Chapter 13 · Oscillations

13.1 Periodic Motion and Simple Harmonic Motion

Motions that repeat have a period $T$ and a frequency $\nu=1/T$. The simplest to-and-fro motion, simple harmonic motion, is $x=A\cos(\omega t+\phi)$ — the shadow of a particle going steadily round a circle.

A great deal of the world repeats itself. The Earth spins once a day and goes round the Sun once a year; a healthy heart beats about once a second; the needle of a sewing machine, the piston of a car engine and the wings of a mosquito go up and down hundreds of times a second; the atoms in a crystal vibrate about their places ten million million times a second. To describe any of these we first need two numbers: how long one repetition takes, and how many repetitions happen each second.

This section defines periodic and oscillatory motion and their period and frequency. It then studies the most important oscillation of all, simple harmonic motion (SHM), whose displacement–time graph is a pure cosine curve: $x=A\cos(\omega t+\phi)$. You will learn what each symbol in that equation means and how to find it from the way a motion starts, and you will see that SHM is exactly what you get when you watch a particle in uniform circular motion edge-on. That picture — the reference circle — turns many SHM problems into simple geometry.

Periodic and oscillatory motion

The hands of a clock, a planet in its orbit and a swinging pendulum all return to the same state — the same position, moving with the same velocity — at regular intervals. Between those returns, each goes through the same sequence of positions again and again.

📘 Definition 13.1 — Periodic motion, period and frequency

A motion that repeats itself exactly at regular intervals of time is periodic. The smallest interval after which it repeats is its period (or time period) $T$, measured in seconds. The number of repetitions per second is its frequency

$$\nu=\frac1T .$$

The SI unit of frequency is the hertz: $1\ \text{Hz}=1$ repetition per second $=1\ \text{s}^{-1}$.

A quantity $f(t)$ that describes a periodic motion satisfies $f(t+T)=f(t)$ for every $t$; so do $f(t+2T)$, $f(t+3T)$… — the period is the smallest such interval. Uniform circular motion (Section 3.4) is periodic, with period $T=\dfrac{2\pi}{\omega}$. Many periodic motions have something extra: the body moves back and forth about a fixed position.

📘 Definition 13.2 — Oscillatory motion, mean position and displacement

A periodic to-and-fro motion of a body about a fixed position is an oscillation (or vibration), and the motion is oscillatory. The fixed position is the mean position or equilibrium position: if the body is placed there at rest, the net force on it is zero and it stays there. When the body is displaced from it, a restoring force pulls it back.

The displacement of an oscillating system is the change, from its equilibrium value, of whatever quantity varies: the position $x$ of a block on a spring, the angle $\theta$ of a pendulum, the excess pressure in a sound wave, the voltage across a capacitor in a radio circuit.

Every oscillation is periodic, but not every periodic motion is an oscillation: the tip of a fan blade repeats its motion every revolution, but it goes round, not back and forth. Real oscillations also die away slowly because of friction, so they are only nearly periodic; over a few cycles that hardly matters. Oscillations are slow for a ship rolling on the sea (several seconds per cycle), fast for a guitar string (hundreds of hertz), and very fast for the quartz crystal in a watch ($32\,768\ \text{Hz}$) or the atoms in a solid (about $10^{13}\ \text{Hz}$).

✏️ Example 1 — Heartbeats, wings and crystals

Find the frequency or the period as appropriate: (a) a heart beating $75$ times a minute; (b) the wings of a hummingbird beating at $50\ \text{Hz}$; (c) a watch crystal vibrating at $32\,768\ \text{Hz}$; (d) a pendulum that completes $20$ oscillations in $36.0\ \text{s}$.

💡 Show solution▶

(a) $\nu=\dfrac{75}{60\ \text{s}}=1.25\ \text{Hz}$; $T=\dfrac1\nu=0.80\ \text{s}$.

(b) $T=\dfrac{1}{50\ \text{Hz}}=0.020\ \text{s}=20\ \text{ms}$.

(c) $T=\dfrac{1}{32\,768}\ \text{s}=3.05\times10^{-5}\ \text{s}\approx30.5\ \mu\text{s}$. ($32\,768=2^{15}$: the watch halves this frequency fifteen times to get one tick per second.)

(d) $T=\dfrac{36.0\ \text{s}}{20}=1.80\ \text{s}$ ($20$ is a counted, exact number); $\nu=\dfrac{1}{1.80}=0.556\ \text{Hz}$.

Check $\nu T=1$ in every case ✓. Timing many oscillations and dividing is how a period is measured accurately in the laboratory: the reaction-time error is shared out over all of them.

Periodic functions. The simplest periodic functions are $\sin\omega t$ and $\cos\omega t$. The sine repeats when its argument increases by $2\pi$, i.e. when $t$ increases by $\dfrac{2\pi}{\omega}$, so their period is $T=\dfrac{2\pi}{\omega}$. A sum of periodic functions is periodic if there is a time that is a whole number of periods of every term; the period of the sum is the smallest such time (the least common multiple of the separate periods). Remarkably, the converse also holds: any periodic function of period $T$, however jagged, can be built up by adding sines and cosines of periods $T,\ T/2,\ T/3,\ldots$ with suitable amplitudes (a result due to Fourier). Sines and cosines are the building blocks of all periodic motion.

✏️ Example 2 — Periodic or not?

