Exercises — Simple harmonic motion — definition, restoring force F = −kx

The one and only test for SHM is: is the force a straight-line push back toward home? In symbols, $F=-kx$ with a positive constant $k$ and a minus sign.

• (a) $F=-5x$: linear in $x$, minus sign present → SHM, with $k=5\text{N/m}$.
• (b) $F=-5x^3$: it does point back home (negative when $x>0$), so it oscillates — but the strength grows as $x^3$, not proportional to $x$. So it is not SHM. Its period would depend on amplitude.
• (c) $F=+5x$: the sign is plus, so the force pushes away from home. This runs away, never oscillates → not SHM (it is an unstable, "anti-restoring" force).

The figure below plots all three force laws so you can see the difference: only the straight line through the origin with negative slope (the blue one) is SHM.

Answer: only (a).

The SHM definition is $a=-{\omega }^{2}x$. So the number multiplying $x$ is ${\omega }^2$. ${\omega }^2=64\Rightarrow \omega =\sqrt{64}=8\text{rad/s}.$ Period is one full turn of the clock, $2\pi$ radians, at $8$ rad/s: $T=\frac{2\pi }{\omega }=\frac{2\pi }{8}=\frac{\pi }{4}\approx 0.785\text{s}.$

Question: "how fast does its clock tick?" → the tool is $\omega =\sqrt{k/m}$. $\omega =\sqrt{\frac{80}{0.20}}=\sqrt{400}=20\text{rad/s}.$ Period (seconds per cycle): $T=\frac{2\pi }{\omega }=\frac{2\pi }{20}=0.3142\text{s}.$ Frequency (cycles per second) is just the reciprocal: $f=\frac{1}{T}=\frac{20}{2\pi }=3.183\text{Hz}.$

Question: "how fast at a given position, without solving for time?" → energy tool $v=\omega \sqrt{A^2-x^2}$. $v=20\sqrt{0.05^2-0.03^2}=20\sqrt{0.0025-0.0009}=20\sqrt{0.0016}=20(0.04)=0.80\text{m/s}.$ Max speed happens at home ($x=0$), where all energy is motion: $v_{\max }=\omega A=20(0.05)=1.0\text{m/s}.$

Acceleration in SHM is $a=-{\omega }^{2}x$, biggest in size when $|x|$ is biggest — that is at the turning points $x=\pm A$. $a_{\max }={\omega }^{2}A=(20{)}^2(0.05)=400(0.05)=20{\text{m/s}}^2,$ occurring at the extremes $x=\pm A$, where the block is momentarily still but the restoring force is strongest.

The total budget is fixed: $E=\frac{1}{2}k{A}^2$, split as $E=\underset{KE}{\underbrace{\frac{1}{2}m{v}^2}}+\underset{PE}{\underbrace{\frac{1}{2}k{x}^2}}$. We want $KE=PE$, which means each holds half the budget: $\frac{1}{2}k{x}^2=\frac{1}{2}E=\frac{1}{2}\left(\frac{1}{2}k{A}^2\right).$ Now we isolate $x$. Why cancel $\frac{1}{2}k$? Both sides are "$\frac{1}{2}k$ times a length-squared," so dividing both sides by the common factor $\frac{1}{2}k$ leaves a clean statement purely about the lengths — the stiffness $k$ was never going to affect where the split happens, so we remove it: $x^2=\frac{1}{2}{A}^2.$ Why take the square root? We know $x^2$ but want $x$ itself, and the square root is the operation that undoes squaring. Since displacement can be either side of home, both signs are physical: $x=\pm \frac{A}{\sqrt{2}}\approx \pm 0.707A.$ Answer: at $x=A/\sqrt{2}$ (either side of home) the energy is split fifty-fifty. The figure below plots the two energies separately against position: the pink potential curve $U=\frac{1}{2}k{x}^2$ rising like a valley, the blue kinetic curve $KE=E-U$ falling the opposite way, and the yellow dashed line their constant total $E$. They cross exactly at $x=A/\sqrt{2}$ — the crossing point is where each holds half the total.

Two springs in parallel both pull the mass; their pushes add. If each gives $-kx$, together they give $-2kx$, so the effective stiffness is $k_{\text{eff}}=2k$. $T_A=2\pi \sqrt{\frac{m}{k}},T_B=2\pi \sqrt{\frac{m}{2k}}=\frac{T_A}{\sqrt{2}}.$ Answer: the parallel pair is stiffer, so B oscillates faster: $T_B=T_A/\sqrt{2}\approx 0.707T_A$.

