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 zero because the displacement from equilibrium is zero.
Acceleration is upward because velocity is upward.
Acceleration is maximum downward because position 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 only if the speed 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 because the displacement 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 acceleration is zero.
The particle is at a turning point ($x=\pm A$).
The particle’s speed is maximum.
The particle is at equilibrium ($x=0$).
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 equals the maximum acceleration magnitude because $x$ is positive.
It is maximum because the mass is displaced from equilibrium.
It is half the maximum acceleration magnitude.
It is zero because the mass is not at $x=0$.
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 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 left because the restoring acceleration points toward equilibrium.
Acceleration is to the right because the cart is moving right.
Acceleration is to the right because $x$ is negative.
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 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?
The sign of $x$ cannot be determined from acceleration.
$x<0$
$x=0$
$x>0$
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 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 is zero because the velocity is momentarily zero.
The acceleration points to the left (negative) with maximum magnitude.
The acceleration points to the right (positive) with maximum magnitude.
The acceleration is zero because the position is at a maximum.
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 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 maximum to the right because the mass is farthest from equilibrium.
Acceleration is zero because the position is momentarily constant.
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.
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?
Speed is zero and acceleration is maximum upward.
Speed is maximum and acceleration is zero.
Speed is maximum and acceleration is maximum upward.
Speed is minimum and acceleration is zero.
Explanation
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 mass on a spring oscillates about equilibrium $x=0$. At time $t_1$, the mass passes through $x=0$ moving right. One-quarter period later, at $t_1+T/4$, what is the position and velocity direction?
$x=0$ with maximum speed to the right.
$x=0$ with maximum speed to the left.
$x=-A$ with zero velocity.
$x=+A$ with zero velocity.
Explanation
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.