Which of these functions of time are periodic? Give the period of each periodic one ($\omega$ is a positive constant). (a) $\sin\omega t+\cos\omega t$ (b) $\sin\omega t+\cos2\omega t$ (c) $\sin^2\omega t$ (d) $\cos\dfrac{\pi t}{2}+\sin\dfrac{\pi t}{3}$, with $t$ in seconds (e) $e^{-\omega t}$

💡 Show solution▶

(a) Both terms have period $\dfrac{2\pi}{\omega}$, so the sum is periodic with $T=\dfrac{2\pi}{\omega}$.

(b) Periods $\dfrac{2\pi}{\omega}$ and $\dfrac{\pi}{\omega}$; $\dfrac{2\pi}{\omega}$ is two periods of the second term, so $T=\dfrac{2\pi}{\omega}$.

(c) Using $\sin^2\theta=\tfrac12(1-\cos2\theta)$ (a double-angle identity from mathematics), $\sin^2\omega t=\tfrac12-\tfrac12\cos2\omega t$, periodic with $T=\dfrac{2\pi}{2\omega}=\dfrac{\pi}{\omega}$ — half the period of $\sin\omega t$ (a negative sine gives the same square).

(d) Periods $\dfrac{2\pi}{\pi/2}=4\ \text{s}$ and $\dfrac{2\pi}{\pi/3}=6\ \text{s}$. The smallest time that is a whole number of both is $12\ \text{s}$ (3 periods of the first, 2 of the second): $T=12\ \text{s}$.

(e) $e^{-\omega t}$ decreases steadily and never returns to a value it has had: not periodic.

Check Interactive 1 tests these by sliding a copy of the graph along by a trial period until it fits on top of the original.

Choose a function of time (here $\omega=\pi\ \text{rad/s}$). The red dashed curve is a copy of the graph shifted left by a trial time $\tau$, i.e. $f(t+\tau)$. When $\tau$ is a period, the copy lands exactly on the original. Press Play to sweep $\tau$, or drag the slider, and watch the mismatch readout.

function:trial shift $\tau$:1.50 s

mismatch $\max|f(t+\tau)-f(t)|$—

is $\tau$ a period?—

smallest period—

simple harmonic?—

Simple harmonic motion

Look back at the trace drawn by the pen on the oscillating block in the chapter introduction. It is not just any wavy line: it is a cosine curve. The block moves fastest as it passes the mean position and slows smoothly to a stop at each end, and the same shape is drawn by a pendulum swinging through a small angle, a tuning-fork prong, or a mass bobbing on a vertical spring. This motion is important enough to have its own name.

📘 Definition 13.3 — Simple harmonic motion

A particle moving along the $x$-axis executes simple harmonic motion (SHM) about the origin if its displacement varies with time as

$$x(t)=A\cos(\omega t+\phi),$$

where $A$, $\omega$ and $\phi$ are constants:

the amplitude $A\ (\gt0)$ is the magnitude of the largest displacement, so $-A\le x\le A$;
the angular frequency $\omega\ (\gt0)$, in $\text{rad/s}$, sets how fast the motion repeats;
the argument $\omega t+\phi$ is the phase of the motion at time $t$ (an angle in radians), and its value at $t=0$, the constant $\phi$, is the phase constant (or initial phase).

The cosine repeats whenever its argument grows by $2\pi$. The phase grows by $2\pi$ when $t$ increases by $\dfrac{2\pi}{\omega}$, so $x\!\left(t+\dfrac{2\pi}{\omega}\right)=A\cos(\omega t+\phi+2\pi)=x(t)$.

📐 Result 13.1 — Period and frequency of SHM

The motion $x=A\cos(\omega t+\phi)$ is periodic, with

$$T=\frac{2\pi}{\omega},\qquad \nu=\frac1T=\frac{\omega}{2\pi},\qquad \omega=2\pi\nu .$$

The period depends only on $\omega$ — not on the amplitude $A$ or the phase constant $\phi$. The particle moves between $x=+A$ and $x=-A$, and in one period it travels a distance $4A$.

What the phase constant does. Two oscillators with the same $A$ and $\omega$ but different $\phi$ make the same motion, shifted in time. The phase constant tells you where the cycle starts: at $t=0$, $x(0)=A\cos\phi$. With $\phi=0$ the particle starts at $+A$; with $\phi=\pi$ at $-A$; with $\phi=-\dfrac\pi2$, $x=A\cos\left(\omega t-\dfrac\pi2\right)=A\sin\omega t$, so it starts at the mean position, heading towards $+x$. A sine is therefore SHM too, with a phase constant $\dfrac\pi2$ less than the cosine form: $A\sin(\omega t+\alpha)=A\cos\left(\omega t+\alpha-\dfrac\pi2\right)$.

Knowing $x(0)$ alone is not enough to fix $\phi$: $\cos\phi=\dfrac{x(0)}{A}$ has two solutions, $\pm\phi_0$. While the phase lies between $0$ and $\pi$, the cosine is decreasing, so the particle moves towards $-x$; while it lies between $-\pi$ and $0$, it moves towards $+x$. So choose $\phi$ in $(0,\pi)$ if the particle is moving towards $-x$ at $t=0$, and in $(-\pi,0)$ if it is moving towards $+x$.

Phase difference. For two SHMs of the same angular frequency, $x_1=A_1\cos(\omega t+\phi_1)$ and $x_2=A_2\cos(\omega t+\phi_2)$, the phase difference $\delta=\phi_2-\phi_1$ stays the same for all time. If $\delta=0$ they are in phase (they reach their maxima together); if $\delta=\pi$ they are in antiphase (one is at $+A_1$ when the other is at $-A_2$); if $\delta\gt0$, oscillator 2 leads, reaching each stage of its cycle a time $\dfrac{\delta}{\omega}=\dfrac{\delta}{2\pi}T$ earlier.

