- A velocity–time graph for SHM shows v crossing zero at t=0 with a positive slope. Equilibrium is at x=0. At t=0, where is the mass relative to equilibrium?
Answer: At x=−A (left turning point)
This question assesses the skill of representing and analyzing simple harmonic motion (SHM) in AP Physics 1. On velocity-time graphs, crossing v=0 with positive slope means transitioning from negative to positive velocity. This occurs at x=-A, where the mass turns from -x to +x direction. Acceleration is positive at x=-A, causing the velocity increase. A common distractor is choice D, which wrongly assumes v=0 only at x=0, but that's maximum velocity. A transferable strategy is to analyze graph slopes at crossings to determine turning point locations, applicable to kinematics in various contexts. - A cart attached to a spring oscillates horizontally. Equilibrium is x=0 and +x is to the right. At some instant, the cart is at x=−A/2 and moving right. Which statement about the acceleration is correct?
Answer: Acceleration is to the right because x is negative.
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, the acceleration follows a = -ω²x, always pointing toward equilibrium with magnitude proportional to displacement. At x=-A/2 (negative position), the acceleration is a = -ω²(-A/2) = +ω²A/2, which is positive (to the right). The direction of velocity doesn't determine acceleration direction; only position does. The acceleration points right because it must restore the cart toward equilibrium at x=0, and since the cart is on the negative side, the restoring acceleration is positive. Choice D incorrectly states the acceleration is to the left, misunderstanding that for negative x, the restoring acceleration is positive. Remember: in SHM, acceleration direction depends solely on position relative to equilibrium. - A particle in SHM moves along a line with equilibrium at x=0 and +x to the right. At an instant, its velocity is zero. Which must be true at that instant?
Answer: The particle is at a turning point (x=±A).
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, velocity is zero only at the turning points where the particle momentarily stops before reversing direction. These turning points occur at the amplitude positions x=±A, where the particle is farthest from equilibrium. At these points, all energy is potential and acceleration is maximum (not zero), pointing toward equilibrium. The particle cannot have zero velocity at equilibrium because it moves fastest there. Choice A incorrectly suggests zero velocity at equilibrium, confusing the conditions for zero velocity and zero acceleration. Remember: in SHM, velocity is zero only at turning points (x=±A) where acceleration is maximum. - A cart oscillates in SHM about equilibrium x=0 (right positive). At an instant when the cart is at x=−A/2, it is moving left. Which statement about velocity and acceleration directions is correct?
Answer: Velocity left; acceleration right.
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, velocity indicates motion direction while acceleration always points toward equilibrium. With the cart at x=-A/2 (left of equilibrium) moving left, velocity is negative (leftward). Since the cart is left of equilibrium, acceleration must point rightward (positive) toward x=0. This creates opposing velocity and acceleration, which causes the cart to slow down before eventually reversing direction. Choice A incorrectly assumes acceleration and velocity must have the same direction. The strategy is to determine acceleration direction by asking "which way is equilibrium?" regardless of velocity direction.
- A mass on a spring undergoes SHM. Equilibrium is x=0 and +x is to the right. At t=0, the mass is at x=+A and begins moving left. Which statement about the acceleration at t=0 is correct?
Answer: The acceleration points to the left (negative) with maximum magnitude.
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, the restoring force and acceleration always point toward equilibrium, with magnitude proportional to displacement: a = -ω²x. At t=0, the mass is at x=+A (maximum positive displacement), so the acceleration must point toward equilibrium at x=0, which is to the left (negative direction). The acceleration magnitude is maximum at the turning points where |x|=A, giving |a|=ω²A. The common misconception in choice A incorrectly assumes that zero velocity implies zero acceleration, but in SHM these quantities are independent—acceleration depends only on position. To solve SHM problems, always remember: acceleration points toward equilibrium with magnitude proportional to displacement.
- A spring-mass system oscillates with equilibrium at x=0 and +x right. At a turning point on the left side, the mass is momentarily at rest. Which statement about acceleration there is correct?
Answer: Acceleration is maximum to the right because the mass is farthest from equilibrium.
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, acceleration follows a = -ω²x and always points toward equilibrium. At the left turning point, x=-A (maximum negative displacement), so acceleration is a = -ω²(-A) = +ω²A, which is positive (to the right). The acceleration magnitude is maximum at turning points because displacement from equilibrium is maximum. At turning points, velocity is zero but acceleration is maximum, demonstrating these quantities are independent. Choice B incorrectly suggests acceleration is to the left at the leftmost position, failing to recognize that acceleration must point toward equilibrium. Remember: in SHM, acceleration always points toward equilibrium with magnitude proportional to displacement.
- A position–time graph for SHM labels equilibrium at x=0. At t=1.0s the graph crosses x=0 with a positive slope (moving +x). What is the sign of the velocity at t=1.0s?
Answer: Positive
This question assesses the skill of representing and analyzing simple harmonic motion (SHM) in AP Physics 1. On position-time graphs, the slope represents velocity, with positive slope indicating positive velocity direction. At equilibrium crossings (x=0), velocity sign matches the slope's sign. Acceleration is zero at x=0, independent of velocity. A common distractor is choice C, which wrongly ties velocity to zero at x=0, but velocity is maximum there. A transferable strategy is to always compute derivatives from graphs: slope for velocity, curvature for acceleration, applicable to any motion graph.
