Chapter 5

Oscillations

Maharashtra HSC · 15 subtopics · textbook pages 109–130

Formula SheetEvery formula this chapter can be examined on, in one table.

Periodic Motion — Period and Frequency

Formula	Symbols, units and conditions
$T=\dfrac{1}{n}\qquad n=\dfrac{1}{T}$	$T$ is the period in second (s), $n$ the frequency in hertz (Hz). $T$ is the smallest time after which the motion repeats.
$\omega=2\pi n=\dfrac{2\pi}{T}$	$\omega$ is the angular frequency in rad/s. Use this to move between $T$ , $n$ and $\omega$ in every numerical of this chapter.

Linear S.H.M. — Restoring Force

Formula	Symbols, units and conditions
$f=-kx$	$f$ is the restoring force (N), $x$ the displacement from the mean position (m), $k$ the force constant (spring constant) in N/m. The minus sign says force and displacement are oppositely directed.
$k=\dfrac{f}{x}=m\omega^2$	Force constant = restoring force per unit displacement, in N/m. Conversion: $1$ dyne/cm $=10^{-3}$ N/m, so $25\times10^{3}$ dyne/cm $=25$ N/m.
$a=\dfrac{f}{m}=-\left(\dfrac{k}{m}\right)x$	Acceleration of a block of mass $m$ (kg) on a spring of force constant $k$ .

Differential Equation of S.H.M.

Formula	Symbols, units and conditions
$\dfrac{d^2x}{dt^2}+\omega^2x=0\qquad\text{with}\qquad \omega^2=\dfrac{k}{m}$	The differential equation of linear S.H.M. Quote it in exactly this form; $\omega$ is the angular frequency in rad/s.
$I\dfrac{d^2\theta}{dt^2}+c\theta=0$	The corresponding differential equation of angular S.H.M.: $I$ is the moment of inertia (kg m $^2$ ) and $c$ the restoring torque per unit angular displacement (N m/rad).

Acceleration, Velocity and Displacement

Formula	Symbols, units and conditions
$a=-\omega^2x$	Acceleration at displacement $x$ ; magnitude $\omega^2x$ , always directed towards the mean position.
$v=\pm\,\omega\sqrt{A^2-x^2}$	Velocity at displacement $x$ ; $A$ is the amplitude (m). Both signs are physical — the particle passes each point twice per oscillation.
$x=A\sin(\omega t+\phi)$	General displacement at time $t$ ; $\phi$ is the initial phase (epoch) in radian.
$x=\pm A\sin\omega t\qquad(\text{particle starts from the mean position, }\phi=0)$	Positive sign if it starts towards the positive side, negative if towards the negative side.
$x=\pm A\cos\omega t\qquad\left(\text{particle starts from an extreme position, }\phi=\dfrac{\pi}{2}\right)$	Positive sign for the positive extreme, negative for the negative extreme.
$v=\dfrac{dx}{dt}=A\omega\cos(\omega t+\phi)\qquad a=\dfrac{dv}{dt}=-A\omega^2\sin(\omega t+\phi)$	Velocity and acceleration at time $t$ for the general starting phase $\phi$ .
$v_{max}=A\omega\quad(\text{at }x=0)\qquad a_{max}=A\omega^2\quad(\text{at }x=\pm A)$	Maximum speed occurs at the mean position; maximum acceleration at the extreme positions.
$\omega=\dfrac{a_{max}}{v_{max}}\qquad A=\dfrac{v_{max}}{\omega}=\dfrac{(v_{max})^2}{a_{max}}$	The quickest route through any "maximum velocity and maximum acceleration given" numerical.

Amplitude, Period and Frequency

Formula	Symbols, units and conditions
$A=x_{max}=\dfrac{\text{path length}}{2}$	Amplitude is the maximum displacement from the mean position; the path length (length of path) of an S.H.M. is $2A$ .
$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{\sqrt{\text{acceleration per unit displacement}}}$	The standard board form of the period. "Acceleration per unit displacement" $=\dfrac{a}{x}=\omega^2$ in magnitude.
$T=2\pi\sqrt{\dfrac{m}{k}}\qquad n=\dfrac{1}{T}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$	Period and frequency of a mass $m$ (kg) oscillating on a spring of force constant $k$ (N/m).
$\dfrac{1}{k_s}=\dfrac{1}{k_1}+\dfrac{1}{k_2}+\ldots\qquad k_s=\dfrac{k_1k_2}{k_1+k_2}\qquad k_s=\dfrac{k}{n}$	Springs in series: effective constant $k_s$ ; product over sum for two springs; $k/n$ for $n$ identical springs.
$k_p=k_1+k_2+\ldots\qquad k_p=mk$	Springs in parallel: effective constant $k_p$ ; $mk$ for $m$ identical springs of constant $k$ .

