Exercises — SHM differential equation — solution - x = A cos(ωt + φ)

Compare term by term with the template $x=A\cos (\omega t+\varphi )$. This is just pattern-matching — the number multiplying the cosine is $A$; the number multiplying $t$ inside is $\omega$; the leftover added angle is $\varphi$.

• $A=0.04$ m
• $\omega =30$ rad/s
• $\varphi =0.5$ rad

What it looks like: $A$ is how far the wiggle reaches from centre; $\omega$ is how fast the internal clock ticks; $\varphi$ is where on the cosine wave the motion sits at $t=0$.

The cosine repeats when its inside grows by $2\pi$. Why $2\pi$? Because cosine is defined on a circle of circumference $2\pi$ radians — one full trip round the reference circle is one oscillation. $T=\frac{2\pi }{\omega }=\frac{2\pi }{30}\approx 0.2094\text{s},f=\frac{1}{T}=\frac{\omega }{2\pi }=\frac{30}{2\pi }\approx 4.775\text{Hz}.$

Step 1 — $\omega$. From Newton + Hooke, $\omega =\sqrt{k/m}$. $\omega =\sqrt{50/0.2}=\sqrt{250}\approx 15.81\text{rad/s}.$ Step 2 — $A$. Released from rest means $v_0=0$, so $A=\sqrt{x_0^2+\frac{v_0^2}{{\omega }^2}}=\sqrt{0.06^2+0}=0.06\text{m}.$ Why: starting at rest means starting at a turning point, and the turning point is the amplitude. Step 3 — $\varphi$. $\tan \varphi =-v_0/(\omega x_0)=0$, and $x_0>0$ forces $\cos \varphi >0$, so $\varphi =0$. Answer: $x(t)=0.06\cos (15.81t)$ m.

$v_{\max }=A\omega =0.06\times 15.81\approx 0.9487\text{m/s (at the centre }x=0).$ $a_{\max }=A{\omega }^2=0.06\times 250=15{\text{m/s}}^2(\text{at the ends }x=\pm A).$ Why opposite locations: speed is greatest where the pull is zero (centre); acceleration is greatest where the pull is strongest (extremes). See Energy in SHM for the energy view of the same fact.

Invert the period relation: $\omega =2\pi /T=2\pi /1.5\approx 4.189$ rad/s. $v_{\max }=A\omega =0.08\times 4.189\approx 0.3351\text{m/s}.$ (For the physics behind a swinging pendulum's $\omega$, see Simple pendulum.)

Amplitude: $A=\sqrt{x_0^2+\frac{v_0^2}{{\omega }^2}}=\sqrt{0.03^2+\frac{(-0.4{)}^2}{10^2}}=\sqrt{0.0009+0.0016}=\sqrt{0.0025}=0.05\text{m}.$ Phase — the delicate part. $\tan \varphi =-v_0/(\omega x_0)=-(-0.4)/(10\times 0.03)=+1.333$. The naive calculator gives $\arctan (1.333)=0.9273$ rad. But $\tan$ repeats every $\pi$, so $\varphi$ could be $0.9273$ or $0.9273-\pi =-2.214$. We must pick using the actual signs of $x_0$ and $v_0$ — see the figure below.

• $x_0=A\cos \varphi >0\Rightarrow \cos \varphi >0$.
• $v_0=-A\omega \sin \varphi <0\Rightarrow \sin \varphi >0$. Both positive $\Rightarrow \varphi$ is in quadrant I, so $\varphi =+0.9273$ rad. Answer: $x(t)=0.05\cos (10t+0.9273)$ m.

$A$ is unchanged: $A=0.05$ m. $\tan \varphi =-v_0/(\omega x_0)=-(+0.4)/(0.3)=-1.333$, giving raw $\arctan =-0.9273$. Sign check: $\cos \varphi >0$ (since $x_0>0$) and $v_0>0\Rightarrow \sin \varphi <0$. Both point to quadrant IV, so $\varphi =-0.9273$ rad. Same magnitude as 3.1 but negative — the direction of motion flipped the phase's sign.

Use the time-free speed relation (from Step 2 of the parent derivation): $v=\pm \omega \sqrt{A^2-x^2}=\pm 8\sqrt{0.10^2-0.06^2}=\pm 8\sqrt{0.01-0.0036}=\pm 8\sqrt{0.0064}=\pm 0.64\text{m/s}.$ At $x=A=0.10$: $v=\pm 8\sqrt{0.10^2-0.10^2}=0$. Why: at the extreme the particle momentarily stops before turning back — the turning point. This matches Energy in SHM: all energy is potential there, none kinetic.

