JEE Corner — Advanced Oscillations

What JEE Adds to This Chapter

Sections 1 to 11 built this chapter properly. $x = A\cos(\omega t + \phi)$, the reference circle, $v = \pm\omega\sqrt{A^2 - x^2}$, $F = -kx$ and $T = 2\pi\sqrt{\dfrac{m}{k}}$, spring combinations, the energy account, the pendulum, damping, resonance — all of it is in place, and for the Board paper that build is complete.

What JEE adds is almost no new physics. It is still one restoring force proportional to one displacement. What changes is that the restoring force is no longer handed to you. The spring is replaced by gravity acting on a rod, by buoyancy on a floating block, by a magnetic torque, by the field inside the Earth. The wall the spring is tied to gets up and walks away. The mass that appears under the square root is not the mass of the block. And very often the whole problem is a potential energy function $U(x)$ with no force written anywhere.

The one idea that organises this whole section

Here it is, up front, because everything below is a variation on it.

Key Point — every oscillator is the same oscillator. Find a coordinate $q$ measured from the equilibrium value, get the equation of motion into the shape $$m_{\text{eff}}\,\ddot{q} + k_{\text{eff}}\,q = 0$$ and you are finished, because then $$\omega = \sqrt{\frac{k_{\text{eff}}}{m_{\text{eff}}}}, \qquad \nu = \frac{\omega}{2\pi}, \qquad T = \frac{2\pi}{\omega}$$ $q$ need not be a length — it can be an angle, a volume, a charge. $m_{\text{eff}}$ is whatever multiplies $\ddot{q}$ and $k_{\text{eff}}$ is whatever multiplies $q$. Getting the equation into that shape is the entire problem; reading $\omega$ off it takes one second.

There are exactly two reliable ways to reach that shape, and both are worked below: the force-and-torque route (displace by $q$, find the net restoring force or torque, show it is $-(\text{constant})\times q$) and the energy route (write the total energy, differentiate it with respect to time, set the derivative to zero). The energy route is usually faster and never drops a term.

Notation for This Section

Symbol	Meaning	Unit
$\omega$	angular frequency	rad/s
$\nu$	frequency (Greek nu, not the italic vee of speed)	Hz
$T$	period	s
$\omega_0$	natural angular frequency, $\sqrt{k/m}$	rad/s
$\omega^{\,\prime}$	damped angular frequency, $\sqrt{\omega_0^2 - \frac{b^2}{4m^2}}$	rad/s
$\omega_d$	driving angular frequency	rad/s
$\delta$	phase difference between two superposed SHMs	rad
$\epsilon$	phase constant of the resultant SHM	rad
$\mu$	reduced mass, $\frac{m_1m_2}{m_1+m_2}$	kg
$I$	moment of inertia about the pivot	kg m$^2$
$b$	damping constant	kg/s
$Q$	quality factor	dimensionless
$\Delta\omega$	half-power bandwidth of the resonance curve	rad/s
$k_{\text{eff}}$, $m_{\text{eff}}$	effective stiffness and effective mass	N/m, kg

Speed is $v$ and frequency is $\nu$; both appear in this section and they are different letters.

The twelve things this section teaches

#	Skill	Why it earns marks
1	Phasor addition of two SHMs along one line	The cosine rule replaces a page of trigonometry
2	Perpendicular SHMs and the phase difference	One equation covers lines, ellipses and circles
3	The physical pendulum, $T = 2\pi\sqrt{\frac{I}{Mgd}}$	A rod or a disc is not a simple pendulum
4	Rolling without slipping in a curved track	The moment of inertia slows the oscillation down
5	Deriving the restoring force before writing a period	Floating bodies, liquid columns, magnets
6	The energy method, $\frac{dE}{dt} = 0$	One differentiation gives the whole equation of motion
7	Effective mass — a heavy spring, a loaded pulley	The mass under the root is rarely the block's mass
8	Reduced mass when both bodies are free	Nothing is bolted to a wall, so $m$ becomes $\mu$
9	Oscillators in accelerating frames	The equilibrium moves; the period does not
10	Small oscillations about a minimum of $U(x)$	$k = U^{\prime\prime}(x_0)$ turns any well into a spring
11	Damped oscillators counted in cycles, not seconds	The answer wanted is a number of swings
12	$Q$ and the half-power bandwidth	Sharpness of resonance compressed into one number

Constants and standard results, fixed now

Quantity	Value
$g$	$9.8$ m/s$^2$
Radius of the Earth $R$	$6.4 \times 10^{6}$ m
$\frac{1}{4\pi\epsilon_0}$	$9 \times 10^{9}$ N m$^2$/C$^2$
Rod of mass $M$, length $L$, about one end	$I = \frac{1}{3}ML^2$
Rod about its centre	$I = \frac{1}{12}ML^2$
Disc or solid cylinder about its axis	$I = \frac{1}{2}MR^2$
Solid sphere about a diameter	$I = \frac{2}{5}MR^2$
Ring or hollow cylinder about its axis	$I = MR^2$

Superposition of perpendicular simple harmonic motions, the energy method, small oscillations about a potential minimum and the sharpness of resonance sit outside the rationalised syllabus body text, and JEE Main and JEE Advanced set them every year, so each is developed here from first principles rather than quoted.

[Exam Tip] Four questions, asked before any algebra, choose the method for almost every problem below. Where is the equilibrium — and am I measuring my displacement from it, or from some natural length? Is anything in this system free to move that I have quietly assumed was fixed? Is my coordinate a length or an angle, and if it is an angle, is it in radians? Does the question want $\omega$, or $\nu$, or $T$? Answer those four and the algebra is usually three lines.

Superposing Two SHMs of the Same Frequency

Two oscillations act on the same particle. What does it do? The answer splits cleanly into two cases, and JEE asks both.