✏️ Example 3 — Reading an SHM equation

A particle moves along the $x$-axis with $x=0.050\cos\left(4\pi t+\dfrac{\pi}{3}\right)$, where $x$ is in metres and $t$ in seconds. Find (a) the amplitude, angular frequency, frequency, period and phase constant; (b) the displacement at $t=0$ and at $t=0.25\ \text{s}$; (c) the first time after $t=0$ at which the particle passes through the mean position.

💡 Show solution▶

(a) Compare with $A\cos(\omega t+\phi)$: $A=0.050\ \text{m}=5.0\ \text{cm}$, $\omega=4\pi=12.6\ \text{rad/s}$, $\nu=\dfrac{\omega}{2\pi}=2.0\ \text{Hz}$, $T=0.50\ \text{s}$, $\phi=\dfrac\pi3$ ($60^\circ$).

(b) $x(0)=0.050\cos60^\circ=0.025\ \text{m}$. At $t=0.25\ \text{s}$ (half a period later) the phase is $\pi+\dfrac\pi3$: $x=0.050\cos240^\circ=-0.025\ \text{m}$ — the mirror-image point.

(c) $x=0$ when the phase reaches $\dfrac\pi2$: $4\pi t+\dfrac\pi3=\dfrac\pi2$, so $t=\dfrac{\pi/6}{4\pi}=\dfrac{1}{24}\ \text{s}=0.042\ \text{s}$.

Check The phase constant $\dfrac\pi3$ lies between $0$ and $\pi$, so at $t=0$ the particle is moving towards $-x$ — consistent with reaching $x=0$ after only $\tfrac{1}{12}$ of a period ($\dfrac{\pi/6}{2\pi}=\dfrac1{12}$) ✓.

✏️ Example 4 — Writing the equation from the start

A block on a spring oscillates in SHM with period $0.40\ \text{s}$ and amplitude $3.0\ \text{cm}$. At $t=0$ it is at $x=+1.5\ \text{cm}$ and moving towards the mean position. (a) Write $x(t)$. (b) When does it first reach $x=-3.0\ \text{cm}$? (c) How would $x(t)$ change if it were moving away from the mean position at $t=0$?

💡 Show solution▶

ω $\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{0.40}=5\pi\ \text{rad/s}$ $(15.7\ \text{rad/s})$.

(a) $\cos\phi=\dfrac{1.5}{3.0}=\dfrac12$, so $\phi=\pm\dfrac\pi3$. Moving towards the mean position from $+1.5\ \text{cm}$ means moving towards $-x$, so $\phi$ lies in $(0,\pi)$: $\phi=\dfrac\pi3$ and $x=3.0\cos\left(5\pi t+\dfrac\pi3\right)\ \text{cm}$.

(b) $x=-A$ when the phase is $\pi$: $5\pi t=\pi-\dfrac\pi3=\dfrac{2\pi}3$, $t=\dfrac{2}{15}\ \text{s}=0.13\ \text{s}$.

(c) Moving away from the mean position (towards $+x$): $\phi=-\dfrac\pi3$, $x=3.0\cos\left(5\pi t-\dfrac\pi3\right)\ \text{cm}$.

Check In (b) the phase has advanced by $\dfrac{2\pi}{3}$, one-third of a cycle: $\tfrac13\times0.40=0.13\ \text{s}$ ✓.

✏️ Example 5 — Sine plus cosine

(a) Show that $x=3.0\sin\omega t+4.0\cos\omega t$ (in cm) is simple harmonic, and find its amplitude and phase constant in the form $A\cos(\omega t+\phi)$. (b) Find the phase difference between $x_1=2\sin\omega t$ and $x_2=2\cos\omega t$, and say which leads.

💡 Show solution▶

Idea Expand the target form: $A\cos(\omega t+\phi)=A\cos\phi\cos\omega t-A\sin\phi\sin\omega t$. Match the coefficients of $\cos\omega t$ and $\sin\omega t$.

(a) $A\cos\phi=4.0$ and $-A\sin\phi=3.0$. Squaring and adding: $A=\sqrt{4.0^2+3.0^2}=5.0\ \text{cm}$. Then $\cos\phi=0.80$, $\sin\phi=-0.60$: $\phi=-37^\circ$ ($-0.64\ \text{rad}$). So $x=5.0\cos(\omega t-0.64)\ \text{cm}$ — a single cosine, hence SHM with the same $\omega$.

(b) $x_1=2\sin\omega t=2\cos\left(\omega t-\dfrac\pi2\right)$, $x_2=2\cos\omega t$: $\delta=0-\left(-\dfrac\pi2\right)=\dfrac\pi2$. The cosine leads the sine by $\dfrac\pi2$, a quarter of a period: $x_2$ reaches its maximum $\dfrac T4$ before $x_1$ does.

Check At $t=0$: $5.0\cos(-37^\circ)=4.0\ \text{cm}$ ✓ (the original gives $0+4.0$). At $\omega t=\dfrac\pi2$: $5.0\cos(90^\circ-37^\circ)=3.0\ \text{cm}$ ✓.

SHM and uniform circular motion

Put a peg on the rim of a turntable and shine a lamp across it so that the peg's shadow falls on a wall. As the turntable spins steadily, the shadow slides back and forth along a straight line — slowly near the ends, fast in the middle — exactly like a block on a spring. Hang a pendulum of the right length next to it and its shadow keeps step with the peg's shadow cycle after cycle. The reason is pure geometry.

