Exercises — Velocity and acceleration in SHM — v = ω√(A² − x²)

(a) Max speed happens at the centre, $x=0$, where the "left-over distance under the root" is largest: $v_{\max }=\omega A=5\times 0.08=0.4\text{ m/s}(\text{at }x=0)$ (b) Max acceleration happens at the extremes, $x=\pm A$, where the restoring pull is strongest: $|a{|}_{\max }={\omega }^{2}A=25\times 0.08=2{\text{ m/s}}^2(\text{at }x=\pm A)$ Notice the swap: speed is biggest where acceleration is zero, and vice versa.

Use the position form directly — no time needed: $v=\omega \sqrt{A^2-x^2}=5\sqrt{0.08^2-0.048^2}=5\sqrt{0.0064-0.002304}$ $=5\sqrt{0.004096}=5\times 0.064=0.32\text{ m/s}$

$a=-{\omega }^{2}x=-25\times 0.048=-1.2{\text{ m/s}}^2$ The minus means it points opposite to $x$. Since $x$ is positive (right of centre), $a$ points left, back toward the centre — a restoring acceleration.

Start from the velocity formula and solve for $x$: $v=\omega \sqrt{A^2-x^2}\Rightarrow \frac{v}{\omega }=\sqrt{A^2-x^2}$ Square both sides (this removes the root): $\frac{v^2}{{\omega }^2}=A^2-x^2\Rightarrow x^2=A^2-\frac{v^2}{{\omega }^2}$ $x^2=0.10^2-\frac{0.48^2}{8^2}=0.01-\frac{0.2304}{64}=0.01-0.0036=0.0064$ $x=\pm \sqrt{0.0064}=\pm 0.08\text{ m}$ Both because the particle reaches speed $0.48$ m/s on the left and on the right of centre.

$\frac{v(\frac{1}{2}A)}{v(0)}=\frac{\omega \sqrt{A^2-(\frac{1}{2}A{)}^2}}{\omega \sqrt{A^2-0}}=\frac{\sqrt{A^2-\frac{1}{4}{A}^2}}{\sqrt{A^2}}=\frac{\sqrt{\frac{3}{4}{A}^2}}{A}=\frac{\sqrt{3}}{2}\approx 0.866$ So halfway out you still keep 87% of top speed — speed sheds slowly near the middle, then plunges near the edge.

(a) The formula uses $x^2$, so the sign of $x$ does not affect speed: $v=10\sqrt{0.06^2-0.036^2}=10\sqrt{0.0036-0.001296}=10\sqrt{0.002304}=10\times 0.048=0.48\text{ m/s}$ (b) Acceleration keeps the sign: $a=-{\omega }^{2}x=-100\times (-0.036)=+3.6{\text{ m/s}}^2$ Positive because $x$ is negative — again pointing back toward centre.

Start from $v^2={\omega }^2(A^2-x^2)$ and rearrange into standard ellipse form (divide so the right side is $1$): $v^2+{\omega }^2{x}^2={\omega }^2{A}^2\Rightarrow \frac{x^2}{A^2}+\frac{v^2}{(\omega A{)}^2}=1$ This is $\frac{x^2}{a^2}+\frac{v^2}{b^2}=1$, an ellipse with

• horizontal semi-axis $a=A=0.10$ m (the max displacement),
• vertical semi-axis $b=\omega A=8\times 0.10=0.80$ m/s (the max speed).

As you move around the ellipse you sweep through every state of the motion — outbound (top half, $v>0$) and inbound (bottom half, $v<0$).

Use $v^2={\omega }^2(A^2-x^2)$ at both points: $v_1^2={\omega }^2(A^2-x_1^2),v_2^2={\omega }^2(A^2-x_2^2)$ Subtract to kill $A^2$ (this is why two points suffice): $v_1^2-v_2^2={\omega }^2(x_2^2-x_1^2)$ $0.09-0.0256={\omega }^2(0.0036-0.0016)\Rightarrow 0.0644={\omega }^2(0.002)$ ${\omega }^2=32.2\Rightarrow \omega \approx 5.675\text{ rad/s}$ Now back-substitute into the first equation to get $A$: $A^2=x_1^2+\frac{v_1^2}{{\omega }^2}=0.0016+\frac{0.09}{32.2}=0.0016+0.002795=0.004395$ $A\approx 0.0663\text{ m}$

Set the two magnitudes equal: $\omega \sqrt{A^2-x^2}={\omega }^2|x|$ Divide by $\omega$ and square: $A^2-x^2={\omega }^2{x}^2\Rightarrow A^2=x^2(1+{\omega }^2)\Rightarrow x=\frac{A}{\sqrt{1+{\omega }^2}}$ $x=\frac{0.10}{\sqrt{1+16}}=\frac{0.10}{\sqrt{17}}=\frac{0.10}{4.1231}\approx 0.02425\text{ m}$ (Note this is not a physically special point — it depends on the units chosen — but it exercises equating the two formulas.)

