In simple harmonic motion, kinetic energy and potential energy trade off perfectly while their sum stays constant. From that single principle drops out the velocity formula $v^2 = n^2(a^2 - x^2)$, the most useful equation in the SHM toolkit. With it you can read off speed at any displacement, locate where the particle is when given a speed, and check answers without integrating $\ddot{x} = -n^2 x$ a second time.
+5 XP warm-up A particle undergoes SHM about $x = 0$ with $\ddot{x} = -n^2 x$.
A particle undergoes SHM about $x = 0$ with $\ddot{x} = -n^2 x$. Before checking where in the motion is the particle moving fastest, and where is it momentarily at rest? Try to give a one-line reason in terms of energy.
+5 XP to read Every SHM energy question rewards two habits: identify the amplitude and angular frequency from the equation...
Every SHM energy question rewards two habits: identify the amplitude $a$ and angular frequency $n$ from the equation of motion, then use $v^2 = n^2(a^2 - x^2)$ rather than re-integrating $\ddot{x} = -n^2 x$. Almost every velocity-at-displacement question collapses to a single substitution into this formula.
The read-identify-substitute workflow: (1) read off $n^2$ from $\ddot{x} = -n^2 x$, (2) identify amplitude $a$ from initial conditions or extremes, (3) substitute into $v^2 = n^2(a^2 - x^2)$.
$\ddot{x} = -n^2 x$ · $v^2 = n^2(a^2 - x^2)$ · $v_{\max} = na$ at $x = 0$
Start from $\ddot{x} = -n^2 x$.
Start from $\ddot{x} = -n^2 x$. Using the chain-rule identity $\ddot{x} = v\,\dfrac{dv}{dx}$:
Energy reading. Rearrange: $\tfrac12 v^2 + \tfrac12 n^2 x^2 = \tfrac12 n^2 a^2$. The left side is KE + PE (per unit mass); the right side is the total mechanical energy. Conservation of energy is exactly the statement that this sum is constant throughout the motion.
Worked through the hook: $a = 4$, $n = 3$.
Derivation: $\ddot{x} = v\,\dfrac{dv}{dx} = -n^2 x \Rightarrow v^2 = n^2(a^2 - x^2)$ · Energy: $\tfrac12 v^2 + \tfrac12 n^2 x^2 = \tfrac12 n^2 a^2$ · $v_{\max} = na$ at $x = 0$; $v = 0$ at $x = \pm a$ · Graph of $v^2$ vs $x$: downward parabola, vertex $(0, n^2 a^2)$, roots $\pm a$
Pause, copy $v^2 = n^2(a^2-x^2)$, its derivation from $a = v\,dv/dx$, $v_{\max} = na$ at $x=0$, and the $v^2$-vs-$x$ parabola (vertex $(0,n^2a^2)$, roots $\pm a$) into your book.
Quick check: A particle moves in SHM with $\ddot{x} = -16x$ and amplitude $a = 5$. What is its speed at $x = 3$?
We just saw that $v^2 = n^2(a^2-x^2)$ comes from integrating $\ddot{x} = -n^2 x$ via $a = v\,dv/dx$, with $v_{\max} = na$ at $x = 0$ and $v = 0$ at the amplitude $x = \pm a$.
We just saw that $v^2 = n^2(a^2-x^2)$ comes from integrating $\ddot{x} = -n^2 x$ via $a = v\,dv/dx$, with $v_{\max} = na$ at $x = 0$ and $v = 0$ at the amplitude $x = \pm a$. That raises a question: what is the energy interpretation of $v^2 = n^2(a^2-x^2)$? This card answers it → KE $= \tfrac{1}{2}v^2$ and PE $= \tfrac{1}{2}n^2x^2$ (per unit mass) sum to the constant total energy $\tfrac{1}{2}n^2a^2$; they are exactly out of phase.
Per unit mass, define
Trade-off through the cycle:
KE = $\tfrac12 v^2$ (per unit mass); PE = $\tfrac12 n^2 x^2$ (per unit mass) · Total energy $E = \tfrac12 n^2 a^2$, constant throughout the motion · KE max at centre; PE max at extremes; they are exactly out of phase · Amplitude from one snapshot: $a^2 = x_1^2 + v_1^2/n^2$
Pause, copy KE $= \tfrac{1}{2}v^2$, PE $= \tfrac{1}{2}n^2x^2$, total $E = \tfrac{1}{2}n^2a^2$ (constant), the KE-max-at-centre / PE-max-at-extremes rule, and the one-snapshot formula $a^2 = x_1^2 + v_1^2/n^2$ into your book.
Did you get this? True or false: at the centre of the motion ($x = 0$) the kinetic energy is at its maximum and the potential energy is zero.