Let a particle $P$ move anticlockwise round a circle of radius $A$, centred at the origin, with constant angular speed $\omega$. Measure its angular position from the $+x$ axis, and let it be at angle $\phi$ at $t=0$. At time $t$ the radius $OP$ has turned to the angle $\theta=\omega t+\phi$ (Section 3.4), so $P$ has coordinates $(A\cos(\omega t+\phi),\ A\sin(\omega t+\phi))$. The foot $Q$ of the perpendicular from $P$ to the $x$-axis — the "shadow" of $P$ — is at $x=A\cos(\omega t+\phi)$.

📐 Result 13.2 — SHM as the projection of uniform circular motion

If a particle moves uniformly round a circle of radius $A$ with angular speed $\omega$, the projection of its position on any diameter executes simple harmonic motion along that diameter, with

amplitude equal to the radius $A$;
angular frequency equal to the angular speed $\omega$ of the particle, and period equal to its period of revolution;
phase $\omega t+\phi$ equal to the angle the radius makes with that diameter.

The circle is called the reference circle of the SHM, and $P$ the reference particle.

This is why $\omega$ is called the angular frequency, and why the phase is an angle. The projection on the $y$-axis, $y=A\sin(\omega t+\phi)=A\cos\left(\omega t+\phi-\dfrac\pi2\right)$, is also SHM, lagging the $x$-projection by $\dfrac\pi2$. So uniform circular motion is the combination of two SHMs at right angles, of equal amplitude and frequency, a quarter-cycle apart. Uniform circular motion itself is periodic but not simple harmonic: it is not a to-and-fro motion along a line.

🧭 Problem-solving strategy — timing with the reference circle

Draw the reference circle of radius $A$ and mark the two positions on the diameter between which you need the time.
For each position $x$, find the point on the circle above (or below) it: the angle $\theta$ from the $+x$ axis with $\cos\theta=\dfrac xA$. Choose the upper half of the circle ($0\lt\theta\lt\pi$) for motion towards $-x$ and the lower half for motion towards $+x$.
Find the angle $\Delta\theta$ the radius turns through (anticlockwise) between the two points.
The time is the same fraction of a period: $\Delta t=\dfrac{\Delta\theta}{2\pi}T=\dfrac{\Delta\theta}{\omega}$.

✏️ Example 6 — Timing a swing with the reference circle

A particle executes SHM with amplitude $10\ \text{cm}$ and period $2.4\ \text{s}$. Find the shortest time it takes to move (a) from the mean position to $x=+5.0\ \text{cm}$; (b) from $x=+5.0\ \text{cm}$ to the extreme $x=+10\ \text{cm}$; (c) from $x=+5.0\ \text{cm}$ to $x=-5.0\ \text{cm}$. (d) Explain why (a) and (b) are not equal although both cover $5.0\ \text{cm}$.

💡 Show solution▶

(a) Moving towards $+x$ (lower half): $x=0$ at $\theta=270^\circ$, $x=+5.0$ at $\cos\theta=\tfrac12$, $\theta=300^\circ$. $\Delta\theta=30^\circ=\dfrac{1}{12}$ turn: $\Delta t=\dfrac{2.4}{12}=0.20\ \text{s}$.

(b) From $300^\circ$ to $360^\circ$: $\Delta\theta=60^\circ=\dfrac16$ turn, $\Delta t=0.40\ \text{s}$.

(c) Moving towards $-x$ (upper half): $x=+5.0$ at $60^\circ$, $x=-5.0$ at $120^\circ$. $\Delta\theta=60^\circ$: $\Delta t=0.40\ \text{s}$.

(d) The particle is fast near the mean position and slow near the extremes, so the outer $5.0\ \text{cm}$ takes twice as long as the inner $5.0\ \text{cm}$.

Check (a) + (b) $=0.60\ \text{s}=\dfrac T4$, the time from the mean position to an extreme ✓. Interactive 2 with $A=10\ \text{cm}$, $T=2.4\ \text{s}$ shows the reference particle sweeping these angles.

The green reference particle goes steadily round the circle. In this picture the oscillation axis ($x$) is drawn vertically, with the angle measured anticlockwise from $+x$, so that the shadow's displacement can be traced against time to the right. The red shadow moves in SHM, and the graph builds up $x=A\cos(\omega t+\phi)$ as it goes. Change $\phi$ to start the cycle at a different place.

amplitude $A$:10 cmperiod $T$:2.4 sphase constant $\phi$:

time $t$—

phase $\omega t+\phi$—

displacement $x$—

$\omega$ and $\nu$—

⚠️ Common mistakes

1. Degrees in the phase. In $x=A\cos(\omega t+\phi)$, $\omega t$ is in radians. Set your calculator to radians, or convert: $\omega t$ in degrees is $\dfrac{360^\circ t}{T}$.

2. Confusing $\omega$ and $\nu$. $\omega=2\pi\nu$. In $x=0.05\cos(4\pi t)$ the frequency is $2\ \text{Hz}$, not $4\pi$ or $4\ \text{Hz}$.

3. Phase constant from $x(0)$ alone. $\cos\phi=\dfrac{x(0)}{A}$ has two answers. The direction of motion at $t=0$ picks one.

4. "Half the distance, half the time." SHM is not uniform motion. From the mean position to $\dfrac A2$ takes $\dfrac{T}{12}$; from $\dfrac A2$ to $A$ takes $\dfrac T6$.

5. Calling every periodic motion SHM. $\sin\omega t+\sin2\omega t$ and $\sin^3\omega t$ are periodic but are not single cosines, so they are not SHM; uniform circular motion is periodic but not SHM.