Reference Circle Method

Formula	Symbols, units and conditions
$y=r\sin(\omega t+\phi)$	Projection of the position vector of a particle in U.C.M. of radius $r$ on the reference diameter — an S.H.M. of amplitude $r$ .
$v_y=r\omega\cos(\omega t+\phi)\qquad a_y=-r\omega^2\sin(\omega t+\phi)=-\omega^2y$	Projections of the tangential velocity $r\omega$ and of the centripetal acceleration $r\omega^2$ on the same diameter.

Phase in S.H.M.

Formula	Symbols, units and conditions
$\theta=\omega t+\phi$	Phase angle at time $t$ (radian); $\phi$ is the initial phase or epoch, i.e. the phase at $t=0$ .
$\theta=2\pi N+\theta_0$	A phase larger than $2\pi$ means a later oscillation: $N$ complete oscillations are over and the particle is in oscillation number $(N+1)$ at phase $\theta_0$ .

Graphical Representation

Formula	Symbols, units and conditions
$x=A\sin\omega t,\quad v=A\omega\cos\omega t,\quad a=-A\omega^2\sin\omega t$	Starting from the mean position ( $\phi=0$ ).
$x=A\cos\omega t,\quad v=-A\omega\sin\omega t,\quad a=-A\omega^2\cos\omega t$	Starting from the positive extreme position $\left(\phi=\dfrac{\pi}{2}\right)$ .

Composition of Two S.H.M.s (same period, same path)

Formula	Symbols, units and conditions
$x=x_1+x_2=R\sin(\omega t+\delta)$	The superposition of $x_1=A_1\sin(\omega t+\phi_1)$ and $x_2=A_2\sin(\omega t+\phi_2)$ is again an S.H.M. of the same period.
$R=\sqrt{A_1^{2}+A_2^{2}+2A_1A_2\cos(\phi_1-\phi_2)}$	Resultant amplitude, in the same unit as $A_1$ and $A_2$ .
$\delta=\tan^{-1}\left[\dfrac{A_1\sin\phi_1+A_2\sin\phi_2}{A_1\cos\phi_1+A_2\cos\phi_2}\right]$	Initial phase (epoch) of the resultant S.H.M.
$\phi_1-\phi_2=0:\ R=A_1+A_2\qquad \phi_1-\phi_2=\dfrac{\pi}{2}:\ R=\sqrt{A_1^{2}+A_2^{2}}\qquad \phi_1-\phi_2=\pi:\ R=\lvert A_1-A_2\rvert$	The three special cases; with $A_1=A_2=A$ they give $2A$ , $\sqrt2\,A$ and $0$ respectively.
$x=a\sin\omega t+b\cos\omega t\ \Rightarrow\ A=\sqrt{a^{2}+b^{2}}$	A sine plus a cosine of the same $\omega$ is a single S.H.M. of this amplitude (the $\phi_1-\phi_2=\dfrac{\pi}{2}$ case).

Energy of a Particle Performing S.H.M.

