7.1 Defining Simple Harmonic Motion (SHM)

1. AEvery periodic motion is SHM, since SHM is any repeating motion.
   A student who thinks any repeating motion is SHM picks this. Repetition makes a motion periodic, but SHM also needs a restoring force proportional to the displacement: a ball bouncing to the same height every time is periodic but not SHM.
2. BSHM is not periodic, since the object's speed changes as it moves.
   A student who links 'periodic' to constant speed, as in uniform circular motion, picks this. Periodic means only that the motion repeats in equal time intervals; the object's speed changes within each cycle of SHM, but the same pattern repeats every period.
3. CSHM need not be periodic, since an oscillation that dies away is SHM.
   A student who calls any real spring or pendulum motion SHM picks this. An oscillation that dies away does not repeat exactly, so it is not periodic and is not SHM; SHM is an idealized motion that repeats with the same amplitude every cycle.
4. DEvery SHM is periodic, but not every periodic motion is SHM. Correct
   SHM is a special case of periodic motion. Every SHM repeats itself exactly in equal time intervals, so it is periodic; but a periodic motion is SHM only if the restoring force is proportional to the displacement from equilibrium, which many periodic motions do not satisfy.

CED 7.1.A.1 · Read this in Fix

1. AThe string's tension exerts a torque about the pivot that pulls the bob back toward the vertical.
   A student who credits the string with the restoring effect picks this. The tension acts along the string, so its line of action passes through the pivot and its torque about the pivot is zero. The restoring torque comes from gravity.
2. BGravity's torque about the pivot is proportional to the angle and turns the bob toward the vertical. Correct
   For small angular displacements the restoring torque exerted by gravity about the pivot is proportional to the angular displacement (τ = mgℓ sin θ, and sin θ is very nearly proportional to θ for small θ) and turns the bob back toward the vertical. That is the rotational form of the SHM condition.
3. CGravity's torque about the pivot acts in the same sense as the bob's motion, which keeps it swinging.
   A student who thinks a force must act along the motion to keep an object moving picks this. Gravity's torque turns the bob toward the vertical: along its motion while it swings down, but against its motion while it swings up and away.
4. DGravity's torque about the pivot acts opposite to the bob's motion, just as friction would.
   A student who thinks a restoring torque resists the motion, like friction, picks this. While the bob swings back toward the vertical, gravity's torque acts in the same sense as its motion and speeds it up; it opposes the motion only while the bob swings away.

CED 7.1.A.2.iii · Read this in Fix

1. AA block oscillating up and down on the end of a vertical ideal spring Correct
   Gravity exerts a constant force on the block, so it only moves the equilibrium position down to where the spring force balances the weight. Measured from that position, the net force on the block is proportional to its displacement and opposite to it, so the motion is SHM.
2. BA ball bouncing on a hard floor to the same height every time
   A student who thinks every periodic motion is SHM picks this. The motion repeats, but between bounces the only force on the ball is its constant weight, which does not depend on the ball's height, and at the floor it receives a brief large push. No force proportional to displacement acts, so it is not SHM.
3. CA block sliding back and forth across the bottom of a V-shaped track
   A student who thinks any force toward the equilibrium point gives SHM picks this. On each straight side the force along the track has constant magnitude and points toward the bottom, however far up the block is. The force is not proportional to the displacement, so the motion is periodic but not SHM.
4. DA simple pendulum released from rest 60° from the vertical
   A student who forgets the small-angle condition picks this. A pendulum is modeled as SHM only for small angles, where the restoring torque is proportional to the angle. At 60° the torque grows noticeably less than in proportion to the angle, so the motion is periodic but not SHM.