Case 1: both along the same line — add them as phasors

Let $$x_1 = A_1\cos\omega t, \qquad x_2 = A_2\cos(\omega t + \delta)$$ so $\delta$ is the phase difference between them. The particle's displacement is $x = x_1 + x_2$. You could expand both cosines and collect terms, and it works, but there is a picture that does it in one line.

Recall from the reference circle that $A\cos(\omega t + \phi)$ is the $x$-projection of a vector of length $A$ turning at $\omega$, starting at angle $\phi$. Such a vector is called a phasor. Projections add, so the two phasors add as vectors — and because both turn at the same rate $\omega$, the triangle they make never changes shape. Freeze it at $t = 0$ and the answer is a piece of geometry:

Key Point — phasor addition. For two SHMs of the same $\omega$ along the same line, $$x_1 + x_2 = A\cos(\omega t + \epsilon)$$ with $$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\delta}, \qquad \tan\epsilon = \frac{A_2\sin\delta}{A_1 + A_2\cos\delta}$$ That is the cosine rule for a triangle with sides $A_1$ and $A_2$ and included angle $180° - \delta$. The resultant has the same frequency as its parts — only the amplitude and the phase constant change.

Three consequences worth memorising:

• $\delta = 0$ gives $A = A_1 + A_2$, the largest possible. Constructive.
• $\delta = \pi$ gives $A = \lvert A_1 - A_2 \rvert$, the smallest possible. Destructive, and exactly zero if the amplitudes are equal.
• $\delta = \frac{\pi}{2}$ gives $A = \sqrt{A_1^2 + A_2^2}$ — the plain Pythagorean sum, which is why $3\cos\omega t + 4\sin\omega t$ has amplitude $5$.

For more than two, add the phasors head to tail, or resolve each into components and add those. Three equal amplitudes with phase constants $0$, $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ close into an equilateral triangle and give exactly zero — a favourite one-line question.

Case 2: along perpendicular lines — the path is a conic

Now the two motions are along $x$ and $y$: $$x = a\cos\omega t, \qquad y = b\cos(\omega t + \delta)$$ Here the question is not "what is the amplitude" but "what curve does the particle trace?" Eliminate $t$. From the first, $\cos\omega t = \frac{x}{a}$ and $\sin\omega t = \sqrt{1 - \frac{x^2}{a^2}}$. Expanding the second and substituting gives, after squaring,

Key Point — the general path. $$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab}\cos\delta = \sin^2\delta$$ This one equation covers every case. It is an ellipse in general, inscribed in the rectangle $\lvert x \rvert \le a$, $\lvert y \rvert \le b$, and it degenerates into a straight line when $\sin\delta = 0$.

Read off the standard cases:

$\delta$	Equation becomes	Path
$0$	$\left(\frac{x}{a} - \frac{y}{b}\right)^2 = 0$	straight line $y = \frac{b}{a}x$
$\frac{\pi}{4}$	full equation, $\sin^2\delta = \frac{1}{2}$	ellipse tilted towards $y = x$
$\frac{\pi}{2}$	$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$	ellipse with axes along $x$ and $y$
$\frac{\pi}{2}$ and $a = b$	$x^2 + y^2 = a^2$	circle
$\frac{3\pi}{4}$	full equation again	ellipse tilted the other way
$\pi$	$\left(\frac{x}{a} + \frac{y}{b}\right)^2 = 0$	straight line $y = -\frac{b}{a}x$

Two details that separate a full-marks answer from a half-marks one.

The circle needs both conditions. $\delta = \frac{\pi}{2}$ alone gives an ellipse; equal amplitudes alone give a tilted ellipse. Only $\delta = \frac{\pi}{2}$ and $a = b$ together give a circle. This is exactly the reference circle of Section 3 read backwards: uniform circular motion is two perpendicular SHMs of equal amplitude a quarter cycle apart.

The area of the traced ellipse is $\pi a b \lvert\sin\delta\rvert$. It is largest at $\delta = \frac{\pi}{2}$ and collapses to zero at $\delta = 0$ or $\pi$, which is another way of saying the ellipse has flattened into a line. And the sense of travel — clockwise or anticlockwise — is fixed by the sign of $\sin\delta$: with $0 < \delta < \pi$ the particle starts at $(a, b\cos\delta)$ and moves so that $y$ is decreasing, that is, clockwise.

[Exam Tip] If a question gives you $x$ and $y$ as sines rather than cosines, do not convert anything. The identity above only needs the two motions written with the same trigonometric function; $\delta$ is then the difference of the two arguments, whatever function you are using.

Oscillators Whose Restoring Force You Have to Find First

In Sections 5 and 6 the spring handed you $F = -kx$ and the rest was arithmetic. Here nothing hands you anything. The recipe never changes:

Key Point — the four-step derivation.

1. Locate the equilibrium and choose a coordinate $q$ measured from it.
2. Displace by a small $q$ and write the net restoring force, or the net restoring torque if $q$ is an angle.
3. Linearise: keep only the term proportional to $q$. This is where $\sin\theta \to \theta$ and $\cos\theta \to 1 - \frac{\theta^2}{2}$ live, and where every angle must be in radians.
4. Compare with $m_{\text{eff}}\ddot{q} = -k_{\text{eff}}q$ and read $\omega = \sqrt{\frac{k_{\text{eff}}}{m_{\text{eff}}}}$ straight off.

The physical pendulum — the workhorse

Any rigid body free to swing about a horizontal axis is a physical pendulum. Let $I$ be its moment of inertia about the pivot, $M$ its mass, and $d$ the distance from the pivot to the centre of mass. Displace it by $\theta$ and gravity supplies a restoring torque $-Mgd\sin\theta$, so $$I\ddot{\theta} = -Mgd\sin\theta \;\xrightarrow{\ \text{small }\theta\ }\; I\ddot{\theta} = -Mgd\,\theta$$ $$\omega = \sqrt{\frac{Mgd}{I}}, \qquad T = 2\pi\sqrt{\frac{I}{Mgd}}$$

Compare with $T = 2\pi\sqrt{\frac{L}{g}}$ and you get the equivalent simple pendulum length $$L_{\text{eq}} = \frac{I}{Md}$$ which is the single most useful line in this whole topic: any rigid swinging body behaves exactly like a simple pendulum of that length.