6. Amplitude and distance. The amplitude is the distance from the mean position to an extreme, half the total swing. In one period the particle travels $4A$, and its displacement is zero.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — adding two SHMs along the same line

In JEE Main, NEET and JEE Advanced (as part of simple harmonic motion); not required by the CBSE syllabus.

Two SHMs of the same angular frequency along the same line, $x_1=A_1\cos\omega t$ and $x_2=A_2\cos(\omega t+\delta)$, add to a single SHM of that frequency. On the reference circle, each is the projection of a rotating radius (a phasor); the two radii, at a fixed angle $\delta$, rotate together, and their vector sum is another radius rotating at $\omega$. Its length is the resultant amplitude and its angle $\varepsilon$ the phase constant:

$$A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\delta},$$ $$\tan\varepsilon=\frac{A_2\sin\delta}{A_1+A_2\cos\delta},\qquad x=A\cos(\omega t+\varepsilon).$$

In phase ($\delta=0$): $A=A_1+A_2$; in antiphase ($\delta=\pi$): $A=|A_1-A_2|$; at $\delta=\dfrac\pi2$: $A=\sqrt{A_1^2+A_2^2}$ (Example 5). Example. $A_1=3.0\ \text{cm}$, $A_2=4.0\ \text{cm}$, $\delta=\dfrac\pi3$: $A=\sqrt{9+16+12}=\sqrt{37}=6.1\ \text{cm}$, $\tan\varepsilon=\dfrac{4.0\times0.866}{3.0+2.0}=0.69$, $\varepsilon=35^\circ$. This is the same law of cosines as for adding vectors (Section 3.2).

Exercises

In every SHM equation $x=A\cos(\omega t+\phi)$ the phase is in radians. Give answers to 2–3 significant figures.

Periodic motion, period and frequency.

1 A pendulum completes $25$ oscillations in $40.0\ \text{s}$. Find its period and frequency.
2 The mains electricity supply in India alternates at $50\ \text{Hz}$. Find its period in milliseconds.
3 A resting heart beats $72$ times a minute. Find the frequency in hertz and the period.
4 Classify each motion as (i) periodic and oscillatory, (ii) periodic but not oscillatory, or (iii) not periodic: (a) the Earth spinning on its axis; (b) a child on a swing (ignoring friction); (c) a car moving at constant speed along a straight road; (d) the piston of an engine running at a steady speed; (e) the Moon going round the Earth.
5 Find the period of $\sin\omega t+\sin2\omega t+\sin4\omega t$.
6 Find the period of $\cos\dfrac{\pi t}{3}+\cos\dfrac{\pi t}{4}$, with $t$ in seconds.
7 Show that $\cos^2\omega t$ is periodic and find its period.
8 Is $\omega t+\sin\omega t$ periodic? Is $\sin\omega t+\sin\sqrt2\,\omega t$? Explain.

💡 Show solutions▶

1 $T=\dfrac{40.0}{25}=1.60\ \text{s}$; $\nu=\dfrac{1}{1.60}=0.625\ \text{Hz}$.

2 $T=\dfrac{1}{50}=0.020\ \text{s}=20\ \text{ms}$.

3 $\nu=\dfrac{72}{60}=1.2\ \text{Hz}$; $T=0.83\ \text{s}$.

4 (a) (ii) — it repeats but turns, it does not move to and fro; (b) (i); (c) (iii); (d) (i) — up and down about a mean position; (e) (ii) — like (a), it goes round rather than back and forth.

5 Periods $\dfrac{2\pi}{\omega}$, $\dfrac{\pi}{\omega}$, $\dfrac{\pi}{2\omega}$; the longest, $\dfrac{2\pi}{\omega}$, is a whole number of each: $T=\dfrac{2\pi}{\omega}$.

6 Periods $6\ \text{s}$ and $8\ \text{s}$; least common multiple $24\ \text{s}$.

7 $\cos^2\omega t=\tfrac12(1+\cos2\omega t)$: periodic with $T=\dfrac{\pi}{\omega}$.

8 $\omega t+\sin\omega t$ keeps growing (the $\omega t$ term never repeats): not periodic. The second has periods $\dfrac{2\pi}{\omega}$ and $\dfrac{2\pi}{\sqrt2\,\omega}$, whose ratio $\sqrt2$ is irrational, so no time is a whole number of both: not periodic (Interactive 1, last option).

The SHM equation and phase.

1 For $x=0.20\cos\left(5t+\dfrac\pi6\right)$ m, find the amplitude, angular frequency, frequency, period and phase constant.
2 Write the equation of an SHM of amplitude $4.0\ \text{cm}$ and period $0.50\ \text{s}$ if the particle is at $x=+4.0\ \text{cm}$ at $t=0$.
3 Repeat part 2 if instead the particle is at the mean position at $t=0$, moving towards $+x$.
4 For $x=6.0\cos\dfrac{\pi t}{2}$ cm, find $x$ at $t=0.50\ \text{s}$, $1.0\ \text{s}$, $2.0\ \text{s}$ and $3.0\ \text{s}$.
5 Show that $x=\sin\omega t-\cos\omega t$ is SHM. Find its amplitude, period and phase constant (cosine form).
6 An SHM has amplitude $8.0\ \text{cm}$. At $t=0$ the particle is at $x=-4.0\ \text{cm}$, moving towards $-x$. Find the phase constant.
7 Find the phase difference between $x_1=A\cos\omega t$ and $x_2=A\sin\left(\omega t+\dfrac\pi3\right)$, and say which leads.
8 Which of these represent SHM? Give the period of each periodic one. (a) $2\cos\left(3\omega t-\dfrac\pi5\right)$ (b) $\sin\omega t+\sin3\omega t$ (c) $\cos^3\omega t$ (use $\cos3\theta=4\cos^3\theta-3\cos\theta$)

💡 Show solutions▶

1 $A=0.20\ \text{m}$, $\omega=5.0\ \text{rad/s}$, $\nu=\dfrac{5}{2\pi}=0.80\ \text{Hz}$, $T=\dfrac{2\pi}{5}=1.26\ \text{s}$, $\phi=\dfrac\pi6$.