CED 7.1.A.1 · Read this in Fix

1. AGraph 4
   A student who thinks the force in SHM points the same way as the displacement picks this. Graph 4 is proportional, but F_x is positive when x is positive: the force pushes the object farther from equilibrium, so it would not oscillate at all.
2. BGraph 3
   A student who thinks any force toward equilibrium gives SHM picks this. The force in Graph 3 does point toward x = 0, but its magnitude is the same at every displacement. It is not proportional to the displacement, so the motion would be periodic but not SHM.
3. CGraph 2 Correct
   SHM requires a net force that is proportional to the displacement and opposite to it, ma_x = −kΔx. Graph 2 is a straight line through the origin with a negative slope: doubling x doubles F_x, and F_x is negative when x is positive, so it supports the claim.
4. DGraph 1
   A student who takes 'proportional' to mean 'increases with' picks this. The force in Graph 1 points toward equilibrium and grows with displacement, but the graph curves: doubling x far more than doubles F_x. Only a straight line through the origin shows proportionality.

Working SHM requires ma_x = −kΔx: the net force must be proportional to the displacement from equilibrium and opposite to it. On an F_x–x graph that is a straight line through the origin with a negative slope (slope −k): Graph 2. Graph 1 points the right way but is not proportional (curved). Graph 3 points the right way but has constant magnitude. Graph 4 is proportional but points away from equilibrium.

CED 7.1.A.2 · Read this in Fix

1. A20 m/s²
   A student who gives the spring's force the direction of the stretch gets +20 m/s². The stretched spring pulls the block back toward equilibrium, opposite to its displacement, which is what the minus sign in ma_x = −kΔx says.
2. B−20 m/s² Correct
   The displacement from equilibrium is Δx = 0.45 m − 0.40 m = +0.050 m. From ma_x = −kΔx, a_x = −(80)(0.050)/(0.20) = −20 m/s²: 20 m/s² back toward the equilibrium position, although the block is momentarily at rest.
3. C0 m/s²
   A student who thinks an object at rest has no acceleration picks this. The block's velocity is zero at the instant of release, but it is 0.050 m from equilibrium, so the spring exerts a net force of 4.0 N on it and it accelerates at 20 m/s².
4. D−180 m/s²
   A student who uses the spring's full length for Δx gets −(80)(0.45)/(0.20) = −180 m/s². Δx is the displacement from equilibrium, 0.45 m − 0.40 m = 0.050 m, not the length of the spring.

Working With no other horizontal forces, the equilibrium position is where the spring has its relaxed length, so Δx = 0.45 m − 0.40 m = +0.050 m. ma_x = −kΔx gives a_x = −kΔx/m = −(80 N/m)(0.050 m)/(0.20 kg) = −20 m/s². The block is at rest at this instant, but its acceleration is not zero: it points back toward equilibrium.

CED 7.1.A.2 · Read this in Fix

1. AOnly at instant Q
   A student who thinks the spring's force points the same way as the displacement picks this: that belief gives a leftward force at P and a rightward force at Q and R. The spring's force points opposite to the displacement, toward equilibrium, so it opposes the velocity at P and R, not at Q.
2. BAt all three instants
   A student who thinks a restoring force opposes the motion, like friction, picks this. At Q the block is to the right of equilibrium and moving left, back toward it: the force points left too, along the velocity, and speeds the block up.
3. CAt P and at R only Correct
   The spring's force always points toward the equilibrium position. At P (left of it) the force points right and at R (right of it) it points left; at both instants the block is moving away from equilibrium, so the force is opposite to the velocity. At Q the block is moving back toward equilibrium, along the force.
4. DAt none of the three
   A student who thinks the force on a moving object points the way it moves picks this. The spring's force points toward equilibrium whichever way the block moves. At P and R the block is moving away from equilibrium, so the force points opposite to its velocity and slows it down.

Working The spring's force is a restoring force: it points opposite to the displacement, toward the equilibrium position. P: displaced left, so the force points right; velocity left, so opposite. Q: displaced right, so the force points left; velocity left, so the same direction. R: displaced right, so the force points left; velocity right, so opposite. Answer: P and R, the instants at which the block is moving away from equilibrium.