For a uniform rod of length $L$ pivoted at one end, $I = \frac{1}{3}ML^2$ and $d = \frac{L}{2}$, so $$L_{\text{eq}} = \frac{\frac{1}{3}ML^2}{M\left(\frac{L}{2}\right)} = \frac{2L}{3}, \qquad T = 2\pi\sqrt{\frac{2L}{3g}}$$ A metre stick swung from its end has a period like a $0.667$ m simple pendulum, not a $1$ m one — a $18\%$ shorter period than the careless answer.

Rolling without slipping in a curved track

A cylinder or a sphere of radius $r$ rolls inside a fixed circular track of radius $R$. Its centre moves on a circle of radius $R - r$, so the centre's speed is $v = (R - r)\dot{\theta}$, and rolling ties the spin to that: $\omega_{\text{spin}} = \frac{v}{r}$. The kinetic energy therefore carries two terms: $$K = \frac{1}{2}mv^2 + \frac{1}{2}I_{\text{cm}}\left(\frac{v}{r}\right)^2 = \frac{1}{2}\left(1 + \frac{I_{\text{cm}}}{mr^2}\right)mv^2$$ The bracket is a pure number: $\frac{3}{2}$ for a solid cylinder, $\frac{7}{5}$ for a solid sphere, $2$ for a ring. Calling it $\beta$, the result is $$\omega = \sqrt{\frac{g}{\beta\,(R-r)}}, \qquad T = 2\pi\sqrt{\frac{\beta\,(R-r)}{g}}$$ So a solid cylinder oscillates $\sqrt{\frac{3}{2}} = 1.22$ times more slowly than a block sliding on the same frictionless track. Rolling always lengthens the period, because part of every joule goes into spin.

The standard derived oscillators, in one table

$g$ is $9.8$ m/s$^2$ throughout, and every angle is in radians.

System	Restoring law	$\omega^2$	Period
Rigid body, pivot at distance $d$ from the centre of mass	torque $-Mgd\,\theta$	$\frac{Mgd}{I}$	$2\pi\sqrt{\frac{I}{Mgd}}$
Uniform rod, length $L$, pivoted at one end	torque $-\frac{MgL}{2}\theta$	$\frac{3g}{2L}$	$2\pi\sqrt{\frac{2L}{3g}}$
Solid cylinder rolling in a track of radius $R$	$\beta = \frac{3}{2}$	$\frac{2g}{3(R-r)}$	$2\pi\sqrt{\frac{3(R-r)}{2g}}$
Solid sphere rolling in the same track	$\beta = \frac{7}{5}$	$\frac{5g}{7(R-r)}$	$2\pi\sqrt{\frac{7(R-r)}{5g}}$
Body floating, cross-section $A$, liquid density $\rho$	$-\rho A g\,x$	$\frac{\rho A g}{m}$	$2\pi\sqrt{\frac{m}{\rho A g}}$
Liquid column of length $L$, arms inclined at $\theta_1$ and $\theta_2$	$-\rho A g(\sin\theta_1 + \sin\theta_2)\,x$	$\frac{g(\sin\theta_1 + \sin\theta_2)}{L}$	$2\pi\sqrt{\frac{L}{g(\sin\theta_1 + \sin\theta_2)}}$
Bar magnet, moment $M_m$, in a field $B$	torque $-M_mB\,\theta$	$\frac{M_mB}{I}$	$2\pi\sqrt{\frac{I}{M_mB}}$
Charge $q$ on the axis of a ring of charge $Q$, radius $R$	$-\frac{Qq}{4\pi\epsilon_0R^3}x$	$\frac{Qq}{4\pi\epsilon_0 m R^3}$	$2\pi\sqrt{\frac{4\pi\epsilon_0 m R^3}{Qq}}$
Particle in a tunnel through the Earth	$-\frac{mg}{R}s$	$\frac{g}{R}$	$2\pi\sqrt{\frac{R}{g}}$

Two of those rows deserve a sentence each.

The U-tube with inclined arms. For arms making angles $\theta_1$ and $\theta_2$ with the horizontal, push the liquid a distance $x$ along the tube. One arm rises by $x\sin\theta_1$, the other falls by $x\sin\theta_2$, so the unbalanced height is $x(\sin\theta_1 + \sin\theta_2)$ and the restoring force is $\rho A g x(\sin\theta_1 + \sin\theta_2)$ acting on a mass $\rho A L$. Set both arms vertical, $\theta_1 = \theta_2 = 90°$, and it collapses to the familiar $T = 2\pi\sqrt{\frac{L}{2g}}$. A tube with one arm vertical and the other at $30°$, holding a $50$ cm column, gives $\omega = \sqrt{\frac{9.8(1 + 0.5)}{0.50}} = 5.42$ rad/s and $T = 1.16$ s.

The magnet. A magnetic moment $M_m$ in a field $B$ feels a torque $-M_mB\sin\theta$, which is identical in form to gravity on a pendulum. That is why a vibration magnetometer is a pendulum in disguise, and why doubling $B$ divides the period by $\sqrt{2}$.

Key Point — the check that catches most errors. Whatever you derive, put it through a dimension check and a limiting check. $\sqrt{\frac{\text{length}}{g}}$ has units of time; $\sqrt{\frac{I}{Mgd}}$ does too. And every rolling result must reduce to the sliding one when the moment of inertia is set to zero.