Compare with the standard form $x=A\cos (\omega t+\varphi )$. Here $\varphi$ is the phase constant — a fixed head-start angle baked into the cosine that says where in its cycle the motion was at the clock-start $t=0$. It does not change with time; it just shifts the whole wave sideways. In this problem $\varphi =\pi /3$.

• (a) Amplitude $A=0.10\text{m}$ (the number in front).
• (b) $\omega =4\text{rad/s}$ (the number multiplying $t$), so $T=2\pi /4=1.571\text{s}$.
• (c) At $t=0$ the angle inside the cosine is just $\varphi =\pi /3$: $x(0)=0.10\cos (\pi /3)=0.10(0.5)=0.05\text{m}.$ Velocity is the time-derivative of position, $v=-A\omega \sin (\omega t+\varphi )$ (differentiating $\cos$ gives $-\sin$, and the chain rule brings down the $\omega$): $v(0)=-0.10(4)\sin (\pi /3)=-0.4(0.8660)=-0.3464\text{m/s}.$ The minus sign says it is heading back toward home at $t=0$.

Start from the tool that links $T$ to $k$: $T=2\pi \sqrt{m/k}$. We know $T$ and $m$, we want $k$ — so we must unwrap $k$ from inside the square root. Why square both sides first? $k$ is trapped under a square root; squaring is the operation that removes a square root, so it frees $k$ from the radical: $T^2=4{\pi }^2\frac{m}{k}.$ Why multiply by $k$ and divide by $T^2$? $k$ is currently in the denominator; multiplying both sides by $k$ lifts it up top, and dividing by $T^2$ leaves $k$ alone on one side: $k=\frac{4{\pi }^{2}m}{T^2}.$ Now substitute the numbers: $k=\frac{4{\pi }^2(0.30)}{(0.50{)}^2}=\frac{4(9.8696)(0.30)}{0.25}=\frac{11.844}{0.25}=47.37\text{N/m}.$ Answer: $k\approx 47.4\text{N/m}$.

The shadow of uniform circular motion is exactly SHM: the shadow's displacement is $x=R\cos (\omega t)$. So the wheel's radius becomes the amplitude, and the wheel's spin rate becomes the SHM $\omega$.

• Amplitude: $A=R=0.20\text{m}$.
• Max speed = the peg's actual rim speed $=\omega R=6(0.20)=1.2\text{m/s}$.
• Max acceleration = the peg's centripetal acceleration $={\omega }^{2}R=36(0.20)=7.2{\text{m/s}}^2$. Answer: $A=0.20\text{m}$, $v_{\max }=1.2\text{m/s}$, $a_{\max }=7.2{\text{m/s}}^2$. In the figure the pink peg rides the circle while the yellow shadow slides on the wall below; the shadow's swing-width is exactly the circle's radius, and it is fastest as it crosses the middle.

(a) All energy sits in the spring at the turning point: $E=\frac{1}{2}k{A}^2$. Why invert for $A$? We know the energy and stiffness and want the reach, so we unwrap $A$: multiply by $2$, divide by $k$, take the root (root undoes the square): $A=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(0.30)}{60}}=\sqrt{0.01}=0.10\text{m}.$ (b) All energy is kinetic at home: $E=\frac{1}{2}m{v}_{\max }^2$, so by the same unwrapping (times $2$, over $m$, root): $v_{\max }=\sqrt{\frac{2E}{m}}=\sqrt{\frac{0.60}{0.40}}=\sqrt{1.5}=1.2247\text{m/s}.$ (c) At $x=A/2=0.05\text{m}$, use $v=\omega \sqrt{A^2-x^2}$ with $\omega =\sqrt{k/m}=\sqrt{150}=12.247\text{rad/s}$: $v=12.247\sqrt{0.10^2-0.05^2}=12.247\sqrt{0.0075}=12.247(0.08660)=1.0607\text{m/s}.$ Answers: $A=0.10\text{m}$, $v_{\max }=1.225\text{m/s}$, $v(A/2)=1.061\text{m/s}$.