Kinetic $KE=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2(A^2-x^2)$; potential $PE=\frac{1}{2}m{\omega }^2{x}^2$ (using $k=m{\omega }^2$). Set them equal: $A^2-x^2=x^2\Rightarrow x^2=\frac{A^2}{2}\Rightarrow x=\frac{A}{\sqrt{2}}=\frac{0.10}{\sqrt{2}}\approx 0.0707\text{m}.$ Why this is neat: the answer $x=A/\sqrt{2}$ is independent of $m$ and $\omega$ — energy splits 50/50 at the same fractional displacement for every SHM.

Starting at $+A$ from rest means $\varphi =0$: $x(t)=A\cos (8t)$. We need $\cos (8t)=0$, whose first positive solution is $8t=\pi /2$. $t=\frac{\pi }{2\omega }=\frac{\pi }{16}\approx 0.1963\text{s}.$ What it looks like: this is a quarter of the full period $T=2\pi /8\approx 0.785$ s — travelling from one extreme to the centre is one-quarter of a cycle. $0.785/4\approx 0.196$ ✓.

(a) $\omega$: $\omega =\sqrt{k/m}=\sqrt{100/0.25}=\sqrt{400}=20$ rad/s. Amplitude: $A=\sqrt{x_0^2+v_0^2/{\omega }^2}=\sqrt{0.05^2+(-1.0{)}^2/20^2}=\sqrt{0.0025+0.0025}=\sqrt{0.005}\approx 0.07071$ m. Phase: $\tan \varphi =-v_0/(\omega x_0)=-(-1.0)/(20\times 0.05)=+1.0$, raw $\arctan (1)=\pi /4=0.7854$. Sign check: $x_0>0\Rightarrow \cos \varphi >0$; $v_0<0\Rightarrow \sin \varphi >0$. Quadrant I → $\varphi =+\pi /4\approx 0.7854$ rad. ✓

(b) $x(t)=0.07071\cos (20t+\frac{\pi }{4})$ m.

(c) We need the first $t>0$ with $x=0$, i.e. $\cos (20t+\frac{\pi }{4})=0$. Cosine is zero at $\pm \pi /2,\pm 3\pi /2,\dots$ The inside starts at $\pi /4$ at $t=0$ and increases, so the first zero is at $20t+\frac{\pi }{4}=\frac{\pi }{2}$: $20t=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}\Rightarrow t=\frac{\pi }{80}\approx 0.03927\text{s}.$

"Leads by $\pi /3$" means B's internal clock is $\pi /3$ radians ahead, so add it inside: $x_B=A\cos (6t+\frac{\pi }{3})$. (See Phase and phase difference.) A reaches $+A$ when its argument $=0$ (at $t=0$). B reaches $+A$ when its argument $6t+\frac{\pi }{3}=0$, i.e. $t=-\frac{\pi }{18}$ — B hit its extreme earlier, by $\Delta t=\frac{\Delta \varphi }{\omega }=\frac{\pi /3}{6}=\frac{\pi }{18}\approx 0.1745\text{s}.$ Why divide phase by $\omega$: $\omega$ converts "radians of phase" into "seconds of time," since the phase advances at rate $\omega$ rad per second.

$x(t)=0.12\cos (5t)$. Set $0.12\cos (5t)=0.06\Rightarrow \cos (5t)=0.5\Rightarrow 5t=\arccos (0.5)=\frac{\pi }{3}$ (first positive value). $t=\frac{\pi }{15}\approx 0.2094\text{s}.$ What it looks like: the projection of the reference-circle point sweeps $60^{\circ }$ to drop from $A$ to $A/2$ — that $60^{\circ }=\pi /3$ is exactly what $\arccos (0.5)$ returns.

Extract $A,\omega ,\varphi$ from a formula ::: match term-by-term; report units (m, rad/s, rad). Get $\omega$ from a spring ::: $\omega =\sqrt{k/m}$. Amplitude from $x_0,v_0$ ::: $A=\sqrt{x_0^2+v_0^2/{\omega }^2}$ (divide velocity by $\omega$ first). Choose the right $\varphi$ quadrant ::: check sign of $\cos \varphi$ (=sign $x_0$) AND sign of $\sin \varphi$ (from $v=-A\omega \sin \varphi$). Speed away from centre ::: $v=\pm \omega \sqrt{A^2-x^2}$, never $A\omega$. Time between two states ::: solve for the argument, then $t=(\text{argument})/\omega$; phase-to-time is $\Delta t=\Delta \varphi /\omega$.