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
A particle moves in SHM with $\ddot{x} = -25x$, amplitude $a = 6$ m. Find its speed when $x = 4$ m, and its maximum speed.
A particle in SHM with $n = 2$ and amplitude $a = 5$ has speed $v = 6$ m/s. Find the displacement $x$ at this instant.
A particle in SHM with $n = 4$ has speed $v = 8$ m/s when $x = 3$ m. Find the amplitude and the maximum speed.
Fill the gap: The SHM velocity formula is $v^2 = n^2(a^2 - $ $)$. The maximum speed is $v_{\max} =$ at $x = 0$.
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for SHM with $n = 3$ and amplitude $a = 4$, the maximum speed is $v_{\max} = 12$ and it occurs at $x = 0$.
Activities · practice with the ideas
Activities · practice with the ideas
A particle in SHM has $\ddot{x} = -9x$ and amplitude $a = 5$. Find its speed at $x = 3$ and its maximum speed.
A particle in SHM with $n = 4$ and amplitude $a = 3$ has what speed at $x = 2$?
For SHM with $n = 2$ and amplitude $a = 6$, find the displacements at which the particle has half its maximum speed.
A particle in SHM with $n = 5$ has speed $v = 20$ m/s at $x = 3$ m. Find the amplitude and the maximum speed.
For SHM with amplitude $a$ and frequency $n$, find the displacement at which KE equals PE.
Odd one out: Three of these statements about SHM are correct. Which one is NOT?
Earlier you predicted the speed at $x = 2$ for SHM with $a = 4$ and $n = 3$, and sketched $v^2$ vs $x$.
The energy formula gave $v_{\max} = na = 12$ at $x = 0$, and $v = 6\sqrt{3} \approx 10.39$ at $x = 2$. The graph of $v^2$ vs $x$ is a downward parabola $v^2 = 9(16 - x^2)$, peaking at $(0, 144)$ and crossing the $x$-axis at $\pm 4$. The whole power of $v^2 = n^2(a^2 - x^2)$ is that it lets you skip $x(t)$ entirely whenever the question is about speed and position rather than time.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A particle undergoes SHM with $\ddot{x} = -36x$ and amplitude $a = 4$. Find its speed at $x = 2$ and its maximum speed. (2 marks)
Q2. A particle in SHM with $n = 3$ has speed $v = 12$ m/s when $x = 3$ m. Find the amplitude and state the speed at $x = 0$. (3 marks)
Q3. A particle moves in SHM with amplitude $a$ and angular frequency $n$. (a) Show, by integrating $\ddot{x} = -n^2 x$, that $v^2 = n^2(a^2 - x^2)$. (b) Find the displacement at which the kinetic energy equals the potential energy. (3 marks)
Activity answers:
1. $n = 3$, $a = 5$. At $x = 3$: $v^2 = 9(25 - 9) = 144$, $v = 12$. $v_{\max} = na = 15$ at $x = 0$.
2. $n = 4$, $a = 3$. At $x = 2$: $v^2 = 16(9 - 4) = 80$, $v = 4\sqrt{5} \approx 8.94$ m/s.
3. $v_{\max} = na = 12$, so half is $v = 6$. Then $36 = 4(36 - x^2)$, giving $x^2 = 27$, $x = \pm 3\sqrt{3}$.
4. $a^2 = x^2 + v^2/n^2 = 9 + 400/25 = 25$, so $a = 5$. $v_{\max} = na = 25$ m/s.
5. KE = PE means $\tfrac12 v^2 = \tfrac12 n^2 x^2$, i.e. $v^2 = n^2 x^2$. Combined with $v^2 = n^2(a^2 - x^2)$ this gives $a^2 = 2x^2$, so $x = \pm a/\sqrt{2}$.
Q1 (2 marks): $n = 6$, $a = 4$. $v^2 = 36(16 - 4) = 432$, so $v = 12\sqrt{3}$ m/s [1]. $v_{\max} = na = 24$ m/s at $x = 0$ [1].
Q2 (3 marks): Use $v^2 = n^2(a^2 - x^2)$: $144 = 9(a^2 - 9)$ [1], $a^2 - 9 = 16$, $a = 5$ m [1]. $v_{\max} = na = 15$ m/s at $x = 0$ [1].
Q3 (3 marks): (a) $v\,dv/dx = -n^2 x \Rightarrow v^2/2 = -n^2 x^2/2 + C$; at $x = a$, $v = 0$ gives $C = n^2 a^2/2$, so $v^2 = n^2(a^2 - x^2)$ [1]. (b) Setting $\tfrac12 v^2 = \tfrac12 n^2 x^2$ gives $v^2 = n^2 x^2$ [1]; combine with $v^2 = n^2(a^2 - x^2)$: $a^2 - x^2 = x^2$, so $x = \pm a/\sqrt{2}$ [1].
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