r2 v1 1. k~v2 k = r1 r1 + r2 2. k~v2 k = v1 2 3. k~v2 k = (r2 − r1 ) v1 r22 v1 r12 r2 5. k~v2 k = 12 v1 r2 r2 − r1 6. k~v2 k = v1 r1 + r2 r2 + r1 7. k~v2 k = v1 r1 − r2 r1 8. k~v2 k = v1 correct r2 Explanation: The angular momentum of the satellite S relative to the planet P remains the same as the satellite travels in the orbit, because the torque on the satellite is zero: 4. k~v2 k = m r1 v1 = m r2 v2 r1 v1 . v2 = r2 AP M 1998 MC 20 020 10.0 points Two identical stars, a fixed distance D apart, revolve in a circle about their mutual center of mass, as shown below. Each star has mass M and speed v. Which of the following is a correct relationship among these quantities? G is the universal gravitational constant. 2 G M2 D GM 2. v 2 = D 4 G M2 3. v 2 = D 4GM 4. v 2 = D 1. v 2 = 5. v 2 = M G D 2GM D GM 7. v 2 = D2 GM 8. v 2 = correct 2D Explanation: From circular orbital movement, the cen2 v2 v2 = . tripetal acceleration is a = D D 2 Using Newton’s second law of motion, the acceleration is 6. v 2 = a= F 1 G M2 GM = · = 2 M M D D2 F GM 2 v2 =a= = D M D2 GM . v2 = 2D AP M 1998 MC 7 8 Version 001 – shmgravityII – holland – (1570) 021 (part 1 of 2) 10.0 points A ball is tossed straight up from the surface of a small, spherical asteroid with no atmosphere. The ball rises to a height equal to the asteroid’s radius and then falls straight down toward the surface of the asteroid. What forces, if any, act on the ball while it is on the way up? 1. Only an increasing gravitational force that acts downward 2. No forces act on the ball. 3. Both a constant gravitational force that acts downward and a decreasing force that acts upward 4. Only a decreasing gravitational force that acts downward correct 5. Only a constant gravitational force that acts downward Explanation: There is no friction in the system, and the ball doesn’t have any contact with other objects, so the only force acting on the ball is the attractive gravitational force, which acts downward. ~ = −G M m r̂, the force will deFrom F r2 crease as the ball rises. 022 (part 2 of 2) 10.0 points The acceleration of the ball at the top of its path is 1. equal to one-half the acceleration at the surface of the asteroid. 2. zero. 3. equal to one-fourth the acceleration at the surface of the asteroid. correct 4. at its maximum value for the ball’s flight. 5. equal to the acceleration at the surface of the asteroid. Explanation: 1 1 F = m a ∝ 2 , so a ∝ 2 and r r 1 1 1 1 = ∝ a. a′ ∝ 2 2 (2 r) 4r 4 11