Simple Harmonic Motion Practice Problems and Answers

Practice problems

Work each one before opening the solution. Every step is shown, with the arithmetic left in so you can find where yours diverged.

1. 1. Period and frequency of a block on a spring

   A 0.90 kg block sits on a frictionless horizontal surface, attached to a spring of spring constant $k = 90\ \text{N/m}$. The block is pulled 0.080 m from equilibrium and released. Find (a) the period of the oscillation and (b) the frequency. (c) State what happens to both if the block is instead pulled 0.16 m aside before release.

   Show the worked solution

   1. (a) The period of a mass on an ideal spring is $T_s = 2\pi\sqrt{m/k}$. Do the fraction first: $m/k = (0.90\ \text{kg})/(90\ \text{N/m}) = 0.010$. The unit works out to $\text{s}^2$, since $\text{kg}/(\text{N/m}) = \text{kg}\cdot\text{m/N} = \text{s}^2$, which is a useful check that you divided the right way round.

   2. $\sqrt{0.010\ \text{s}^2} = 0.10\ \text{s}$, so $T = 2\pi(0.10) = 0.628\ \text{s}$.

   3. (b) $f = 1/T = 1/0.628 = 1.59\ \text{Hz}$. Check the pair against each other: $(1.59\ \text{Hz})(0.628\ \text{s}) = 1.00$, as $T = 1/f$ requires.

   4. (c) Nothing happens to either one. The amplitude does not appear in $T_s = 2\pi\sqrt{m/k}$, so it cannot change the period, and the frequency is the period's reciprocal. Essential knowledge 7.3.A.2 states this directly: changing the amplitude of a system exhibiting simple harmonic motion will not change the period of that system.

   5. It is worth being clear about what the bigger amplitude does change. The block now travels twice as far in each half cycle, so it must also move faster: at double the amplitude the maximum speed doubles and the total energy quadruples. The extra distance and the extra speed cancel exactly, which is the physical reason the period holds still.

   6. Sanity check on the size. A period of 0.628 s is about two thirds of a second per full there-and-back trip, so the block crosses the middle roughly three times a second. That is a plausible pace for a kilogram on a spring you could stretch by hand.

   (a) $T = 0.628\ \text{s}$. (b) $f = 1.59\ \text{Hz}$. (c) Both are unchanged. Amplitude is absent from $T_s = 2\pi\sqrt{m/k}$, so doubling it changes the maximum speed and quadruples the energy while leaving the timing alone.

2. 2. Counting cycles: period, frequency, and which oscillator is faster

   A student times an oscillating mass and counts 45 complete oscillations in 18.0 s. (a) Find its period and frequency. (b) A second oscillator has a period of 0.250 s. Which has the higher frequency, and by what factor? (c) How many complete oscillations does the first one make in 2.0 minutes?

   Show the worked solution

   1. (a) The period is the time for one full cycle, so divide the total time by the number of cycles: $T = (18.0\ \text{s})/45 = 0.400\ \text{s}$.

   2. The frequency is cycles per second, which is the same division the other way up: $f = 45/(18.0\ \text{s}) = 2.50\ \text{Hz}$. Check with the printed relation: $f = 1/T = 1/0.400 = 2.50\ \text{Hz}$.

   3. Timing many cycles and dividing is not just convenience. A stopwatch reaction error of about 0.2 s spread over 45 cycles becomes 0.004 s per cycle, so the count is what buys the precision.

   4. (b) The second oscillator: $f_2 = 1/T_2 = 1/(0.250\ \text{s}) = 4.00\ \text{Hz}$. That is higher, and the factor is $4.00/2.50 = 1.60$.

   5. The trap here is reading a shorter period as a lower frequency. Period and frequency are reciprocals, so they move in opposite directions: the oscillator that finishes each cycle faster has more cycles per second, not fewer. Confirm with the times themselves, $0.400/0.250 = 1.60$, the same factor upside down.

   6. (c) Two minutes is 120 s, so $120/0.400 = 300$ complete oscillations. Or go through the frequency: $(2.50\ \text{Hz})(120\ \text{s}) = 300$. Both routes are the same statement.

