Simple harmonic motion

SHM is the centrepiece of the CBSE Class 11 Physics chapter "Oscillations," sitting right after periodic motion and displacement-as-a-function-of-time (built in the prerequisite article) and right before the formal link to uniform circular motion (the next article in this sequence). The prescribed scope here is precise: the SHM equations $x=A\cos(\omega t+\phi)$, $v=-A\omega\sin(\omega t+\phi)$ and $a=-\omega^2x$; the restoring force law $F=-kx$ and the loaded-spring period $T=2\pi\sqrt{m/k}$; the simple pendulum's $T=2\pi\sqrt{L/g}$; and energy in SHM. Damped, forced and driven oscillations and resonance have been removed from the 2025-26 CBSE prescription — they are mentioned only in passing below as "studied in Class 12 / higher classes," never developed.

The defining equation of SHM describes displacement as a single sinusoid in time. For a particle oscillating about a mean position taken as the origin,

where: $x(t)$ is the displacement from the mean position at time $t$; $A$ is the amplitude, the maximum displacement on either side; $\omega$ is the angular frequency in radians per second, related to the time period $T$ and ordinary frequency $\nu$ by $\omega = 2\pi/T = 2\pi\nu$; and $\phi$ is the phase constant, fixing where in the cycle the motion starts at $t=0$. The quantity $(\omega t+\phi)$ itself is called the phase of the motion at time $t$ — it tells you exactly which point of the repeating cycle the particle occupies. Differentiating once gives the velocity, and again gives the acceleration:

The last step is the heart of the whole chapter: $a = -\omega^2x$. Acceleration is always directed back toward the mean position (the minus sign) and its size grows in direct proportion to how far the particle has moved away (the $x$). This single relation is both necessary and sufficient for SHM — any motion obeying it is SHM, and any true SHM must obey it. A physical reading helps fix the picture. At the extreme positions $x=\pm A$, the particle is momentarily at rest ($v=0$) but is being pulled back with the largest possible acceleration, $a_{max}=\omega^2A$. At the mean position $x=0$, the pull vanishes ($a=0$) but the particle is moving at its fastest, $v_{max}=A\omega$. Everywhere in between, speed and acceleration trade off smoothly — as one grows the other shrinks — which is exactly why the motion keeps overshooting the mean position, decelerating, stopping at the far extreme, and reversing, cycle after cycle.

Watch a peg fixed on the rim of a turntable spinning at a steady angular speed $\omega$, but instead of watching the peg itself, watch only its shadow cast on a screen along one diameter. That shadow does not sweep round smoothly — it slows at the two edges and speeds up in the middle, oscillating back and forth exactly like a mass on a spring. If the turntable has radius $A$ and the peg starts at angle $\phi$ from the reference diameter, the shadow's position along that diameter is precisely $x = A\cos(\omega t + \phi)$ — the SHM equation, with no extra assumptions. This is why $\omega$ in SHM is called angular frequency even though nothing is visibly rotating in a swinging pendulum or a bobbing spring: it is literally the angular speed of the imaginary reference circle whose projection reproduces the same $x(t)$.

A particle in SHM continuously exchanges energy between kinetic and potential forms, but the sum of the two is fixed for a given amplitude. Since $v = -A\omega\sin(\omega t+\phi)$ and $x=A\cos(\omega t+\phi)$, using $\sin^2+\cos^2=1$ gives $v^2 = \omega^2(A^2-x^2)$, so the kinetic energy at displacement $x$ is

The restoring force $F=-kx=-m\omega^2x$ (using $k=m\omega^2$) is conservative, and the potential energy stored against it, taken as zero at the mean position, works out to

where: $m$ is the oscillating mass, $\omega$ its angular frequency, $A$ the amplitude and $x$ the instantaneous displacement. Adding the two, the $x^2$ terms cancel:

The total mechanical energy is constant throughout the motion, proportional to the square of the amplitude and to the square of the angular frequency — double the amplitude and the energy quadruples. At the mean position ($x=0$) all the energy is kinetic ($KE=E$, $PE=0$); at either extreme ($x=\pm A$) all of it is potential ($PE=E$, $KE=0$); everywhere in between the two share the fixed total, tracing out complementary parabolas when plotted against $x$.