The Energy Method, Effective Mass and Frames That Accelerate

Differentiate the energy: one line, no free-body diagram

For any conservative oscillator the total mechanical energy is constant. Write it in terms of one coordinate $q$ and its rate $\dot{q}$: $$E = \frac{1}{2}m_{\text{eff}}\dot{q}^{\,2} + \frac{1}{2}k_{\text{eff}}q^2 = \text{constant}$$ Differentiate with respect to time and use $\frac{d}{dt}\left(\dot{q}^{\,2}\right) = 2\dot{q}\ddot{q}$: $$m_{\text{eff}}\dot{q}\ddot{q} + k_{\text{eff}}q\dot{q} = 0$$ Cancel the common $\dot{q}$ (legitimate everywhere except at the two instants when the body is momentarily at rest) and you have the equation of motion:

Key Point — the energy method. $$\frac{dE}{dt} = 0 \qquad\Longrightarrow\qquad m_{\text{eff}}\ddot{q} + k_{\text{eff}}q = 0 \qquad\Longrightarrow\qquad \omega = \sqrt{\frac{k_{\text{eff}}}{m_{\text{eff}}}}$$ Everything that multiplies $\frac{1}{2}\dot{q}^{\,2}$ is the effective mass; everything that multiplies $\frac{1}{2}q^2$ is the effective stiffness. No forces, no torques, no signs to lose.

This is worth doing even when the force method would work, because it automatically collects contributions you might forget — the rotational kinetic energy of a pulley, the kinetic energy of the spring itself, the kinetic energy of a liquid in the arms of a tube.

Effective mass: the mass under the square root is not the block's mass

A spring that is not light. A uniform spring of mass $m_s$ and natural length $\ell$, fixed at one end, carries a block $m$ at the other. When the block moves at speed $v$, the element at distance $y$ from the fixed end moves at $\frac{y}{\ell}v$. Its kinetic energy is $$\int_0^{\ell}\frac{1}{2}\left(\frac{m_s}{\ell}\,dy\right)\left(\frac{yv}{\ell}\right)^2 = \frac{m_sv^2}{2\ell^3}\int_0^{\ell}y^2\,dy = \frac{1}{2}\left(\frac{m_s}{3}\right)v^2$$ So the spring behaves as though one third of it were stuck to the block: $$T = 2\pi\sqrt{\frac{m + \frac{m_s}{3}}{k}}$$

A pulley in the loop. A block $m$ hangs from a string that runs over a pulley of moment of inertia $I$ and radius $R$; the other end of the string is tied to a spring of constant $k$. With $v = R\omega_{\text{spin}}$, $$E = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{R}\right)^2 + \frac{1}{2}kx^2 \qquad\Longrightarrow\qquad m_{\text{eff}} = m + \frac{I}{R^2}$$

Reduced mass: when nothing is bolted to a wall

Two blocks $m_1$ and $m_2$ on a frictionless surface, joined by a spring of constant $k$. There is no wall. Write Newton's law for each, using the extension $x = x_2 - x_1 - \ell_0$: $$m_1\ddot{x}_1 = kx, \qquad m_2\ddot{x}_2 = -kx$$ Subtract, after dividing each by its own mass: $$\ddot{x} = \ddot{x}_2 - \ddot{x}_1 = -k\left(\frac{1}{m_1} + \frac{1}{m_2}\right)x = -\frac{k}{\mu}x$$

Key Point — the two-body oscillator. The extension of the spring performs SHM with $$\mu = \frac{m_1m_2}{m_1 + m_2}, \qquad \omega = \sqrt{\frac{k}{\mu}}, \qquad T = 2\pi\sqrt{\frac{\mu}{k}}$$ Both blocks oscillate at that same $\omega$ about the centre of mass, which never moves. Their amplitudes divide in the inverse ratio of their masses: $$\frac{A_1}{A_2} = \frac{m_2}{m_1}, \qquad A_1 + A_2 = (\text{amplitude of the extension})$$ and their momenta are always equal and opposite, so $m_1v_1 = m_2v_2$ at every instant.

The same $\mu$ appears whenever the "wall" is really a body free to recoil — a spring between a block and a plank that can slide, a spring between two carts, a diatomic molecule. Set $m_2 \to \infty$ and $\mu \to m_1$, recovering the fixed-wall answer, which is the check to run every time.

Oscillators in an accelerating frame

Put the whole apparatus on a trolley with constant acceleration $a$. Work in the trolley's frame and add a pseudo-force $-ma$ on every mass. A constant force does exactly one thing to a spring oscillator: it moves the equilibrium and leaves the period alone.

For a block on a horizontal spring on such a trolley, $$\text{new equilibrium is shifted by } \; x_0 = \frac{ma}{k}, \qquad T = 2\pi\sqrt{\frac{m}{k}} \; \text{ unchanged}$$ And if the block happens to be sitting at the old (natural-length) position when the trolley starts, then it is $x_0$ away from its new equilibrium with zero velocity — so the amplitude is exactly $\frac{ma}{k}$.

This is the same argument as the vertical spring in Section 5, where gravity shifts the equilibrium down by $\frac{mg}{k}$ and cancels out of the equation of motion entirely. In a lift accelerating upward at $a$ the extension at equilibrium becomes $\frac{m(g+a)}{k}$ and the period is still $2\pi\sqrt{\frac{m}{k}}$, with no $g$ and no $a$ in it.

Key Point — the rule for any constant extra force. Add a constant force $F_0$ to a spring oscillator and the equilibrium moves by $\frac{F_0}{k}$; $\omega$, $\nu$ and $T$ do not change at all. Measure your displacement from the new equilibrium and the constant force disappears from the algebra. This is not true for a pendulum, where a constant sideways force changes the effective gravity and therefore does change the period.

Why SHM Is Universal: Small Oscillations About Any Minimum

This is the deepest idea in the chapter, it is absent from the rationalised syllabus, and it is asked at JEE almost every year — usually as a problem that gives you a potential energy function and nothing else.