   (a) $T = 0.400\ \text{s}$ and $f = 2.50\ \text{Hz}$. (b) The second oscillator, at $4.00\ \text{Hz}$, is faster by a factor of 1.60. (c) 300 complete oscillations in 2.0 minutes.

3. 3. A simple pendulum, and the bob that does not matter

   A simple pendulum has a string of length 0.60 m and a bob of mass 0.25 kg. It swings with a small amplitude of about $5^\circ$. Use $g = 9.8\ \text{m/s}^2$. Find (a) the period and (b) the frequency. (c) State what happens to the period if the bob is replaced by a 0.75 kg one, and (d) what length would double the period.

   Show the worked solution

   1. (a) For a simple pendulum displaced by a small angle, $T_p = 2\pi\sqrt{\ell/g}$. Inside the root: $(0.60\ \text{m})/(9.8\ \text{m/s}^2) = 0.06122\ \text{s}^2$.

   2. $\sqrt{0.06122} = 0.2474\ \text{s}$, so $T = 2\pi(0.2474) = 1.5547\ \text{s}$, which is $1.55\ \text{s}$.

   3. (b) $f = 1/T = 1/1.5547 = 0.643\ \text{Hz}$, a little under two thirds of a cycle per second.

   4. (c) The period does not change at all. Look at what is inside $T_p = 2\pi\sqrt{\ell/g}$: the length and the local gravitational field strength. The mass of the bob is not there, so tripling it changes nothing, and the period stays $1.55\ \text{s}$.

   5. That is not a coincidence of the formula. A heavier bob is pulled back toward the middle by a proportionally larger gravitational force, and it also has proportionally more inertia resisting that pull. The two effects cancel exactly, and the mass falls out of the physics before it can reach the answer.

   6. (d) The period goes as $\sqrt{\ell}$, so doubling $T$ needs $\ell$ multiplied by $2^2 = 4$: a length of $4(0.60) = 2.40\ \text{m}$.

   7. Check that directly rather than trusting the scaling: $T = 2\pi\sqrt{2.40/9.8} = 2\pi\sqrt{0.24490} = 2\pi(0.49487) = 3.109\ \text{s}$, and $2 \times 1.5547 = 3.109\ \text{s}$. They agree.

   8. One assumption worth naming. Essential knowledge 7.2.A.1.ii gives this period for a pendulum displaced by a small angle, so the $5^\circ$ in the problem is doing real work. The formula is a small-angle result, and it drifts low as the swing gets wide.

   (a) $T = 1.55\ \text{s}$. (b) $f = 0.643\ \text{Hz}$. (c) Unchanged at $1.55\ \text{s}$; the bob's mass is not in the expression, because the extra gravitational force on a heavier bob is exactly matched by its extra inertia. (d) $2.40\ \text{m}$, four times the original length.

4. 4. Finding a spring constant from a measured period

   A 0.48 kg block on a horizontal spring is timed at 25 complete oscillations in 20.0 s. (a) Find the period. (b) Find the spring constant. (c) What mass on the same spring would double the period?

   Show the worked solution

   1. (a) $T = (20.0\ \text{s})/25 = 0.800\ \text{s}$.

   2. (b) Start from $T = 2\pi\sqrt{m/k}$ and rearrange for $k$ rather than guessing at the algebra. Divide by $2\pi$: $T/(2\pi) = \sqrt{m/k}$. Square both sides: $T^2/(4\pi^2) = m/k$. Then cross-multiply:

      $$k = \frac{4\pi^2 m}{T^2}$$

   3. Substitute: $k = 4\pi^2(0.48\ \text{kg})/(0.800\ \text{s})^2 = (39.478)(0.48)/(0.640) = 18.950/0.640 = 29.6\ \text{N/m}$.

   4. Always run a rearrangement forwards again before you trust it: $T = 2\pi\sqrt{0.48/29.609} = 2\pi\sqrt{0.016211} = 2\pi(0.12732) = 0.800\ \text{s}$. That is the measured period back, so the rearrangement was right.

   5. A 29.6 N/m spring stretches about 0.16 m under the block's own weight of $(0.48)(9.8) = 4.7\ \text{N}$, which is a spring you could stretch easily by hand. Sanity checks like that catch a squared term dropped from the denominator, which is the usual failure here.