2 $\omega=\dfrac{2\pi}{0.50}=4\pi\ \text{rad/s}$, $\phi=0$: $x=4.0\cos4\pi t\ \text{cm}$.

3 $\phi=-\dfrac\pi2$: $x=4.0\cos\left(4\pi t-\dfrac\pi2\right)=4.0\sin4\pi t\ \text{cm}$.

4 $t=0.50$: $6.0\cos\dfrac\pi4=4.2\ \text{cm}$; $t=1.0$: $6.0\cos\dfrac\pi2=0$; $t=2.0$: $6.0\cos\pi=-6.0\ \text{cm}$; $t=3.0$: $6.0\cos\dfrac{3\pi}2=0$.

5 Match $A\cos\phi=-1$, $-A\sin\phi=1$: $A=\sqrt2$, $\cos\phi=-\dfrac1{\sqrt2}$, $\sin\phi=-\dfrac1{\sqrt2}$, so $\phi=-\dfrac{3\pi}{4}$: $x=\sqrt2\cos\left(\omega t-\dfrac{3\pi}4\right)$, a single cosine, SHM with $T=\dfrac{2\pi}{\omega}$.

6 $\cos\phi=-\dfrac12$, so $\phi=\pm\dfrac{2\pi}3$; moving towards $-x$ needs $\phi$ in $(0,\pi)$: $\phi=\dfrac{2\pi}3$ $(120^\circ)$.

7 $x_2=A\cos\left(\omega t+\dfrac\pi3-\dfrac\pi2\right)=A\cos\left(\omega t-\dfrac\pi6\right)$. Phase difference $\dfrac\pi6$; $x_1$ leads by $\dfrac\pi6$ (a time $\dfrac{T}{12}$).

8 (a) SHM, $T=\dfrac{2\pi}{3\omega}$. (b) Periodic, $T=\dfrac{2\pi}\omega$, not SHM (two frequencies). (c) $\cos^3\omega t=\tfrac34\cos\omega t+\tfrac14\cos3\omega t$: periodic, $T=\dfrac{2\pi}\omega$, not SHM.

III

The reference circle.

1 A particle moves round a circle of radius $15\ \text{cm}$ at $120$ revolutions per minute. Find the amplitude, angular frequency and period of its projection on a diameter.
2 A reference particle moves anticlockwise on a circle of radius $0.10\ \text{m}$ at $2.0\ \text{rad/s}$ and is at $30^\circ$ from the $+x$ axis at $t=0$. Write the equations of its projections on the $x$- and $y$-axes.
3 An SHM has period $1.2\ \text{s}$. Find the shortest time to go from $x=+A$ to $x=+\dfrac A2$.
4 For the same SHM, find the shortest time to go from the mean position to $x=\dfrac{A}{\sqrt2}$.
5 For the same SHM, find the shortest time to go from $x=+\dfrac A2$ to $x=-\dfrac A2$.
6 A particle in SHM starts at $x=+A$. Where is it after $\dfrac T3$, and what distance has it travelled?
7 A particle in SHM takes $0.50\ \text{s}$ to move from an extreme position to half the amplitude. Find the period.
8 Explain why uniform circular motion is periodic but not simple harmonic, while its projection on any diameter is simple harmonic.

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1 $A=15\ \text{cm}$; $2.0\ \text{rev/s}$, so $\omega=4\pi=12.6\ \text{rad/s}$ and $T=0.50\ \text{s}$.

2 $x=0.10\cos\left(2.0t+\dfrac\pi6\right)\ \text{m}$, $y=0.10\sin\left(2.0t+\dfrac\pi6\right)\ \text{m}$.

3 From $0^\circ$ to $60^\circ$: $\dfrac T6=0.20\ \text{s}$.

4 From $270^\circ$ to $315^\circ$: $45^\circ$, $\dfrac T8=0.15\ \text{s}$.

5 From $60^\circ$ to $120^\circ$: $\dfrac T6=0.20\ \text{s}$.

6 The radius turns $120^\circ$: $x=A\cos120^\circ=-\dfrac A2$. Distance: $A$ to the mean position, then $\dfrac A2$ more, $\dfrac{3A}{2}$ in all.

7 Extreme to $\dfrac A2$ is $60^\circ$, i.e. $\dfrac T6=0.50\ \text{s}$, so $T=3.0\ \text{s}$.

8 The circling particle returns to the same point with the same velocity every revolution (periodic), but it does not move back and forth along a line about a mean point, so it cannot be described by a single $x=A\cos(\omega t+\phi)$. Its projection on a diameter is $A\cos(\omega t+\phi)$ exactly (Result 13.2), so the projection is SHM.

Applications and harder problems.