Effective spring constant $k=m{\omega }^2=0.20\times 64=12.8$ N/m. See Energy in SHM — kinetic and potential. (a) Total energy: $E=\frac{1}{2}k{A}^2=\frac{1}{2}(12.8)(0.10^2)=0.064\text{ J}$ (b) Speed at $x=0.06$: $v=8\sqrt{0.10^2-0.06^2}=8\sqrt{0.0064}=8(0.08)=0.64$ m/s. $KE=\frac{1}{2}m{v}^2=\frac{1}{2}(0.20)(0.64^2)=\frac{1}{2}(0.20)(0.4096)=0.04096\text{ J}$ (c) Potential energy: $PE=\frac{1}{2}k{x}^2=\frac{1}{2}(12.8)(0.06^2)=\frac{1}{2}(12.8)(0.0036)=0.02304\text{ J}$ Check: $KE+PE=0.04096+0.02304=0.064$ J $=E$. ✔

Max speed occurs when $\cos (\omega t)=\pm 1$, i.e. $\omega t=0$. Max displacement occurs when $\sin (\omega t)=\pm 1$, i.e. $\omega t=\frac{\pi }{2}$. The gap in phase is $\frac{\pi }{2}$ — a quarter turn. Since a full period is $2\pi$ of phase: $\frac{\Delta t}{T}=\frac{\pi /2}{2\pi }=\frac{1}{4}$ So velocity leads displacement by a quarter period ($90^{\circ }$), exactly as in the parent table. See Phase and phase difference.

Average speed $=$ (distance travelled) $÷$ (time taken). The distance from centre to extreme is $A=0.10$ m. The time for a quarter cycle is $t=\frac{T}{4}=\frac{1}{4}\cdot \frac{2\pi }{\omega }=\frac{\pi }{2\omega }=\frac{\pi }{16}\approx 0.19635\text{ s}$ (using Angular frequency ω and time period T). Therefore $\bar{v}=\frac{A}{T/4}=\frac{0.10}{0.19635}\approx 0.5093\text{ m/s}$ Compare to $v_{\max }=\omega A=0.80$ m/s: the average is about $64\%$ of the peak — because the particle spends more time crawling near the slow extreme.

The acceleration pins down $\omega$ immediately, because $|a|={\omega }^2|x|$ at a known $x$: $16={\omega }^2(0.04)\Rightarrow {\omega }^2=400\Rightarrow \omega =20\text{ rad/s}$ Now use the speed formula: $v^2={\omega }^2(A^2-x^2)\Rightarrow 0.6^2=400(A^2-0.04^2)$ $0.36=400A^2-400(0.0016)=400A^2-0.64$ $400A^2=1.00\Rightarrow A^2=0.0025\Rightarrow A=0.05\text{ m}$

$KE=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2(A^2-x^2)$ and $PE=\frac{1}{2}m{\omega }^2{x}^2$. Set them equal: $A^2-x^2=x^2\Rightarrow A^2=2x^2\Rightarrow x=\frac{A}{\sqrt{2}}$ For $A=0.10$ m: $x=0.10/\sqrt{2}\approx 0.07071$ m. At this point $PE=\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2\cdot \frac{A^2}{2}=\frac{1}{4}m{\omega }^2{A}^2=\frac{1}{2}E$. So each holds exactly half the total. ✔

Compare with the general ellipse $\frac{x^2}{A^2}+\frac{v^2}{(\omega A{)}^2}=1$ from L3-Q1.

• $A^2=0.09\Rightarrow A=0.30$ m.
• $(\omega A{)}^2=9\Rightarrow \omega A=3\Rightarrow \omega =3/0.30=10$ rad/s.
• $v_{\max }=\omega A=3$ m/s (the vertical semi-axis).
• Acceleration at $x=-0.15$ m: $a=-{\omega }^{2}x=-100(-0.15)=+15$ m/s² (pointing back toward centre).

Can you, without notes: (1) state where $v$ and $a$ are max/min? (2) invert $v=\omega \sqrt{A^2-x^2}$ for $x$? (3) recognise the $(x,v)$ ellipse and read $A$, $\omega$ off it? (4) find $\omega$ from acceleration alone? (5) locate where $KE=PE$? If any is shaky, redo that level.