   6. (c) $T \propto \sqrt{m}$ at fixed $k$, so doubling the period needs four times the mass: $m = 4(0.48) = 1.92\ \text{kg}$.

   7. Confirm: $T = 2\pi\sqrt{1.92/29.609} = 2\pi\sqrt{0.064845} = 2\pi(0.25465) = 1.600\ \text{s}$, twice 0.800 s. Note that adding mass to slow an oscillator is expensive, because the square root gives you back only half of every factor you spend.

   (a) $T = 0.800\ \text{s}$. (b) $k = 29.6\ \text{N/m}$. (c) $1.92\ \text{kg}$, four times the original mass, because the period depends on the square root of the mass.

5. 5. Six changes, and which of them move the period

   Two oscillators are set up side by side. A simple pendulum has a string 1.00 m long, a 0.20 kg bob, and an amplitude of $6^\circ$. A separate 0.20 kg block oscillates on a spring with $k = 50\ \text{N/m}$ and an amplitude of 0.050 m. Use $g = 9.8\ \text{m/s}^2$. First find each period. Then give the new period after each of these changes, made one at a time from the original setup: (a) the pendulum bob is replaced by a 0.60 kg one; (b) the pendulum's amplitude is doubled to $12^\circ$; (c) the pendulum's string is lengthened to 2.00 m; (d) the block on the spring is replaced by a 0.80 kg one; (e) the spring's amplitude is doubled to 0.10 m; (f) both oscillators are moved to a place where the gravitational field strength is $1.6\ \text{N/kg}$.

   Show the worked solution

   1. Baselines first. Pendulum: $T_p = 2\pi\sqrt{1.00/9.8} = 2\pi\sqrt{0.10204} = 2\pi(0.31944) = 2.007\ \text{s}$. Spring: $T_s = 2\pi\sqrt{0.20/50} = 2\pi\sqrt{0.0040} = 2\pi(0.063246) = 0.397\ \text{s}$.

   2. Before touching any of the six, write down what each formula actually contains. $T_p$ contains $\ell$ and $g$. $T_s$ contains $m$ and $k$. Any change to a quantity that is not on those lists cannot move the period, and that single observation answers three of the six parts immediately.

   3. (a) Bob mass tripled: no change, $2.007\ \text{s}$. Mass is not in $T_p$.

   4. (b) Pendulum amplitude doubled from $6^\circ$ to $12^\circ$: no change, $2.007\ \text{s}$. Amplitude is not in $T_p$, and $12^\circ$ is still comfortably inside the small-angle range the formula assumes. Essential knowledge 7.3.A.2 covers this in general: changing the amplitude of a system exhibiting simple harmonic motion will not change the period of that system.

   5. (c) Length doubled to 2.00 m: the period goes as $\sqrt{\ell}$, so it rises by $\sqrt{2} = 1.414$. New period $= (2.007)(1.414) = 2.838\ \text{s}$. Direct check: $2\pi\sqrt{2.00/9.8} = 2\pi\sqrt{0.20408} = 2\pi(0.45175) = 2.838\ \text{s}$.

   6. (d) Block mass quadrupled to 0.80 kg: the period goes as $\sqrt{m}$, so it doubles. New period $= 2(0.397) = 0.795\ \text{s}$. Direct check: $2\pi\sqrt{0.80/50} = 2\pi\sqrt{0.016} = 2\pi(0.12649) = 0.795\ \text{s}$.

   7. (e) Spring amplitude doubled: no change, $0.397\ \text{s}$. Amplitude is not in $T_s$ either.

   8. (f) Weaker gravity, $g = 1.6\ \text{N/kg}$. The pendulum slows, because $g$ sits under the root in the denominator: $T_p = 2\pi\sqrt{1.00/1.6} = 2\pi(0.79057) = 4.967\ \text{s}$, a factor of $\sqrt{9.8/1.6} = 2.475$ longer. The spring oscillator does not care at all: $g$ is nowhere in $T_s$, so it stays at $0.397\ \text{s}$. A mass on a spring keeps time in a place where a pendulum clock runs slow.