The argument

Let a particle of mass $m$ move in a potential energy $U(x)$ with a minimum at $x = x_0$. Expand $U$ in a Taylor series about that point: $$U(x) = U(x_0) + \left.\frac{dU}{dx}\right\rvert_{x_0}(x - x_0) + \frac{1}{2}\left.\frac{d^2U}{dx^2}\right\rvert_{x_0}(x - x_0)^2 + \dots$$

The first term is a constant and does nothing. The second term vanishes, because $x_0$ is a minimum and the slope there is zero — that is exactly what "equilibrium" means. So the first surviving term is the quadratic one. Write $X = x - x_0$ for the displacement from equilibrium and it reads $$U = U(x_0) + \frac{1}{2}kX^2 \qquad \text{with} \qquad k = \left.\frac{d^2U}{dx^2}\right\rvert_{x_0}$$ which is precisely the potential energy of a spring. The force follows: $$F = -\frac{dU}{dx} = -kX$$

Key Point — every smooth minimum is a spring. $$x_0 \; \text{from} \; \left.\frac{dU}{dx}\right\rvert_{x_0} = 0, \qquad k = \left.\frac{d^2U}{dx^2}\right\rvert_{x_0}, \qquad \omega = \sqrt{\frac{1}{m}\left.\frac{d^2U}{dx^2}\right\rvert_{x_0}}$$ A positive second derivative means a minimum and a genuine oscillation; a negative one means a maximum, an unstable equilibrium and no oscillation at all. This is why simple harmonic motion is everywhere in physics — atoms in a crystal, the two atoms of a molecule, a satellite nudged off a stable orbit, a ball in a valley. Zoom in far enough on any smooth minimum and it is a parabola.

The approximation is good only while the cubic term is small compared with the quadratic one, which is what "small oscillations" means. Push the amplitude up and the period starts to drift — exactly as the simple pendulum's period drifts once the swing is no longer small.

The recipe, in three lines

Given $U(x)$ and a mass $m$:

1. Solve $\frac{dU}{dx} = 0$ for $x_0$. Discard any root where $\frac{d^2U}{dx^2}$ is negative.
2. Evaluate $k = \left.\frac{d^2U}{dx^2}\right\rvert_{x_0}$ in N/m.
3. $\omega = \sqrt{\frac{k}{m}}$, then $\nu = \frac{\omega}{2\pi}$ and $T = \frac{2\pi}{\omega}$.

Two families come up so often they are worth knowing:

Potential	Minimum at	$k = U^{\prime\prime}(x_0)$
$U = \frac{a}{x^2} - \frac{b}{x}$	$x_0 = \frac{2a}{b}$	$\frac{b^4}{8a^3}$
$U = \frac{a}{x^{12}} - \frac{b}{x^{6}}$	$x_0 = \left(\frac{2a}{b}\right)^{1/6}$	$\frac{72a}{x_0^{14}} = \frac{36b}{x_0^{8}}$

The tunnel through the Earth

Treat the Earth as a uniform sphere of radius $R$ and surface gravity $g$. Only the mass inside your radius pulls you, and that mass grows as $r^3$ while the inverse-square law divides by $r^2$, so the field inside is linear: $$g(r) = g\,\frac{r}{R}$$ Bore a tunnel straight through the centre and drop a ball in. The force on it is $$F = -\frac{mg}{R}\,r \qquad\Longrightarrow\qquad k_{\text{eff}} = \frac{mg}{R}, \qquad \omega = \sqrt{\frac{g}{R}}, \qquad T = 2\pi\sqrt{\frac{R}{g}}$$ Notice that the mass cancels — a marble and a truck take the same time. Putting in $R = 6.4 \times 10^{6}$ m and $g = 9.8$ m/s$^2$ gives $T = 5077$ s, that is $84.6$ minutes for the round trip and $42.3$ minutes to fall to the other side of the planet.

Now the part that surprises everybody. Bore the tunnel along a chord instead, at perpendicular distance $d$ from the centre. When the ball is at distance $r$ from the centre and $s$ from the tunnel's midpoint, the pull is $\frac{mgr}{R}$ towards the centre, and the component along the tunnel is $$F_{\parallel} = -\frac{mgr}{R}\cdot\frac{s}{r} = -\frac{mg}{R}\,s$$ The $r$ cancels. The period is $2\pi\sqrt{\frac{R}{g}}$ for every chord, no matter how short — $84.6$ minutes from Delhi to Mumbai or from pole to pole. Only the amplitude changes, and with it the maximum speed, $v_{\max} = \omega\sqrt{R^2 - d^2}$. For the diametric tunnel that is $\sqrt{gR} = 7.92$ km/s, which is the orbital speed at the Earth's surface — not a coincidence, but that is a story for gravitation.

A charged bead on the axis of a ring

A ring of radius $R$ carries a total charge $Q$, spread uniformly. On its axis, at distance $x$ from the centre, the field is $$E = \frac{1}{4\pi\epsilon_0}\cdot\frac{Qx}{\left(R^2 + x^2\right)^{3/2}}$$ For $x \ll R$ the denominator is just $R^3$, so $E$ is linear in $x$. A bead of mass $m$ carrying an opposite charge $-q$ is therefore pulled back towards the centre with $F = -\frac{Qq}{4\pi\epsilon_0R^3}x$, giving $$\omega = \sqrt{\frac{Qq}{4\pi\epsilon_0 m R^3}}$$ With $Q = 1.0\times10^{-7}$ C, $q = 1.0\times10^{-8}$ C, $m = 1.0\times10^{-6}$ kg and $R = 0.1$ m this comes to $\omega = 94.9$ rad/s, so $\nu = 15.1$ Hz. The whole problem is the linearisation of one denominator.

[Exam Tip] Whenever a problem gives you $U(x)$, a field, or a force that is not obviously linear, your first move is to expand about the equilibrium and keep the linear term. Every one of these problems is the same problem.

Counting Cycles, and the Sharpness of Resonance

From "how long" to "how many"

Section 9 gave the damped amplitude $A(t) = A_0e^{-bt/2m}$. JEE rarely asks for the time; it asks for the number of complete oscillations before the amplitude falls to some fraction. That is a two-step calculation and both steps are easy to fumble.