A simple pendulum — a point mass $m$ on a light, inextensible string of length $L$ — is not an ideal spring, yet for small swings it obeys the very same $a=-\omega^2x$ pattern, and hence is SHM to a very good approximation. When the string makes a small angle $\theta$ with the vertical, gravity's component along the arc is $-mg\sin\theta$, restoring the bob toward the bottom. For angles small enough that $\theta$ is measured in radians and $\sin\theta \approx \theta$ (true to better than one per cent below about $10$ degrees), the restoring force becomes proportional to the arc-length displacement $x = L\theta$, giving $a = -(g/L)x$. Matching this against $a=-\omega^2x$ gives $\omega=\sqrt{g/L}$ and hence the CBSE headline result

Notice this period depends only on the length $L$ and on $g$ — not on the mass of the bob and not on the amplitude of swing (as long as the small-angle condition holds), which is exactly why pendulum clocks kept reliable time for centuries regardless of small changes in how hard they were wound.

Board level - amplitude, angular frequency and displacement at a given time

A particle executes SHM with amplitude $0.2\ \text{m}$ and angular frequency $5\ \text{rad/s}$. At $t=0$ the particle is at its mean position moving in the positive direction. Write the equation for $x(t)$, and find the displacement, velocity and acceleration at $t=0.3\ \text{s}$.

Step 1. Choose the correct form. Since $x(0)=0$ with $v(0)>0$, using $x = A\sin(\omega t)$ satisfies both: $x(0)=0$, and $v(0)=A\omega\cos(0)=A\omega>0$. So $x(t) = 0.2\sin(5t)\ \text{m}$, equivalent to $A\cos(\omega t+\phi)$ with $\phi=-\pi/2$.

Step 2. Displacement at $t=0.3\ \text{s}$: $x = 0.2\sin(5\times0.3) = 0.2\sin(1.5\ \text{rad})$. Since $\sin(1.5) \approx 0.997$, $x \approx 0.2\times0.997 \approx 0.199\ \text{m}$ — almost at the extreme $A=0.2\ \text{m}$.

Step 3. Velocity at the same instant: $v = A\omega\cos(5t) = 0.2\times5\times\cos(1.5) = 1.0\times0.0707 \approx 0.071\ \text{m/s}$, small, consistent with the particle being nearly stationary near an extreme.

Step 4. Acceleration: $a=-\omega^2x = -(5)^2\times0.199 \approx -4.98\ \text{m/s}^2$, close to the maximum magnitude $\omega^2A = 25\times0.2=5\ \text{m/s}^2$, directed back toward the mean position. Answer: $x(t)=0.2\sin(5t)\ \text{m}$; at $t=0.3\ \text{s}$, $x\approx0.199\ \text{m}$, $v\approx0.071\ \text{m/s}$, $a\approx-4.98\ \text{m/s}^2$.

Competitive level - spring-mass period, maximum speed and energy split

A block of mass $0.5\ \text{kg}$ is attached to a spring of force constant $k = 200\ \text{N/m}$ and set oscillating with amplitude $A = 0.10\ \text{m}$ on a smooth horizontal table. Find (a) the angular frequency and time period, (b) the maximum speed, and (c) the kinetic and potential energy when the block is at $x = 0.05\ \text{m}$.

Step 1. Angular frequency from $\omega=\sqrt{k/m} = \sqrt{200/0.5} = \sqrt{400} = 20\ \text{rad/s}$. Time period $T = 2\pi\sqrt{m/k} = 2\pi/\omega = 2\pi/20 \approx 0.314\ \text{s}$.