   9. Summary of the six: three no-changes (a, b, e), and three real ones (c up by 1.414, d up by 2, f up by 2.475 for the pendulum only). Every no-change is a quantity absent from the formula, and every change is a quantity present in it, softened by a square root.

   Baselines: pendulum $2.007\ \text{s}$, spring $0.397\ \text{s}$. (a) unchanged, $2.007\ \text{s}$. (b) unchanged, $2.007\ \text{s}$. (c) $2.838\ \text{s}$, longer by $\sqrt{2}$. (d) $0.795\ \text{s}$, exactly doubled. (e) unchanged, $0.397\ \text{s}$. (f) the pendulum becomes $4.967\ \text{s}$ while the spring stays at $0.397\ \text{s}$. Amplitude never mattered, the bob's mass never mattered, and $g$ mattered only to the pendulum.

6. 6. Energy in a spring oscillator: total, maximum speed, and speed part way out

   A 0.64 kg block oscillates on a frictionless horizontal surface, attached to a spring with $k = 100\ \text{N/m}$, with an amplitude of 0.16 m. Find (a) the total mechanical energy of the oscillation, (b) the maximum speed of the block, (c) the period and frequency, and (d) the block's speed when it is 0.10 m from equilibrium.

   Show the worked solution

   1. (a) At the turning point the block is momentarily at rest, so every joule in the system is spring potential energy. That instant is the easiest place to count the total: $E = \frac{1}{2}kA^2 = \frac{1}{2}(100\ \text{N/m})(0.16\ \text{m})^2 = (50)(0.0256) = 1.28\ \text{J}$.

   2. (b) At $x = 0$ the spring is relaxed, so the potential energy is zero and all 1.28 J is kinetic: $\frac{1}{2}mv_{\text{max}}^2 = 1.28\ \text{J}$, so $v_{\text{max}}^2 = 2(1.28)/0.64 = 4.00\ \text{m}^2/\text{s}^2$ and $v_{\text{max}} = 2.00\ \text{m/s}$.

   3. Check with the shortcut those two lines produce in general, $v_{\text{max}} = A\sqrt{k/m} = (0.16)\sqrt{100/0.64} = (0.16)\sqrt{156.25} = (0.16)(12.5) = 2.00\ \text{m/s}$. Same number, and now you know where the shortcut comes from.

   4. (c) $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.64/100} = 2\pi\sqrt{0.0064} = 2\pi(0.080) = 0.503\ \text{s}$, and $f = 1/0.503 = 1.99\ \text{Hz}$.

   5. (d) At $x = 0.10\ \text{m}$ the energy is split. Potential: $U_s = \frac{1}{2}(100)(0.10)^2 = (50)(0.010) = 0.50\ \text{J}$. Kinetic is whatever is left of the 1.28 J: $K = 1.28 - 0.50 = 0.78\ \text{J}$.

   6. $v = \sqrt{2K/m} = \sqrt{2(0.78)/0.64} = \sqrt{2.4375} = 1.56\ \text{m/s}$.

   7. Check with the general form of the same statement, $v = v_{\text{max}}\sqrt{1 - (x/A)^2} = (2.00)\sqrt{1 - (0.625)^2} = (2.00)\sqrt{0.60938} = (2.00)(0.78063) = 1.56\ \text{m/s}$.

   8. Look at the split before moving on, because this is where the intuition usually fails. The block is 62.5 percent of the way out to the turning point, yet it still holds $0.78/1.28 = 61$ percent of the energy as kinetic and is still moving at 78 percent of its top speed. The potential energy goes as $x^2$, so it stays small until quite late and then climbs steeply. Speed does not fall off in step with distance.

   (a) $E = 1.28\ \text{J}$. (b) $v_{\text{max}} = 2.00\ \text{m/s}$. (c) $T = 0.503\ \text{s}$ and $f = 1.99\ \text{Hz}$. (d) $1.56\ \text{m/s}$, still 78 percent of the maximum even though the block is 62.5 percent of the way out.

7. 7. Where the kinetic and potential energies are equal

   A 0.25 kg glider oscillates on a frictionless track attached to a spring with $k = 40\ \text{N/m}$ and an amplitude of 0.20 m. Find (a) the total energy, (b) the maximum speed, (c) the displacement at which the kinetic and spring potential energies are equal, and (d) the glider's speed there.

