7.1 Defining Simple Harmonic Motion (SHM)

1. AIt is SHM, as any motion that repeats in equal time intervals is SHM.
   A student who thinks every periodic motion is SHM picks this. Repeating in equal time intervals makes the motion periodic; SHM is the special case in which the restoring force is proportional to the displacement, which is not true here, since the force along each side is mg sin θ at every point.
2. BIt is SHM, since the net force on it points back toward the bottom at all times.
   A student who thinks any force directed back toward equilibrium produces SHM picks this. The force does point toward the bottom, but it has the same magnitude, mg sin θ, at every point on a side; SHM also needs the magnitude to be proportional to the displacement.
3. CIt is periodic but not SHM: along each side the net force has a constant magnitude. Correct
   On either side the net force on the block is the component of its weight along the incline, mg sin θ, directed down the slope toward the bottom; the normal force cancels the other component. That force has the same magnitude whether the block is near the bottom or far up the side, so it is not proportional to the displacement from the bottom. The motion repeats, so it is periodic, but it is not SHM.
4. DIt is periodic but not SHM: the constant force along each side keeps its speed constant.
   A student who thinks a constant force produces a constant speed picks this. The net force mg sin θ along each side gives the block a constant acceleration, g sin θ, so its speed changes steadily; the motion is not SHM because that force is not proportional to the displacement, not because the speed is constant.

CED 7.1.A.1 · Read this in Fix

1. AThe block is at rest for an instant, since it is in equilibrium there.
   A student who thinks equilibrium means being at rest picks this. Equilibrium means zero net force. The block is momentarily at rest at the top and bottom of its motion, where the net force is largest.
2. BThe net force on the block is zero, although the block is moving. Correct
   By definition, the equilibrium position is where the net force is zero: the spring force balances the weight. An oscillating block is not at rest there; it moves through it, and that is where its speed is greatest.
3. CThe spring is at its natural length, so it exerts no force on the block.
   A student who thinks the equilibrium position is where the spring is unstretched picks this. At the natural length the only force on the block is its weight, so the net force is mg. At equilibrium the spring is stretched by mg/k.
4. DThe net force on the block is largest there, as its speed is largest there.
   A student who thinks the force is largest where the speed is largest picks this. The block's speed is largest at equilibrium, but the net force there is zero; the net force is largest at the turning points.

CED 7.1.A.2.ii · Read this in Fix

1. AGraph 1
   A student who thinks the magnitude of the force only has to be proportional to the displacement, whatever its direction, picks the straight line with positive slope. That force points the same way as the displacement and pushes the object farther from x = 0, so the object does not oscillate about x = 0 at all.
2. BGraph 2
   A student who thinks any restoring force that grows with displacement gives SHM picks the curve that steepens away from x = 0. That force is opposite to the displacement, but doubling x more than doubles it, so it is not proportional to x and the oscillation is not SHM.
3. CGraph 3
   A student who thinks a restoring force of constant magnitude produces SHM picks the pair of horizontal segments. That force always points back toward x = 0, but it has the same size at every distance, so it is not proportional to the displacement.
4. DGraph 4 Correct
   The straight line through the origin with negative slope is F = −kx: the force is proportional to the displacement from x = 0 and opposite to it, so it is a restoring force of the kind that produces SHM about x = 0.

Working SHM about x = 0 needs F = −kx with k > 0: a straight line through the origin with negative slope. The positive-slope line pushes the object away from x = 0; the constant-magnitude force and the steepening curve point back toward x = 0 but are not proportional to x.

CED 7.1.A.2 · Read this in Fix

1. A7.0 N
   A student who measures Δx from the spring's natural length picks this: (50 N/m)(0.14 m) = 7.0 N. That is the spring force alone; the block's weight of 5.0 N acts in the opposite direction, so the net force is 2.0 N.
2. B2.0 N Correct
   At equilibrium the spring is stretched 0.10 m, where kΔx balances mg. Pulled 0.040 m lower, the spring force is (50)(0.14) = 7.0 N up and the weight is 5.0 N down, so the net force is 2.0 N up: k times the displacement from equilibrium, (50)(0.040) = 2.0 N.
3. C3.0 N
   A student who measures the displacement from equilibrium and then also subtracts the weight picks this: |(50)(0.040) − 5.0| = 3.0 N. Measuring from the equilibrium position already includes gravity, so counting the weight again is a double count.
4. D0.0 N
   A student who thinks an object at rest has no net force on it picks this. The block is at rest only for an instant; it is 0.040 m from equilibrium, so the net force on it is k(0.040 m) = 2.0 N toward equilibrium.

Working At the equilibrium position the spring is stretched by mg/k = (0.50 kg)(10 m/s²)/(50 N/m) = 0.10 m. Just after release the stretch is 0.10 m + 0.040 m = 0.14 m, so the spring force is (50 N/m)(0.14 m) = 7.0 N upward, and the weight is (0.50 kg)(10 m/s²) = 5.0 N downward. Net force = 7.0 N − 5.0 N = 2.0 N upward. Equivalently, F = k × (displacement from equilibrium) = (50 N/m)(0.040 m) = 2.0 N.