Step 2. Maximum speed occurs at the mean position: $v_{max} = A\omega = 0.10\times20 = 2.0\ \text{m/s}$.

Step 3. Total energy: $E = \tfrac12 m\omega^2A^2 = \tfrac12\times0.5\times400\times(0.10)^2 = \tfrac12\times0.5\times400\times0.01 = 1.0\ \text{J}$.

Step 4. At $x=0.05\ \text{m}$ (half the amplitude): $PE = \tfrac12 m\omega^2x^2 = \tfrac12\times0.5\times400\times0.0025 = 0.25\ \text{J}$, and by conservation $KE = E-PE = 1.0-0.25 = 0.75\ \text{J}$. Answer: $\omega=20\ \text{rad/s}$, $T\approx0.314\ \text{s}$, $v_{max}=2.0\ \text{m/s}$, and at $x=0.05\ \text{m}$: $PE=0.25\ \text{J}$, $KE=0.75\ \text{J}$ (total $1.0\ \text{J}$). Timing note: under ninety seconds once the pattern is drilled. Get $\omega$ from $\sqrt{k/m}$ first — everything else (period, $v_{max}=A\omega$, total energy $\tfrac12m\omega^2A^2$) follows directly by substitution, and the energy split at any $x$ is just $E$ minus $\tfrac12m\omega^2x^2$; there is no need to ever go back to solving the equation of motion from scratch.

• Calling any repeating motion SHM. A motion is SHM only if it is a pure sine/cosine of time and, equivalently, only if $a=-\omega^2x$ holds exactly — a bouncing ball or a person pacing back and forth is periodic but not SHM.
• Writing $x=A\cos(\omega t)$ by habit and dropping $\phi$. That special case is only correct if the particle starts at the positive extreme; general starting conditions (like starting at the mean position, as in the board example above) need a nonzero phase constant.
• Confusing angular frequency $\omega$ (rad/s) with ordinary frequency $\nu$ (Hz) — they are related by $\omega=2\pi\nu$, and mixing them up throws every numerical answer off by a factor of $2\pi$.
• Believing the loaded-spring period depends on $g$ or on amplitude. It does not — $T=2\pi\sqrt{m/k}$ contains only $m$ and $k$; gravity merely relocates the mean position when the spring hangs vertically.
• Applying $T=2\pi\sqrt{L/g}$ at large swing angles. The formula is only valid for small $\theta$ (roughly under $10$ degrees, where $\sin\theta\approx\theta$); at large amplitude the period actually increases slightly, a correction outside this syllabus.
• Sign errors differentiating $\cos$ to get $v$ and $a$: remember $\frac{d}{dt}\cos(\omega t+\phi)=-\omega\sin(\omega t+\phi)$, not $+\omega\sin(\cdot)$.

• Write and interpret $x=A\cos(\omega t+\phi)$, identifying amplitude, angular frequency and phase constant from an equation or a physical situation.
• Derive $v=-A\omega\sin(\omega t+\phi)$ and $a=-\omega^2x$ by differentiation, and explain why $a=-\omega^2x$ is the defining test for SHM.
• Explain SHM as the projection of uniform circular motion onto a diameter.
• Derive $T=2\pi\sqrt{m/k}$ from $F=-kx$ and Newton's second law, and state which quantities the period does and does not depend on.
• Derive $T=2\pi\sqrt{L/g}$ for a simple pendulum under the small-angle approximation, and know why the approximation is needed.
• Compute $KE=\tfrac12m\omega^2(A^2-x^2)$, $PE=\tfrac12m\omega^2x^2$ and the constant total $E=\tfrac12m\omega^2A^2$ at any displacement.

Ask the tutor: Not sure how to pick the right form of $x(t)$ (sine vs. cosine, and the correct phase constant $\phi$) from a word problem's starting conditions, or want to see the pendulum's small-angle approximation checked numerically? Ask the AI tutor to walk through a few initial-condition and energy-split problems with you.