CED 7.1.A.2.i · Read this in Fix

1. ADiagram 2 Correct
   An equilibrium position is a location at which the net force is zero. For a hanging block that is where the stretched spring's upward force balances the block's weight, so the two arrows have equal lengths. The block's upward velocity does not change this.
2. BDiagram 1
   A student who thinks the equilibrium position is where the spring has its relaxed length leaves out the spring force. For a hanging block the spring is stretched at equilibrium; with only F_g acting, the net force would be downward and the position would not be an equilibrium position.
3. CDiagram 3
   A student who thinks the net force on a moving object points the way it moves draws a larger upward force because the block is moving up. At the equilibrium position the net force is zero whatever the velocity, so F_s and F_g are equal in magnitude.
4. DDiagram 4
   A student who thinks a restoring force always opposes the motion draws a net downward force because the block is moving up. At the equilibrium position the displacement is zero, so the restoring (net) force is zero: F_s and F_g are equal in magnitude.

Working The equilibrium position is where the net force on the block is zero. At that position the spring is stretched just enough that the upward spring force has the same magnitude as the downward gravitational force, whichever way the block is moving: Diagram 2.

CED 7.1.A.2.ii · Read this in Fix

1. Ag + gA/d
   A student who measures the displacement from the spring's relaxed length, as if that were the equilibrium position, uses Δx = d + A and gets a = k(d + A)/m = g + gA/d. That is the spring's force alone divided by m; the weight, mg = kd, acts downward and cancels the kd part. The equilibrium position, where the net force is zero, is a distance d below the relaxed position.
2. Bg(A − d)/d
   A student who takes the spring's force to be kA and then subtracts the weight gets a net force kA − mg (negative, i.e. downward, if A < d) and a = g(A − d)/d. Because A is measured from the equilibrium position, where the spring already balances the weight, kA is already the net force; subtracting mg counts the weight twice.
3. C0
   A student who thinks that an object momentarily at rest has no acceleration picks 0. At release the block's velocity is zero, but the net force on it, kA, is not, so it accelerates upward at gA/d from that instant.
4. DgA/d Correct
   At the equilibrium position the net force is zero, so kd = mg and k = mg/d. At release the spring pulls up with k(d + A) while the weight pulls down with mg, so the net force is kA and a = kA/m = gA/d, directed toward the equilibrium position.

Working At the equilibrium position the net force is zero: kd = mg, so k = mg/d. At release the spring is stretched d + A, so it exerts an upward force k(d + A); the weight mg acts downward. Net upward force = k(d + A) − mg = kd + kA − mg = kA. Acceleration a = kA/m = (mg/d)A/m = gA/d, directed upward toward the equilibrium position. (Equivalently, ma_y = −kΔy with Δy = −A, the displacement from the equilibrium position, so a_y = +kA/m.)

CED 7.1.A.2.ii · Read this in Fix

1. A1.50
   A student who takes the acceleration to depend on displacement alone picks this: 0.15/0.10 = 1.50. The spring sets the force, k|Δx|; dividing by twice the mass halves the acceleration, giving 0.75.
2. B0.19
   A student who uses the spring's length for Δx picks this: (0.15/0.40) × 1/2 ≈ 0.19. Δx is the displacement from equilibrium, 0.10 m and 0.15 m, not the length of the spring; the factor is (0.15/0.10) × 1/2 = 0.75.
3. C0.00
   A student who thinks a block that is momentarily at rest has no net force and no acceleration picks this. At release the spring is compressed by 0.15 m and pushes the block with force k(0.15 m), so it accelerates away from the wall; the factor is 0.75.
4. D0.75 Correct
   From ma_x = −kΔx, the magnitude of the acceleration is k|Δx|/m, with Δx measured from equilibrium (the relaxed length here). The displacement grows from 0.10 m to 0.15 m (×1.5) and the mass doubles (×1/2): 1.5 × 1/2 = 0.75.

Working The net horizontal force is the spring force, so m|a_x| = k|Δx| and |a_x| = k|Δx|/m, with Δx measured from the relaxed length (the equilibrium position here). First block: |Δx| = 0.40 − 0.30 = 0.10 m, so a = k(0.10)/m. New block: |Δx| = 0.30 − 0.15 = 0.15 m and mass 2m, so |a| = k(0.15)/(2m). Ratio = (0.15/0.10)/2 = 0.75. Being momentarily at rest does not make the force or the acceleration zero.

CED 7.1.A.2 · Read this in Fix