- A block in SHM about equilibrium x=0 is observed at x=+A/2 while moving to the right (positive direction). Which statement about its acceleration is correct at that instant?
Answer: Acceleration is to the left and nonzero.
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, acceleration always points toward equilibrium with magnitude proportional to displacement: a = -ω²x. When the block is at x=+A/2 (right of equilibrium), the acceleration must point leftward (negative direction) toward equilibrium. The velocity direction (rightward) doesn't determine acceleration direction; the block is moving away from equilibrium but will eventually slow down and return. Choice B incorrectly states acceleration points right, which would push the block further from equilibrium. The strategy is to always draw acceleration vectors pointing toward equilibrium, regardless of velocity direction.
- A velocity–time graph for SHM crosses v=0 at t=0.30s while the mass is at x=+A (rightmost turning point from equilibrium). Immediately after t=0.30s, which direction is the mass moving?
Answer: Toward −x, because it must reverse direction at a turning point
This question assesses the skill of representing and analyzing simple harmonic motion (SHM) in AP Physics 1. In SHM representations, position reaches extremes at +A or -A where velocity is zero, marking turning points. Velocity-time graphs show sinusoidal behavior, with zeros corresponding to these turning points and the direction after indicating the reversal. Acceleration, always toward equilibrium, causes the velocity to change sign at these points. A common distractor is choice C, which incorrectly suggests the mass stops permanently at v=0, ignoring the ongoing oscillation driven by the restoring force. A transferable strategy is to interpret graph crossings and slopes to infer direction changes, applying this to any oscillatory motion analysis.
- A mass oscillates in SHM about equilibrium x=0. At some instant, the velocity is at its maximum positive value (moving right fastest). What is the position at that instant?
Answer: x=0, because speed is greatest at equilibrium.
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, velocity reaches maximum magnitude when passing through equilibrium (x=0), where all energy is kinetic. At the turning points (x=±A), velocity is zero as all energy is potential. Maximum positive velocity (rightward at greatest speed) occurs when crossing equilibrium while moving right. Choice A incorrectly claims maximum speed occurs at endpoints, where the object is actually momentarily at rest. The strategy is to remember the energy relationship: maximum kinetic energy (and thus maximum speed) occurs at equilibrium where potential energy is zero. - A velocity-time graph for SHM shows v at a maximum positive value at t=0 while x=0 is equilibrium. Where is the mass at t=0?
Answer: At x=0
This question assesses the skill of representing and analyzing simple harmonic motion (SHM) in AP Physics 1. Velocity-time graphs in SHM are sinusoidal, with maximum velocity occurring at equilibrium where position x=0. Position relates to velocity via phase: max positive velocity corresponds to crossing x=0 moving toward +x. Acceleration, the slope of velocity-time, is zero at max velocity points. A common distractor is choice A, which incorrectly places max velocity at x=+A, but that's where velocity is zero. A transferable strategy is to recall phase relationships: velocity leads position by π/2 in SHM, helping locate positions from velocity data. - A mass-spring oscillator has equilibrium at x=0 and +x right. At an instant, the mass is at x=+A/2. Which statement about the acceleration magnitude is correct?
Answer: It is half the maximum acceleration magnitude.
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, acceleration magnitude is |a| = ω²|x|, proportional to displacement from equilibrium. At x=+A/2, the acceleration magnitude is |a| = ω²(A/2) = (1/2)ω²A. The maximum acceleration magnitude occurs at the amplitude positions (x=±A) where |a|max = ω²A. Therefore, at x=A/2, the acceleration magnitude is half the maximum. The sign of position doesn't affect the magnitude calculation. Choice B incorrectly suggests any displacement gives maximum acceleration, not recognizing the proportional relationship. To find acceleration magnitude in SHM, use |a| = ω²|x| where |x| is the distance from equilibrium.
- A mass-spring oscillator has equilibrium at x=0 and +x to the right. At t=0 the mass is at x=0 moving right. Which is true about its acceleration at t=0?
Answer: Acceleration is zero because the displacement is zero.
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, acceleration is determined by position according to a = -ω²x, where x is displacement from equilibrium. At t=0, the mass is at equilibrium (x=0), so the acceleration is a = -ω²(0) = 0. The fact that the mass is moving right with maximum speed doesn't affect the acceleration at this instant. At equilibrium, all energy is kinetic and there's no restoring force. Choice B incorrectly suggests acceleration is maximum because velocity is maximum, but these quantities are 90° out of phase in SHM. To analyze SHM motion, always remember: acceleration depends only on position, reaching zero at equilibrium and maximum at turning points.
- A block in SHM has equilibrium at x=0 and +x upward. At one instant the block passes through x=0 moving upward. Which describes the acceleration at that instant?
Answer: Acceleration is zero because the displacement from equilibrium is zero.