Key Point — cycles to a given amplitude fraction. To fall from $A_0$ to $A$, $$t = \frac{2m}{b}\ln\!\left(\frac{A_0}{A}\right), \qquad n = \frac{t}{T^{\,\prime}} \quad\text{with}\quad T^{\,\prime} = \frac{2\pi}{\omega^{\,\prime}}, \quad \omega^{\,\prime} = \sqrt{\omega_0^2 - \frac{b^2}{4m^2}}$$ Divide by the damped period $T^{\,\prime}$, not by $\frac{2\pi}{\omega_0}$. For light damping the two are nearly equal, and for heavy damping they are not.

Useful fractions, all from the same logarithm: $\ln 2 = 0.693$ for half, $\ln 4 = 1.386$ for a quarter, $\ln 10 = 2.303$ for a tenth, and $\ln e = 1$ for the $\frac{1}{e}$ point. And since the energy goes as the square of the amplitude, $E(t) = E_0e^{-bt/m}$, the energy reaches half in exactly half the time the amplitude does.

The quality factor

One number measures how good an oscillator is at oscillating:

Key Point — the quality factor. $$Q = \frac{m\omega_0}{b} = \frac{\sqrt{km}}{b} = \frac{\omega_0}{2\gamma}, \qquad \gamma = \frac{b}{2m}$$ $Q$ is dimensionless. A high $Q$ means light damping: the free oscillation rings for a long time, and the resonance peak is tall and narrow. A car suspension has $Q \approx 2$; a tuning fork about $10^3$; a quartz crystal $10^5$ or more.

The number of cycles before the amplitude drops to $\frac{1}{e}$ follows immediately: the amplitude time constant is $\frac{1}{\gamma} = \frac{2m}{b}$, and dividing by the period $\frac{2\pi}{\omega_0}$ gives $$n_{1/e} = \frac{\omega_0}{2\pi\gamma} = \frac{Q}{\pi}$$ So "$Q = 25$" already tells you the thing will manage about $8$ swings before it has faded to $37\%$.

The sharpness of resonance: half-power bandwidth

Section 10 drew the resonance curve and showed that lower damping makes it taller and sharper. How much sharper is what $Q$ measures. Start from the steady-state amplitude, $$A(\omega_d) = \frac{F_0/m}{\sqrt{\left(\omega_0^2 - \omega_d^2\right)^2 + \left(\frac{b\omega_d}{m}\right)^2}}$$ Near resonance write $\omega_d = \omega_0 + \Delta$ with $\Delta$ small. Then $\omega_0^2 - \omega_d^2 \approx -2\omega_0\Delta$, and the peak value is $A_{\max} \approx \frac{F_0}{b\omega_0}$. The amplitude falls to $\frac{A_{\max}}{\sqrt{2}}$ — the half-power points, so called because the energy stored goes as $A^2$ and has halved there — when the two terms under the root are equal: $$2\omega_0\lvert\Delta\rvert = \frac{b\omega_0}{m} \qquad\Longrightarrow\qquad \lvert\Delta\rvert = \frac{b}{2m}$$

Key Point — bandwidth and $Q$. The full width of the resonance curve between the half-power points is $$\Delta\omega = \frac{b}{m} = \frac{\omega_0}{Q} \qquad\Longleftrightarrow\qquad Q = \frac{\omega_0}{\Delta\omega} = \frac{\nu_0}{\Delta\nu}$$ A high-$Q$ oscillator responds to a narrow band of driving frequencies and ignores everything else. That is what makes a radio tuner select one station out of hundreds.

One more identity worth carrying: at resonance the amplitude is $Q$ times the amplitude the same force would produce if applied statically. $$\frac{A_{\text{res}}}{A_{\text{static}}} = \frac{F_0/(b\omega_0)}{F_0/k} = \frac{k}{b\omega_0} = \frac{m\omega_0}{b} = Q$$ So $Q$ is simultaneously the resonant amplification, the reciprocal of the fractional bandwidth, and $\pi$ times the number of free cycles to $\frac{1}{e}$. Three questions, one number.

The four traps this chapter sets, every single year

1. Measuring displacement from the natural length instead of the equilibrium. For a hanging spring the mean position is the stretched equilibrium, $\frac{mg}{k}$ below the natural length. Release a block from the natural length of an unstretched vertical spring and its amplitude is $\frac{mg}{k}$, so it falls a total of $\frac{2mg}{k}$ before coming back up — not $\frac{mg}{k}$, and not "it just hangs there". The gravity term cancels exactly once you measure from the right place, which is why $T = 2\pi\sqrt{\frac{m}{k}}$ has no $g$ in it at all. The same applies to the trolley, the lift and the incline: find the shifted equilibrium first, then start the clock.

2. Degrees where the small-angle approximation needs radians. $\sin\theta \approx \theta$ is true for $\theta$ in radians and nowhere else. At $\theta = 6°$ the sine is $0.10453$ and the radian measure is $0.10472$, agreeing to two parts in a thousand; the number $6$ agrees with nothing. Every $\theta$ inside a $\sin$, every angular amplitude, every phase constant you substitute — radians. A phase quoted with no unit is in radians.

3. Forgetting that the energy varies at twice the frequency. With $x = A\cos\omega t$, both $U = \frac{1}{2}kA^2\cos^2\omega t$ and $K = \frac{1}{2}kA^2\sin^2\omega t$ contain $\cos 2\omega t$. Their angular frequency is $2\omega$ and their period is $\frac{T}{2}$. Their sum is flat. If a question mentions kinetic or potential energy and a frequency in the same breath, the factor of $2$ is the whole question.

4. Quoting $\omega$ when the question asked for $\nu$. They differ by $2\pi = 6.28$, and the wrong one is always on the option list. "Frequency" means $\nu$ in hertz. "Angular frequency" means $\omega$ in radians per second. Combine this trap with the previous one and a block with $\omega = 10$ rad/s has a potential energy that varies at $\omega_U = 20$ rad/s and $\nu_U = 3.18$ Hz — and the answers $10$, $20$ and $1.59$ are all wrong for different reasons.