Substitute the trial solution into Newton's law. First we need its second time-derivative. Differentiating once, $\dot{x}=-A\omega \sin (\omega t+\varphi )$; differentiating again, $\ddot{x}=-A{\omega }^2\cos (\omega t+\varphi )=-{\omega }^{2}x.$ Put this into $m\ddot{x}=-kx$: $m(-{\omega }^{2}x)=-kx.$ Why can we match the coefficients? This equation must hold at every instant, and $x(t)$ sweeps through many different values as the block moves. The only way $-m{\omega }^{2}x=-kx$ can be true for all those different values of $x$ simultaneously is if the constants multiplying $x$ are equal. Hence: $m{\omega }^2=k\Rightarrow \omega =\sqrt{\frac{k}{m}},T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}.$ Look at the result: the amplitude $A$ cancelled out entirely — it appeared on both sides of $m\ddot{x}=-kx$ (inside $x$ and inside $\ddot{x}$) and divided away, so it never touches the relation between $m$, $k$, and $\omega$. Therefore $T$ cannot depend on $A$.

Physical reason the intuition fails: doubling $A$ doubles the distance the block must travel, so naively you expect a longer trip. But doubling $A$ also doubles the displacement at every corresponding point, and since $F=-kx$ is linear, it doubles the restoring force and hence doubles the top speed ($v_{\max }=\omega A$). Longer path, proportionally faster trip — the two effects cancel exactly. This exact cancellation happens only because $F$ is strictly proportional to $x$; for a nonlinear law like $F=-k{x}^3$ the cancellation is imperfect and the period would depend on amplitude.

• (a) $A=0$: the block sits exactly at home with $x(t)=0$ for all $t$. Every quantity — speed, force, energy — is zero. This is the trivial equilibrium; it is technically SHM with zero amplitude but nothing moves.
• (b) $m\to \infty$: $\omega =\sqrt{k/m}\to 0$, so $T=2\pi \sqrt{m/k}\to \infty$. An enormously heavy mass is so sluggish that one oscillation takes forever — it barely moves. The limit is a stationary block.
• (c) $k\to 0$: the spring becomes limp, no restoring force. $\omega \to 0$, $T\to \infty$. With no push-back, a nudged block just drifts at constant velocity forever — no oscillation. This is the boundary where SHM ceases to exist: a restoring force needs $k>0$.

These limits confirm the formula's honesty: it predicts "no oscillation" precisely when the physics loses its restoring force or its ability to accelerate.

Near a minimum at $x=0$, expand the potential as a power series (a Taylor series — the tool for "approximate any smooth curve by a polynomial"): $U(x)\approx U(0)+U^{\prime }(0)x+\frac{1}{2}{U}^{\prime \prime }(0)x^2+\cdots$ Now use two facts that are true because $x=0$ is a stable minimum:

1. The slope is flat at the bottom of a valley, so $U^{\prime }(0)=0$. This kills the linear term.
2. The valley bends upward, so the curvature $U^{\prime \prime }(0)>0$.

The leftover constant $U(0)$ is just a baseline — a fixed height we can set to zero, because only changes in energy produce force (force is the negative slope, and the slope of a constant is zero). Dropping the killed linear term and the inert constant leaves: $U(x)\approx \frac{1}{2}{U}^{\prime \prime }(0)x^2.$ That is a parabola — exactly the shape of spring potential $\frac{1}{2}k{x}^2$. Now get the force by taking the negative slope (force is minus the derivative of energy): $F=-\frac{dU}{dx}\approx -U^{\prime \prime }(0)x=-k_{\text{eff}}x,\boxed{k_{\text{eff}}=U^{\prime \prime }(0)}.$ Conclusion: every stable equilibrium behaves like a spring for small nudges, with effective stiffness equal to the curvature of the energy valley at its bottom, $k_{\text{eff}}=U^{\prime \prime }(0)$. That single fact is why SHM appears everywhere — pendulums, atoms in crystals, molecules vibrating — all share the parabolic valley bottom.

Level 1 — What is the single test for SHM? ::: The force must be $F=-kx$: linear in $x$ with a minus sign and $k>0$. Level 2 — Speed at $x=0.03$ for $\omega =20$, $A=0.05$? ::: $v=20\sqrt{0.05^2-0.03^2}=0.80\text{m/s}$. Level 3 — Where does $KE=PE$? ::: At $x=A/\sqrt{2}\approx 0.707A$. Level 4 — Design $k$ for $T=0.5\text{s}$, $m=0.3\text{kg}$? ::: $k=4{\pi }^{2}m/T^2\approx 47.4\text{N/m}$. Level 5 — What is $k_{\text{eff}}$ from a general potential? ::: $k_{\text{eff}}=U^{\prime \prime }(0)$, the curvature of the energy valley at its minimum.