Representing and Analyzing SHM
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AP Physics 1 › Representing and Analyzing SHM
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?
Acceleration is maximum upward because speed is maximum.
Acceleration is maximum downward because position is zero.
Acceleration is upward because velocity is upward.
Acceleration is zero because the displacement from equilibrium is zero.
Explanation
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-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 $t=0$?
Acceleration is zero because the displacement is zero.
Acceleration is maximum to the left because velocity is maximum.
Acceleration is maximum to the right because velocity is to the right.
Acceleration is zero only if the speed is zero.
Explanation
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 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?
The particle is at a turning point ($x=\pm A$).
The particle is at equilibrium ($x=0$).
The acceleration is zero.
The particle’s speed is maximum.
Explanation
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 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?
It is maximum because the mass is displaced from equilibrium.
It equals the maximum acceleration magnitude because $x$ is positive.
It is zero because the mass is not at $x=0$.
It is half the maximum acceleration magnitude.
Explanation
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 oscillates in SHM along the $x$-axis with equilibrium at $x=0$ and +$x$ right. At an instant, the acceleration is positive. Which must be true about the position $x$ at that instant?
$x>0$
$x<0$
The sign of $x$ cannot be determined from acceleration.
$x=0$
Explanation
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, acceleration is given by a = -ω²x, which means acceleration and position have opposite signs. If acceleration is positive (to the right), then -ω²x > 0, which requires x < 0. This means the mass must be on the negative (left) side of equilibrium for the restoring acceleration to point right (positive). The acceleration always points toward equilibrium, so positive acceleration occurs when the mass is displaced to the left. Choice A incorrectly suggests x > 0, which would give negative acceleration. To solve SHM problems, use the fundamental relationship: acceleration and displacement always have opposite signs.
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?
Acceleration is zero because the cart is not at a turning point.
Acceleration is to the right because $x$ is negative.
Acceleration is to the right because the cart is moving right.
Acceleration is to the left because the restoring acceleration points toward equilibrium.
Explanation
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 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?
The acceleration points to the left (negative) with maximum magnitude.
The acceleration is zero because the velocity is momentarily zero.
The acceleration is zero because the position is at a maximum.
The acceleration points to the right (positive) with maximum magnitude.
Explanation
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 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?
Acceleration is maximum and directed to the right.
Acceleration is minimum (most negative) because the cart is farthest left.
Acceleration is maximum and directed to the left.
Acceleration is zero because the cart is momentarily at rest.
Explanation
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.
A mass–spring system oscillates about equilibrium $x=0$. At a certain time, the position is $x=0$ and the mass is moving to the left (negative direction). Which is the direction of the acceleration at that time?
Right, because acceleration always points opposite the direction of motion.
Cannot be determined without the amplitude.
Left, because acceleration always points in the direction of motion.
Zero, because the displacement from equilibrium is zero.
Explanation
This question tests understanding of representing and analyzing simple harmonic motion (SHM). In SHM, acceleration is proportional to displacement from equilibrium: a = -ω²x. When the mass is at equilibrium position (x=0), the displacement is zero, making the acceleration zero regardless of velocity direction. The velocity being leftward tells us about the motion's direction but doesn't affect the acceleration at equilibrium. Choice A incorrectly assumes acceleration points in the direction of motion, which violates SHM principles. The key strategy is to remember that acceleration in SHM depends only on position (displacement from equilibrium), not on velocity.
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?
Acceleration is zero because the position is momentarily constant.
Acceleration is maximum to the right because the mass is farthest from equilibrium.
Acceleration is zero because the velocity is zero.
Acceleration is maximum to the left because the mass is farthest left.
Explanation
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.