1 (Modelling) In a strong wind the top of a tall building sways east–west in SHM with period $6.0\ \text{s}$ and amplitude $0.30\ \text{m}$. (a) Write $x(t)$ (east positive) if at $t=0$ the top passes through its mean position moving east. (b) Find its displacement $1.0\ \text{s}$ later. (c) For what fraction of each cycle is the top more than $0.15\ \text{m}$ from its mean position?
2 (Modelling) The piston of an engine running at $3000\ \text{rpm}$ has a stroke (the full distance between its extreme positions) of $8.0\ \text{cm}$. Treating its motion as SHM, find its amplitude, angular frequency and period.
3 A particle in SHM is at $x=3.0\ \text{cm}$ at $t=0$ and at $x=4.0\ \text{cm}$ a quarter of a period later. Find the amplitude and the phase constant.
4 Two particles oscillate along the same line about the same point with the same amplitude $A$ and period $T$. One starts at $x=+A$; the other starts at the mean position moving towards $+x$. When and where do they first meet?
5 The phases of the Moon repeat every $29.5$ days. Express this frequency in hertz.
6 A particle in SHM of period $T$ is at $x=+\dfrac A2$, moving towards $+x$. How long is it until it next passes through $x=+\dfrac A2$, and how long until it next passes through $x=-\dfrac A2$?

💡 Show solutions▶

1 (a) $\omega=\dfrac{2\pi}{6.0}=\dfrac\pi3\ \text{rad/s}$, $\phi=-\dfrac\pi2$: $x=0.30\sin\dfrac{\pi t}{3}\ \text{m}$. (b) $x=0.30\sin60^\circ=0.26\ \text{m}$ east. (c) $|x|\gt\dfrac A2$ when the radius is within $60^\circ$ of either end of the diameter: $4\times60^\circ=240^\circ$ of each $360^\circ$, i.e. $\dfrac23$ of the time.

2 $A=\dfrac{8.0}{2}=4.0\ \text{cm}$; $\nu=50\ \text{Hz}$, $\omega=100\pi=314\ \text{rad/s}$, $T=0.020\ \text{s}$.

3 $x(0)=A\cos\phi=3.0$ and $x\!\left(\tfrac T4\right)=A\cos\left(\phi+\tfrac\pi2\right)=-A\sin\phi=4.0$. So $A=5.0\ \text{cm}$, $\cos\phi=0.60$, $\sin\phi=-0.80$: $\phi=-53^\circ$ ($-0.93\ \text{rad}$).

4 $x_1=A\cos\omega t$, $x_2=A\sin\omega t$; equal when $\tan\omega t=1$, $\omega t=\dfrac\pi4$: $t=\dfrac T8$, at $x=\dfrac{A}{\sqrt2}\approx0.71A$.

5 $\nu=\dfrac{1}{29.5\times86\,400\ \text{s}}=3.92\times10^{-7}\ \text{Hz}$.

6 It is at $300^\circ$ on the reference circle. Next at $+\dfrac A2$: $60^\circ$ (i.e. $420^\circ$), $120^\circ$ later, $\dfrac T3$. Next at $-\dfrac A2$: $120^\circ$ (i.e. $480^\circ$), $180^\circ$ later, $\dfrac T2$.

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Define simple harmonic motion. Distinguish between periodic and oscillatory motion, giving one example of a motion that is periodic but not oscillatory.

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SHM A particle moving along a line executes SHM if its displacement from the mean position varies with time as $x=A\cos(\omega t+\phi)$, with $A$, $\omega$, $\phi$ constant. [1]

Difference A periodic motion repeats itself at regular intervals; an oscillatory motion is a periodic to-and-fro motion about a mean (equilibrium) position. Every oscillation is periodic, but not conversely: uniform circular motion (a fan blade, the hands of a clock) is periodic but not oscillatory. [1]

Exam tip Name all three constants in the SHM equation; "motion that repeats" alone defines only periodic motion.

E2. NEET The period of the function $y=\sin^2\omega t$ is (a) $\dfrac{2\pi}{\omega}$ (b) $\dfrac{\pi}{\omega}$ (c) $\dfrac{\pi}{2\omega}$ (d) $\dfrac{4\pi}{\omega}$

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Idea $\sin^2\omega t=\tfrac12-\tfrac12\cos2\omega t$ oscillates at angular frequency $2\omega$.

Answer $T=\dfrac{2\pi}{2\omega}=\dfrac\pi\omega$: (b).

Exam tip (a) is the period of $\sin\omega t$ itself; squaring makes the negative half-cycles positive, so the pattern repeats twice as often.

E3. JEE Main A particle executes SHM of amplitude $A$ and period $T$. The shortest time it takes to go from $x=\dfrac{A}{\sqrt2}$ to $x=-\dfrac A2$ is (a) $\dfrac T8$ (b) $\dfrac{5T}{24}$ (c) $\dfrac T3$ (d) $\dfrac{7T}{24}$

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Set-up Reference circle, moving towards $-x$ (upper half): $\cos\theta_1=\dfrac1{\sqrt2}$, $\theta_1=45^\circ$; $\cos\theta_2=-\dfrac12$, $\theta_2=120^\circ$.

Time $\Delta\theta=75^\circ$: $\Delta t=\dfrac{75}{360}T=\dfrac{5T}{24}$.

Answer (b).

Exam tip Split it at the mean position: $\dfrac{A}{\sqrt2}\to0$ takes $\dfrac T8$, $0\to-\dfrac A2$ takes $\dfrac T{12}$, and $\dfrac T8+\dfrac T{12}=\dfrac{5T}{24}$ ✓. (a) stops at the mean position; (c) starts from the extreme.

E4. JEE Main (Numerical value) The displacement of a particle is $x=3\sin(10\pi t)+4\cos(10\pi t)$, with $x$ in centimetres and $t$ in seconds. The motion is simple harmonic. Find its amplitude in centimetres.

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Idea $a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\cos(\theta-\alpha)$ with $\tan\alpha=\dfrac ab$ — a single cosine of the same angular frequency.

Answer $A=\sqrt{3^2+4^2}=5$ (cm). (Also $\nu=5\ \text{Hz}$.)