   Show the worked solution

   1. (a) $E = \frac{1}{2}kA^2 = \frac{1}{2}(40)(0.20)^2 = (20)(0.040) = 0.80\ \text{J}$.

   2. (b) $v_{\text{max}} = \sqrt{2E/m} = \sqrt{2(0.80)/0.25} = \sqrt{6.40} = 2.53\ \text{m/s}$. Check: $A\sqrt{k/m} = (0.20)\sqrt{160} = (0.20)(12.649) = 2.53\ \text{m/s}$.

   3. (c) Equal shares means each holds half the total, so $U_s = E/2 = 0.40\ \text{J}$. Set $\frac{1}{2}kx^2 = 0.40$: $x^2 = 2(0.40)/40 = 0.020\ \text{m}^2$, so $x = \sqrt{0.020} = 0.1414\ \text{m}$.

   4. Notice what that is as a fraction of the amplitude: $0.1414/0.20 = 0.707$, which is $1/\sqrt{2}$. Setting $\frac{1}{2}kx^2 = \frac{1}{2}\left(\frac{1}{2}kA^2\right)$ gives $x = A/\sqrt{2}$ for any spring oscillator, whatever the mass, the spring constant or the amplitude. The glider is 71 percent of the way out, not half way.

   5. (d) The kinetic energy there is also 0.40 J, so $v = \sqrt{2(0.40)/0.25} = \sqrt{3.20} = 1.79\ \text{m/s}$.

   6. Check that against the maximum: $v_{\text{max}}/\sqrt{2} = 2.53/1.4142 = 1.79\ \text{m/s}$. At the point where the energy splits evenly, the speed is $1/\sqrt{2}$ of the maximum, because kinetic energy goes as $v^2$ and half the energy means $1/\sqrt{2}$ of the speed.

   7. The trap this problem exists to expose: at half the amplitude, $x = 0.10\ \text{m}$, the potential energy is only $\frac{1}{2}(40)(0.010) = 0.20\ \text{J}$, one quarter of the total, not half. Both energies depend on a square, so nothing in this motion splits where you would first guess.

   (a) $E = 0.80\ \text{J}$. (b) $v_{\text{max}} = 2.53\ \text{m/s}$. (c) At $x = 0.141\ \text{m}$, which is $A/\sqrt{2}$, or 71 percent of the amplitude. (d) $1.79\ \text{m/s}$, which is $v_{\text{max}}/\sqrt{2}$.

8. 8. A block launched by a compressed spring

   A 0.30 kg block is pressed against a horizontal spring of spring constant $k = 500\ \text{N/m}$, compressing it 0.080 m from its relaxed length, and released on a frictionless surface. The block stays attached to the spring, so it oscillates rather than flying off. Find (a) the amplitude, (b) the total energy, (c) the maximum speed, (d) the period, (e) the block's speed when the spring is compressed 0.040 m, and (f) the time from release until the block first reaches maximum speed.

   Show the worked solution

   1. (a) The block is released from rest at maximum compression, and by definition the amplitude is the largest displacement from equilibrium, so $A = 0.080\ \text{m}$.

   2. (b) $E = \frac{1}{2}kA^2 = \frac{1}{2}(500)(0.080)^2 = (250)(0.0064) = 1.6\ \text{J}$. All of it is spring potential energy at the instant of release.

   3. (c) That entire 1.6 J becomes kinetic energy at $x = 0$: $v_{\text{max}} = \sqrt{2E/m} = \sqrt{2(1.6)/0.30} = \sqrt{10.667} = 3.27\ \text{m/s}$. Check with the shortcut: $A\sqrt{k/m} = (0.080)\sqrt{500/0.30} = (0.080)\sqrt{1666.7} = (0.080)(40.825) = 3.27\ \text{m/s}$.

   4. (d) $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.30/500} = 2\pi\sqrt{0.00060} = 2\pi(0.024495) = 0.1539\ \text{s}$. A stiff spring and a light block make a fast oscillator, about 6.5 cycles a second.