CED 7.1.A.2 · Read this in Fix

1. AAt Q, but not at P Correct
   The restoring force is opposite to the block's displacement from equilibrium. At Q the block is left of the equilibrium position, so the force points to the right; at P it is right of the equilibrium position, so the force points to the left. The direction of the velocity does not matter.
2. BAt P, but not at Q
   A student who takes the force to point along the displacement, ignoring the minus sign in F_x = −kΔx, picks this: at P the displacement is to the right. A restoring force is opposite to the displacement, so at P it points left and at Q it points right.
3. CAt both P and at Q
   A student who thinks a restoring force always opposes the velocity picks this, since the block moves left at both instants. The restoring force opposes the displacement, not the velocity: at P it points left, in the direction of motion, speeding the block up toward equilibrium.
4. DAt neither P nor Q
   A student who thinks the force on a moving object points along its velocity picks this, since the block moves left at both instants. At Q the block is left of equilibrium and moving away from it, so the restoring force points right, opposite to its velocity, and slows it.

CED 7.1.A.2.i · Read this in Fix

1. AF_x = 0
   A student who thinks springs on opposite sides of a block always cancel picks this. They balance at x = 0, but when the block moves right, one spring stretches and pulls left while the other compresses and pushes left; the two forces add.
2. BF_x = −kx/2
   A student who treats the two springs as a series combination picks this: k_eq = k·k/(k + k) = k/2. In series the springs share one force and their stretches add; here each spring changes length by the full displacement x, so their forces add.
3. CF_x = −2kx Correct
   The left spring is stretched by x and pulls the block left with force kx; the right spring is compressed by x and pushes the block left with force kx. Both act toward x = 0, so F_x = −2kx, a restoring force proportional to the displacement.
4. DF_x = −kx
   A student who thinks a compressed spring exerts no force counts only the stretched left spring. The compressed right spring pushes on the block with a force of magnitude kx, also to the left, so the net force is −2kx.

Working Left spring: stretched by x, so it pulls the block toward the left wall: F₁ = −kx. Right spring: compressed by x, so it pushes the block away from the right wall, also to the left: F₂ = −kx. Net: F_x = F₁ + F₂ = −2kx. The net force is proportional to the displacement and opposite to it, so the block undergoes SHM about x = 0 with k_eq = 2k.

CED 7.1.A.2 · Read this in Fix

1. A0.24 m
   A student who thinks the equilibrium position is where the block is at rest picks the lowest point of the motion, where it momentarily stops: mg sin 37° · s = (1/2)ks² gives s = 0.24 m. There the spring force is 24 N up the slope and the weight component is 12 N down it, so the net force is 12 N, not zero.
2. B0.16 m
   A student who takes the component of the weight along the incline as mg cos θ picks this: (2.0)(10)(0.80)/100 = 0.16 m. The component along the incline is mg sin θ = 12 N; mg cos θ is the component perpendicular to it, balanced by the normal force.
3. C0.20 m
   A student who thinks the spring must support the block's whole weight picks this: (2.0)(10)/100 = 0.20 m. The normal force balances the perpendicular component of the weight, so the spring balances only mg sin 37° = 12 N.
4. D0.12 m Correct
   The equilibrium position is where the net force is zero. Along the incline the spring force must balance the component of the weight down the slope: kΔx = mg sin 37° = (2.0)(10)(0.60) = 12 N, so Δx = 12 N/(100 N/m) = 0.12 m.

Working At the equilibrium position the net force is zero. Perpendicular to the incline the normal force balances mg cos 37°; along the incline the spring force balances the component of the weight down the slope: kΔx = mg sin 37°. Δx = (2.0 kg)(10 m/s²)(0.60)/(100 N/m) = 12 N/(100 N/m) = 0.12 m.

CED 7.1.A.2.ii · Read this in Fix

1. AR > Q > P
   A student who ranks by the signed displacement picks this: +0.05 > +0.02 > −0.08. The magnitude of the restoring force depends on |Δx|; P, 0.08 m from O, has the largest restoring force.
2. BP > R > Q Correct
   The magnitude of the restoring force is k|Δx|, so it is ranked by distance from O: P is 0.08 m away, R is 0.05 m away and Q is 0.02 m away. The negative sign at P gives only the direction of the force (to the right).
3. CQ > R > P
   A student who thinks the force is largest where the block moves fastest, near O, picks this: Q is closest to O. The restoring force is proportional to the distance from O, so it is smallest at Q and largest at P.
4. DP = Q = R
   A student who thinks the restoring force in SHM has the same magnitude at every position picks this. For SHM the magnitude is proportional to the displacement, k|Δx|, so it differs at the three positions.

CED 7.1.A.2 · Read this in Fix

1. AWherever it is, twice the distance from equilibrium gives twice the force, pointing back. Correct
   This is the defining condition for SHM, F_x = −kΔx, in the form of evidence: a force that doubles when the distance from equilibrium doubles is proportional to the displacement, and pointing back makes it a restoring force.
2. BIt passes the same point, moving the same way, at equal intervals, cycle after cycle.
   A student who thinks any periodic motion is SHM picks this. It shows only that the motion is periodic; a ball bouncing elastically on a floor also does this, and its motion is not SHM.
3. CThe force on it points back toward the point midway between its turning points.
   A student who thinks any force directed back toward equilibrium gives SHM picks this. A force of constant magnitude toward that point also fits this observation and gives periodic motion that is not SHM.
4. DThe force on it is larger whenever it is farther from equilibrium, on either side of it.
   A student who reads 'proportional' as 'increases with' picks this. A force that grows as Δx³, for example, also fits this observation, but it is not proportional to the displacement, so the motion would not be SHM.

CED 7.1.A.2 · Read this in Fix