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, acceleration is given by a = -ω²x, where x is the displacement from equilibrium. When the block passes through equilibrium (x=0), the acceleration must be zero regardless of the velocity. At equilibrium, the restoring force is zero because there's no displacement to restore. The velocity is maximum at equilibrium because all the energy is kinetic, but this doesn't affect the acceleration. Choice C incorrectly assumes that maximum speed implies maximum acceleration, confusing the phase relationship between these quantities. To analyze SHM, always use the fundamental relationship: acceleration depends only on position, not velocity.
- A mass on a spring oscillates about equilibrium x=0. At time t1, the mass passes through x=0 moving right. One-quarter period later, at t1+T/4, what is the position and velocity direction?
Answer: x=+A with zero velocity.
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, one complete cycle takes period T, with quarter-period intervals marking key positions. Starting at equilibrium (x=0) moving right, the mass reaches maximum positive displacement (x=+A) after T/4, where velocity becomes zero at the turning point. The motion follows a sinusoidal pattern: from equilibrium with maximum velocity to maximum displacement with zero velocity takes exactly one-quarter period. Choice D incorrectly suggests the mass returns to equilibrium in T/4, which would only be a half-cycle. The strategy is to visualize SHM as circular motion projected onto a line, where T/4 represents a 90° rotation. - A cart oscillates in SHM along the x-axis with equilibrium at x=0 (right is positive). At some instant it is at x=−A and begins moving right. Which statement about acceleration at that instant is correct?
Answer: Acceleration is maximum and directed to the right.
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, acceleration always points toward equilibrium and is proportional to displacement: a = -ω²x. When the cart is at x=-A (leftmost position), the displacement is negative, making acceleration positive (rightward). At the turning points (±A), acceleration reaches its maximum magnitude because displacement is maximum. Choice A incorrectly assumes acceleration is zero at turning points, confusing zero velocity with zero acceleration. The strategy is to remember that in SHM, acceleration always points toward equilibrium and has maximum magnitude at the endpoints.
- An object in SHM has equilibrium at x=0 and +x upward. At some instant it is at x=0 moving downward. Which describes its speed and acceleration at that instant?
Answer: Speed is maximum and acceleration is zero.
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, position and velocity vary sinusoidally with a 90° phase difference. At equilibrium (x=0), acceleration is a = -ω²(0) = 0 because there's no displacement to restore. Speed is maximum at equilibrium because all energy is kinetic—the object moves fastest as it passes through the center. The direction of motion (downward) doesn't affect these magnitudes. Choice A incorrectly suggests zero speed at equilibrium, confusing equilibrium with turning points where velocity is zero. Remember: at equilibrium in SHM, speed is maximum and acceleration is zero.
- A position–time graph of SHM (equilibrium at x=0) shows the mass at x=0 and moving in the −x direction at t=0. Which best describes the acceleration at t=0?
Answer: Zero, because x=0 implies net force is zero
This question assesses the skill of representing and analyzing simple harmonic motion (SHM) in AP Physics 1. In SHM, position-time graphs show equilibrium at x=0, where the restoring force and thus acceleration are zero. Velocity can be non-zero at x=0, as it's the slope of the position graph, and direction is independent of acceleration at that instant. Acceleration depends solely on position via a = - (ω²)x, making it zero regardless of velocity. A common distractor is choice D, which confuses maximum speed with maximum acceleration, but acceleration is zero while speed is max at equilibrium. A transferable strategy is to separate kinematic variables: use position for acceleration and its derivative for velocity in graph interpretations. - A mass on a spring undergoes SHM. The position–time graph shows x=0 (equilibrium) at t=0 with the mass moving in the +x direction, reaching a maximum at t=0.50s. What is the sign of the acceleration at t=0.50s?
Answer: Negative, because acceleration points toward equilibrium when x>0
This question assesses the skill of representing and analyzing simple harmonic motion (SHM) in AP Physics 1. In SHM, the position-time graph is sinusoidal, with the mass oscillating between +A and -A around equilibrium at x=0. Velocity is the derivative of position, zero at maximum displacements where the mass turns around, and maximum at equilibrium. Acceleration is - (ω²)x, so it is maximum in magnitude and opposite to the displacement, pointing toward equilibrium. A common distractor is choice A, which wrongly claims acceleration is zero at turning points, but acceleration is actually at its peak magnitude there since the restoring force is greatest. A transferable strategy is to use the SHM equations like a = - (k/m) x to determine signs and magnitudes directly from position. - A position–time graph for SHM shows equilibrium at x=0. At t=0, the mass is at x=−A and then moves in the +x direction. At what position is the speed greatest?
Answer: x=0
This question assesses the skill of representing and analyzing simple harmonic motion (SHM) in AP Physics 1. Position in SHM varies sinusoidally, with velocity peaking in magnitude at x=0 due to conservation of energy, where potential energy is minimum and kinetic is maximum. At extremes like x=±A, velocity is zero as all energy is potential. Acceleration is zero at equilibrium and maximum at amplitudes, unrelated directly to speed's location. A common distractor is choice D, which falsely claims constant speed, but speed varies throughout SHM, maximum only at equilibrium. A transferable strategy is to link energy conservation to velocity maxima at equilibrium positions in any conservative oscillatory system.