   5. (e) Halfway back, at $x = 0.040\ \text{m}$: $U_s = \frac{1}{2}(500)(0.040)^2 = (250)(0.0016) = 0.40\ \text{J}$, so $K = 1.6 - 0.40 = 1.2\ \text{J}$ and $v = \sqrt{2(1.2)/0.30} = \sqrt{8.0} = 2.83\ \text{m/s}$.

   6. That is $2.83/3.27 = 86.6$ percent of the top speed at half the amplitude, which is the same $\sqrt{1 - (x/A)^2} = \sqrt{0.75} = 0.866$ pattern as problem 6. Half the distance out is nowhere near half the speed.

   7. (f) Maximum speed happens at $x = 0$, and getting from a turning point to the middle is a quarter of a full cycle: $t = T/4 = 0.1539/4 = 0.0385\ \text{s}$.

   8. A closing check on part (b) that catches sign and factor errors. The spring force at maximum compression is $|F_s| = k|\Delta x| = (500)(0.080) = 40\ \text{N}$, and the energy stored is the average force over the compression times the distance, $\frac{1}{2}(40)(0.080) = 1.6\ \text{J}$. The factor of one half in $\frac{1}{2}k(\Delta x)^2$ is that average, not a decoration.

   (a) $A = 0.080\ \text{m}$. (b) $E = 1.6\ \text{J}$. (c) $v_{\text{max}} = 3.27\ \text{m/s}$. (d) $T = 0.154\ \text{s}$. (e) $2.83\ \text{m/s}$. (f) $T/4 = 0.0385\ \text{s}$ after release.

9. 9. A pendulum through energy, then checked against the SHM model

   A 0.40 kg bob hangs from a light string 0.90 m long. It is pulled aside until the string makes an angle of $8.0^\circ$ with the vertical, then released from rest. Use $g = 9.8\ \text{m/s}^2$. Find (a) the height the bob rises above its lowest point, (b) the total energy of the swing measured from that lowest point, (c) the bob's maximum speed, (d) the period, and (e) check the maximum speed a second way, using the simple harmonic motion model.

   Show the worked solution

   1. Take the lowest point of the swing as the zero of gravitational potential energy, and take up as positive.

   2. (a) Geometry first. With the string at angle $\theta$ from the vertical, the bob hangs a vertical distance $\ell\cos\theta$ below the pivot, so it sits $h = \ell(1 - \cos\theta)$ above its lowest point. With $\cos 8.0^\circ = 0.99027$: $h = (0.90)(1 - 0.99027) = (0.90)(0.0097319) = 0.008759\ \text{m}$, about 8.8 mm.

   3. That is a small number and it should be. An $8^\circ$ swing on a 0.90 m string moves the bob about 0.125 m sideways but lifts it less than a centimetre, because near the bottom the arc is almost horizontal.

   4. (b) At release the bob is at rest, so all the energy is gravitational: $E = mgh = (0.40)(9.8)(0.008759) = 0.0343\ \text{J}$.

   5. (c) At the lowest point all of it is kinetic, so $\frac{1}{2}mv^2 = mgh$. The mass divides out, leaving $v = \sqrt{2gh} = \sqrt{2(9.8)(0.008759)} = \sqrt{0.17167} = 0.414\ \text{m/s}$.

   6. Notice that the mass cancelled. The 0.40 kg was needed for part (b), the energy in joules, but not for the speed, which is exactly why a pendulum's timing and speed pattern are independent of its bob.

   7. Check part (b) forwards from the speed: $K = \frac{1}{2}(0.40)(0.414)^2 = (0.20)(0.17167) = 0.0343\ \text{J}$, matching the potential energy at the top of the swing.

   8. (d) $T = 2\pi\sqrt{\ell/g} = 2\pi\sqrt{0.90/9.8} = 2\pi\sqrt{0.091837} = 2\pi(0.30305) = 1.904\ \text{s}$.

   9. (e) Now the independent check. Treated as simple harmonic motion, the bob's amplitude measured along its arc is $A = \ell\theta$ with $\theta$ in radians: $8.0^\circ = 0.13963\ \text{rad}$, so $A = (0.90)(0.13963) = 0.1257\ \text{m}$. The maximum speed of any simple harmonic oscillator is $v_{\text{max}} = A(2\pi/T) = (0.1257)(2\pi/1.904) = (0.1257)(3.2998) = 0.4147\ \text{m/s}$.