Exam tip Adding the coefficients ($7$) is the trap: the two terms are a quarter-cycle apart and add like perpendicular vectors.

E5. JEE Advanced (One or more correct) A particle moves anticlockwise round a circle of radius $R$, centred at the origin, with constant angular speed $\omega$ and period $T$. Which statements are correct? (A) The projection of its position on the $x$-axis executes SHM of amplitude $R$. (B) The projection on the $y$-axis leads the projection on the $x$-axis by $\dfrac\pi2$. (C) The $x$-projection takes $\dfrac T4$ to go from $x=R$ to $x=0$. (D) In one period the $x$-projection travels a distance $4R$.

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(A) $x=R\cos(\omega t+\phi)$: SHM with amplitude $R$ ✓.

(B) $y=R\sin(\omega t+\phi)=R\cos\left(\omega t+\phi-\dfrac\pi2\right)$: the $y$-projection lags by $\dfrac\pi2$ ✗. (Check at $\phi=0$: $x$ is at its maximum at $t=0$, $y$ only a quarter-period later.)

(C), (D) The radius turns $90^\circ$, a quarter-turn ✓. In a period the shadow goes $R\to-R\to R$, distance $4R$ ✓.

Answer (A), (C), (D).

Exam tip For clockwise motion the answer to (B) would reverse — always check the sense of rotation before quoting a lead or a lag.

E6. Board · Case study A music teacher sets a mechanical metronome to $120$ ticks per minute. It ticks each time its swinging arm reaches an extreme position, $20^\circ$ either side of the vertical, and the arm's angle varies in SHM. (i) Find the period and the frequency of the arm's oscillation. [1] (ii) Write $\theta(t)$ in degrees, taking $t=0$ at a tick on the $+20^\circ$ side. [1] (iii) Find the angle at $t=\tfrac16\ \text{s}$, and the shortest time between the arm being at $+20^\circ$ and at $-10^\circ$. [2]

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(i) Two ticks (one at each extreme) per oscillation: $60$ oscillations per minute, so $T=1.0\ \text{s}$ and $\nu=1.0\ \text{Hz}$. [1]

(ii) $\omega=2\pi\ \text{rad/s}$, $\phi=0$: $\theta=20^\circ\cos(2\pi t)$. [1]

(iii) $\theta\!\left(\tfrac16\right)=20^\circ\cos\dfrac\pi3=10^\circ$. From $+20^\circ$ ($0^\circ$ on the reference circle) to $-10^\circ$ ($\cos\theta=-\tfrac12$, $120^\circ$): $\tfrac13$ of a period, $0.33\ \text{s}$. [2]

Exam tip "Ticks per minute" is not oscillations per minute: read the statement to see how many ticks make one cycle.

🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. How periodic is a human heartbeat — and does the "period" of your heart change with every breath?

Why it matters. Doctors and sports scientists across the world measure heart-rate variability, the beat-to-beat change in the interval between heartbeats. A healthy heart is not perfectly periodic: its period shortens as you breathe in and lengthens as you breathe out. Low variability is linked to stress, fatigue and heart disease, and many fitness watches now report it.

What to do. Work in pairs, sitting quietly.

Find your partner's pulse at the wrist. Count beats for $15\ \text{s}$, $30\ \text{s}$ and $60\ \text{s}$; from each count find the frequency in beats per minute and in hertz, and the mean period $T$.
Repeat the $60\ \text{s}$ count three times. How much do the results differ? Which counting time gives the most reliable $T$, and why?
Ask your partner to breathe slowly — $5\ \text{s}$ in, $5\ \text{s}$ out — while you tap a key on a phone stopwatch (lap mode) at every beat for one minute. Find each beat-to-beat interval.
Plot interval against beat number. Is there a slow rise and fall in time with the breathing? Estimate its period.
Repeat after one minute of brisk stepping. How do the mean period and the spread of intervals change?

Tools: A phone stopwatch with lap times (or a free pulse app), a spreadsheet for the intervals, mean and standard deviation.

Hand in: A table of counts and periods; the interval–beat graph; the mean period and its spread at rest and after exercise; and a paragraph on whether the heartbeat should be called periodic, nearly periodic, or neither.

Going further. Find out how an electrocardiogram (ECG) measures the R–R interval, and why the variability is often analysed by splitting it into slow and fast components — a version of Fourier's idea that any repeating signal is a sum of sines.

Instructor note: resting rates of $60$–$80$ beats per minute give $T\approx0.75$–$1.0\ \text{s}$ ($72$ per minute: $T=0.83\ \text{s}$). A $15\ \text{s}$ count multiplied by $4$ has a resolution of $\pm4$ beats per minute (one missed beat), a $60\ \text{s}$ count $\pm1$. Hand-tapped intervals carry roughly $\pm0.05\ \text{s}$ of reaction-time scatter, so encourage students to look for the slow breathing pattern (typically a swing of several beats per minute at $6$ breaths per minute) rather than single intervals. Young healthy adults show beat-to-beat standard deviations of several tens of milliseconds.

🧪 Self-check quiz

0/5 correct

Q1. Which motion is periodic but not oscillatory?

Q2. For $x=5\cos(3t+1)$, with $x$ in metres and $t$ in seconds, the angular frequency is

Q3. An SHM has angular frequency $10\pi\ \text{rad/s}$. Its period is

Q4. A particle moves uniformly round a circle of radius $r$ with angular speed $\omega$. Its projection on a diameter executes

Q5. A particle in SHM of period $T$ goes from the mean position to half its amplitude in the shortest time of

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