Key Point — the last line of every solution. Write the unit. $\omega$ is rad/s, $\nu$ is Hz, $T$ is s, $k$ is N/m, $b$ is kg/s, $Q$ is a pure number. If the unit does not match what the question asked for, the answer is wrong however good the algebra was.

Solved Examples, Part 1: Superposing Oscillations

Values used throughout, unless a problem says otherwise: $g = 9.8$ m/s$^2$, $R_{\text{Earth}} = 6.4\times10^{6}$ m, $\frac{1}{4\pi\epsilon_0} = 9\times10^{9}$ N m$^2$/C$^2$. Every angle inside a sine or cosine is in radians. The standard form is $x = A\cos(\omega t + \phi)$.

Example 1: Two oscillations on one line, added the fast way

A particle is acted on by two simple harmonic motions along the same straight line: $$x_1 = 6\cos\omega t \ \text{cm}, \qquad x_2 = 8\cos\left(\omega t + \frac{\pi}{3}\right)\ \text{cm}$$ (a) Write the resultant in the form $A\cos(\omega t + \epsilon)$. (b) Keeping both amplitudes fixed, what are the largest and smallest resultant amplitudes obtainable by changing $\delta$? (c) For what phase difference would the resultant amplitude be exactly $10$ cm?

Solution:

1. Draw the phasors and use the cosine rule. With $A_1 = 6$, $A_2 = 8$ and $\delta = \frac{\pi}{3}$, so $\cos\delta = 0.5$: $$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\delta} = \sqrt{36 + 64 + 2(6)(8)(0.5)} = \sqrt{148} = 12.17\ \text{cm}$$

2. The phase constant of the resultant. $$\tan\epsilon = \frac{A_2\sin\delta}{A_1 + A_2\cos\delta} = \frac{8\left(\frac{\sqrt{3}}{2}\right)}{6 + 8(0.5)} = \frac{6.928}{10} = 0.6928$$ $$\epsilon = 0.606\ \text{rad} = 34.7°$$ So $x = 12.17\cos\left(\omega t + 0.606\right)$ cm, and it is one SHM of the same angular frequency $\omega$.

3. Sanity check on the phase constant. $\epsilon$ must lie between $0$ and $\delta$, because the resultant phasor lies inside the angle between the two. It does: $0 < 34.7° < 60°$. And it is nearer the larger amplitude's phasor than the smaller one's, as the parallelogram demands.

4. (b) The extremes. $\cos\delta$ can only run from $+1$ to $-1$: $$A_{\max} = A_1 + A_2 = 14\ \text{cm} \quad \text{at } \delta = 0, \qquad A_{\min} = \lvert A_1 - A_2 \rvert = 2\ \text{cm} \quad \text{at } \delta = \pi$$ Every possible resultant lies between $2$ cm and $14$ cm.

5. (c) Work backwards for $\delta$. Set $A = 10$: $$100 = 36 + 64 + 96\cos\delta \qquad\Longrightarrow\qquad 96\cos\delta = 0 \qquad\Longrightarrow\qquad \delta = \frac{\pi}{2}$$ Which is the Pythagorean case, $\sqrt{6^2 + 8^2} = 10$, as it had to be.

Final Answer: (a) $x = 12.17\cos\left(\omega t + 0.606\right)$ cm, that is amplitude $12.17$ cm with phase constant $0.606$ rad or $34.7°$. (b) Between $2$ cm and $14$ cm. (c) $\delta = \frac{\pi}{2}$.

Takeaway: Two SHMs of the same frequency on the same line always give a third SHM of that same frequency. Only the amplitude and the phase constant are new, and one triangle delivers both.

Example 2: Three phasors, and one that vanishes

(a) Three SHMs act along the same line on one particle: $$x_1 = 3\cos\omega t, \qquad x_2 = 4\cos\left(\omega t + \frac{\pi}{2}\right), \qquad x_3 = 5\cos\left(\omega t + \pi\right)$$ all in centimetres. Find the resultant. (b) Three SHMs of equal amplitude $A$ have phase constants $0$, $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$. Find their resultant.

Solution:

1. (a) With more than two phasors, resolve into components. Take the $x$-axis along the first phasor:

Phasor	Amplitude	Phase	Along axis	Perpendicular
$x_1$	$3$	$0$	$3$	$0$
$x_2$	$4$	$\frac{\pi}{2}$	$0$	$4$
$x_3$	$5$	$\pi$	$-5$	$0$
Sum			$-2$	$4$

2. Recombine. $$A = \sqrt{(-2)^2 + 4^2} = \sqrt{20} = 4.47\ \text{cm}$$ $$\epsilon = \arctan\left(\frac{4}{-2}\right) \; \text{in the second quadrant} \; = \pi - \arctan 2 = 2.034\ \text{rad} = 116.6°$$ The signs matter: the along-axis component is negative and the perpendicular one positive, so $\epsilon$ is in the second quadrant. A calculator's $\arctan(-2) = -63.4°$ is the wrong branch.

3. (b) Three equal phasors at $120°$ to each other. Resolve: $$\text{along} = A\left[\cos 0 + \cos\frac{2\pi}{3} + \cos\frac{4\pi}{3}\right] = A\left[1 - \frac{1}{2} - \frac{1}{2}\right] = 0$$ $$\text{perpendicular} = A\left[\sin 0 + \sin\frac{2\pi}{3} + \sin\frac{4\pi}{3}\right] = A\left[0 + \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right] = 0$$ The resultant is exactly zero. The three phasors form a closed equilateral triangle, so the particle does not move at all.

Final Answer: (a) $x = 4.47\cos\left(\omega t + 2.034\right)$ cm. (b) Zero — the particle stays at rest.

Takeaway: Beyond two phasors, always go to components; the cosine rule only handles a pair. And any set of equal phasors spread evenly around the circle sums to zero, which is why three-phase electricity has no return current.