   10. Compare the two: 0.4143 m/s from energy conservation, which makes no small-angle assumption, against 0.4147 m/s from the simple harmonic motion model, which does. They differ by 0.08 percent. That gap is the entire cost of the small-angle approximation at $8^\circ$, and watching it stay this small is the best evidence that the model is doing honest work here. Repeat the same comparison for a $40^\circ$ release on this pendulum and the gap widens to about 2 percent, with the simple harmonic motion value now visibly too high.

   (a) $h = 0.00876\ \text{m}$, under a centimetre. (b) $E = 0.0343\ \text{J}$. (c) $v_{\text{max}} = 0.414\ \text{m/s}$. (d) $T = 1.904\ \text{s}$. (e) The simple harmonic motion model gives $0.4147\ \text{m/s}$ against energy conservation's $0.4143\ \text{m/s}$, a difference of 0.08 percent, which is how good the small-angle approximation is at $8^\circ$.

10. 10. Where in the cycle is the speed largest, and where is the acceleration?

    A block oscillates on a horizontal spring between $x = -A$ and $x = +A$, passing through the equilibrium position at $x = 0$. Without calculating anything, say where in the cycle the speed is largest and where it is zero, where the acceleration is largest in magnitude and where it is zero, and where the kinetic and potential energies each peak. Then explain what is wrong with a student's claim that the speed and the acceleration must both peak at the ends of the motion, since that is where the block has travelled furthest.

    Show the worked solution

    1. Start from the force, because everything else follows from it. The spring obeys $\vec{F}_s = -k\Delta\vec{x}$, so the net force is proportional to the displacement and always points back toward $x = 0$. Newton's second law then gives an acceleration proportional to displacement and opposite in sign to it.

    2. Acceleration is therefore largest in magnitude exactly where the displacement is largest, at $x = \pm A$, and it is zero exactly where the displacement is zero, at $x = 0$. The spring is relaxed at the middle, so at that one instant nothing is pushing the block at all.

    3. Speed does the opposite. Energy is conserved and split between $U_s = \frac{1}{2}kx^2$ and $K = \frac{1}{2}mv^2$. At $x = \pm A$ the potential energy holds the whole budget, so the kinetic energy is zero and the block is momentarily at rest. At $x = 0$ the potential energy is zero, so the kinetic energy holds everything and the speed is at its maximum.

    4. So the energies follow their partners: kinetic energy peaks at $x = 0$ alongside the speed, and spring potential energy peaks at $x = \pm A$ alongside the acceleration. Essential knowledge 7.3.A.1.i is the general statement that minima, maxima and zeros of displacement, velocity and acceleration are features of harmonic motion, and 7.3.A.1.ii points out that knowing where they fall is often enough to describe the motion without solving anything.

    5. Now the student's error. Distance travelled is not the issue; the ends of the motion are turning points, and a turning point is by definition where the velocity passes through zero on its way to changing sign. The block is not fast there, it is stationary there.

    6. Underneath the mistake is a conflation of velocity with acceleration. Acceleration says how quickly the velocity is changing, not how large it is, and the two hit their extremes a quarter of a cycle apart in this motion. At the ends the block is momentarily still while being flung back hardest; at the middle it is moving fastest while feeling no force at all. Neither instant has both.

    7. A useful mental test for any oscillation question: if the block had maximum speed and maximum restoring force at the same instant, the motion would never turn around, because nothing would be slowing it before the turning point. The offset between speed and acceleration is what makes the motion repeat.

    Speed is largest at $x = 0$ and zero at $x = \pm A$. Acceleration is largest in magnitude at $x = \pm A$ and zero at $x = 0$. Kinetic energy peaks at the middle with the speed; spring potential energy peaks at the ends with the acceleration. The student has confused travelling furthest with moving fastest: the ends are turning points, where the velocity is passing through zero, and they are also where the spring is stretched or compressed most and so pushes hardest. Speed and acceleration reach their extremes a quarter of a cycle apart, and that offset is what makes the motion turn around and repeat.