Example 3: One pair of perpendicular equations, four different paths

A particle's coordinates are $x = 4\cos\omega t$ cm and $y = 3\cos\left(\omega t + \delta\right)$ cm. Find the path, and the area it encloses, for $\delta = 0$, $\frac{\pi}{2}$, $\pi$ and $\frac{\pi}{4}$.

Solution:

1. Write the master equation once, with $a = 4$ and $b = 3$: $$\frac{x^2}{16} + \frac{y^2}{9} - \frac{2xy}{12}\cos\delta = \sin^2\delta$$

2. $\delta = 0$. The right side is zero and the left is a perfect square: $$\left(\frac{x}{4} - \frac{y}{3}\right)^2 = 0 \qquad\Longrightarrow\qquad y = \frac{3}{4}x$$ A straight line through the origin, travelled back and forth between $(4, 3)$ and $(-4, -3)$. The motion along that line is itself SHM of amplitude $\sqrt{4^2 + 3^2} = 5$ cm. Enclosed area zero.

3. $\delta = \frac{\pi}{2}$. Now $\cos\delta = 0$ and $\sin^2\delta = 1$: $$\frac{x^2}{16} + \frac{y^2}{9} = 1$$ An ellipse with semi-axes $4$ cm and $3$ cm lying along the coordinate axes. It is not a circle, because $a \ne b$. Area $= \pi(4)(3) = 37.7$ cm$^2$.

4. $\delta = \pi$. Right side zero again, cross term flips sign: $$\left(\frac{x}{4} + \frac{y}{3}\right)^2 = 0 \qquad\Longrightarrow\qquad y = -\frac{3}{4}x$$ The same line reflected — the other diagonal of the rectangle. Area zero.

5. $\delta = \frac{\pi}{4}$. With $\cos\delta = \sin\delta = \frac{1}{\sqrt{2}}$: $$\frac{x^2}{16} + \frac{y^2}{9} - \frac{xy}{6\sqrt{2}} = \frac{1}{2}$$ A tilted ellipse, still inscribed in the rectangle $\lvert x \rvert \le 4$, $\lvert y \rvert \le 3$, touching its sides. Its area is $\pi a b \lvert\sin\delta\rvert = 37.7 \times \frac{1}{\sqrt{2}} = 26.7$ cm$^2$.

6. The pattern. As $\delta$ runs from $0$ to $\pi$ the path opens from a line, fattens to the upright ellipse at $\frac{\pi}{2}$, and closes back to the opposite line. The area $\pi ab\lvert\sin\delta\rvert$ tracks it exactly.

Final Answer: $\delta = 0$: line $y = \frac{3}{4}x$; $\delta = \frac{\pi}{2}$: ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ of area $37.7$ cm$^2$; $\delta = \pi$: line $y = -\frac{3}{4}x$; $\delta = \frac{\pi}{4}$: tilted ellipse of area $26.7$ cm$^2$.

Takeaway: Learn one equation, not four cases. And note that none of these is a circle: with $a = 4$ and $b = 3$ no choice of $\delta$ can make one, because a circle needs equal amplitudes and a quarter-cycle lag.

Example 4: How far from the origin does a tilted ellipse reach?

A particle moves with $$x = 2\sin\omega t\ \text{m}, \qquad y = 2\sin\left(\omega t + \frac{\pi}{3}\right)\ \text{m}$$ (a) Find the equation of its path. (b) Find its greatest and least distances from the origin. (c) A student answers (b) with $2\sqrt{2}$ m. What went wrong?

Solution:

1. (a) The master equation works with sines too, since both motions use the same function and $\delta = \frac{\pi}{3}$. With $a = b = 2$ and $\cos\delta = \frac{1}{2}$, $\sin^2\delta = \frac{3}{4}$: $$\frac{x^2}{4} + \frac{y^2}{4} - \frac{2xy}{4}\cdot\frac{1}{2} = \frac{3}{4}$$ Multiply through by $4$: $$x^2 + y^2 - xy = 3$$

2. Recognise the shape. The cross term means it is tilted. Because the $x^2$ and $y^2$ coefficients are equal, the symmetry axes are the lines $y = x$ and $y = -x$.

3. (b) Look along each axis of symmetry. On $y = x$, substitute: $$x^2 + x^2 - x^2 = 3 \qquad\Longrightarrow\qquad x = \pm\sqrt{3}$$ That point is $\left(\sqrt{3}, \sqrt{3}\right)$, at distance $\sqrt{3 + 3} = \sqrt{6} = 2.45$ m. On $y = -x$: $$x^2 + x^2 + x^2 = 3 \qquad\Longrightarrow\qquad x = \pm 1$$ That point is $(1, -1)$, at distance $\sqrt{2} = 1.41$ m.

4. So the semi-major axis is $\sqrt{6}$ m along $y = x$, and the semi-minor axis is $\sqrt{2}$ m along $y = -x$. The greatest distance from the origin is $2.45$ m and the least is $1.41$ m.

5. (c) Where $2\sqrt{2}$ comes from. It is $\sqrt{2^2 + 2^2}$ — the corner of the bounding square, obtained by assuming $x$ and $y$ reach their maxima at the same instant. They do not: they are $\frac{\pi}{3}$ out of step, so when $x = 2$ the value of $y$ is $2\cos\frac{\pi}{3} = 1$, giving only $\sqrt{5} = 2.24$ m. The true maximum, $\sqrt{6} = 2.45$ m, occurs at neither extreme.

Final Answer: (a) $x^2 + y^2 - xy = 3$. (b) Greatest $\sqrt{6} = 2.45$ m, least $\sqrt{2} = 1.41$ m. (c) $2\sqrt{2}$ assumes both coordinates peak together, which a phase difference forbids.

Takeaway: With perpendicular SHMs, the maximum of the distance is not the combination of the two separate maxima. Get the path equation, then look along its symmetry axes.