Formula	Symbols, units and conditions
$E_k=\dfrac{1}{2}mv^2=\dfrac{1}{2}m\omega^2(A^2-x^2)=\dfrac{1}{2}k(A^2-x^2)$	Kinetic energy (J) at displacement $x$ .
$E_k=\dfrac{1}{2}kA^2\cos^2(\omega t+\phi)$	Kinetic energy at time $t$ — it varies as $\cos^2\theta$ .
$E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}m\omega^2x^2$	Potential energy (J) at displacement $x$ .
$E_p=\dfrac{1}{2}kA^2\sin^2(\omega t+\phi)$	Potential energy at time $t$ — it varies as $\sin^2\theta$ .
$E=E_k+E_p=\dfrac{1}{2}m\omega^2A^2=\dfrac{1}{2}kA^2=\dfrac{1}{2}m(v_{max})^2$	Total energy — constant, independent of $x$ and $t$ .
$E=2\pi^2mn^2A^2=\dfrac{2\pi^2mA^2}{T^2}$	Total energy in terms of frequency and period: $E\propto m$ , $E\propto A^2$ , $E\propto n^2$ , $E\propto k$ , $E\propto \dfrac{1}{T^2}$ .
$E_k=E_p=\dfrac{E}{2}\ \text{at}\ x=\pm\dfrac{A}{\sqrt2}$	The displacement at which kinetic and potential energies are equal.
$\text{At }x=\pm\dfrac{A}{2}:\ E_p=\dfrac{E}{4},\quad E_k=\dfrac{3E}{4}$	That is, 25% potential and 75% kinetic.
$\dfrac{E_k}{E}=1-\dfrac{x^2}{A^2}\qquad \dfrac{E_p}{E}=\dfrac{x^2}{A^2}$	Handy fraction forms — e.g. at $x=\dfrac{A}{3}$ , $E_k=\dfrac{8}{9}E$ .
$\text{At }v=\dfrac{v_{max}}{2}:\ x=\pm\dfrac{\sqrt3}{2}A\qquad \text{At }x=\pm\dfrac{A}{2}:\ v=\dfrac{\sqrt3}{2}v_{max}$	Two ratios the board sets repeatedly.
$\omega^2A\le g\ \Rightarrow\ n_{max}=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{A}}$	Condition for a block resting on a vertically vibrating platform not to lose contact; the energy per unit mass is then $\dfrac{E}{m}=\dfrac{1}{2}\omega^2A^2$ .

Simple Pendulum

Formula	Symbols, units and conditions
$F=-mg\sin\theta\approx-mg\theta=-\dfrac{mg}{L}x$	Restoring force on the bob; valid for small angular amplitude ( $\theta<10^\circ$ ), where $\theta=\dfrac{x}{L}$ in radian.
$T=2\pi\sqrt{\dfrac{L}{g}}\qquad n=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}$	Period and frequency of a simple pendulum; $L$ is the length in metre (point of suspension to the centre of gravity of the bob), $g=9.8$ m/s $^2$ .
$T\propto\sqrt{L}\qquad \dfrac{T_2}{T_1}=\sqrt{\dfrac{L_2}{L_1}}$	The working form for every "length changed by …, period changes by …" problem.
$L_s=\dfrac{g}{\pi^2}\approx0.993\ \text{m}$	Length of a second's pendulum — a simple pendulum whose period is $T=2$ s.
$\text{Number of oscillations per day}=\dfrac{86400}{T}\qquad \text{Time gained / lost per day}=\dfrac{86400\times\Delta T}{T}$	Clock problems: $86400$ s in a day; $\Delta T$ is the change in the pendulum's period.
$(E_p)_{max}=\dfrac{1}{2}m\omega^2A^2=\dfrac{mgA^2}{2L}$	Potential energy of a pendulum bob at the extreme position, using $\omega^2=\dfrac{g}{L}$ .

Angular S.H.M.

Formula	Symbols, units and conditions
$\tau\propto-\theta\qquad \tau=-c\theta$	Restoring torque (N m); $c$ is the torque per unit angular displacement (N m/rad), $\theta$ the angular displacement (rad).
$T=\dfrac{2\pi}{\sqrt{\text{angular acceleration per unit angular displacement}}}=2\pi\sqrt{\dfrac{I}{c}}$	Period of angular (torsional) S.H.M.; $I$ is the moment of inertia about the axis of oscillation.
$\tau_{max}=c\,\theta_{max}\qquad c=\dfrac{4\pi^2I}{T^2}$	Maximum restoring torque in a twisted wire, with $\theta_{max}$ in radian.
$\tau=-\mu B\sin\theta\approx-\mu B\theta\qquad \alpha=-\left(\dfrac{\mu B}{I}\right)\theta$	Bar magnet of magnetic moment $\mu$ (A m $^2$ ) in a uniform field $B$ (T or Wb/m $^2$ ).
$T=2\pi\sqrt{\dfrac{I}{\mu B}}\qquad n=\dfrac{1}{2\pi}\sqrt{\dfrac{\mu B}{I}}$	Period and frequency of a magnet vibrating in a uniform magnetic field.
$I=m\left(\dfrac{l^2+b^2}{12}\right)\qquad I_{disc}=\dfrac{1}{2}MR^2$	Moment of inertia of a rectangular bar magnet (length $l$ , breadth $b$ , axis through the centre perpendicular to the $l\times b$ face) and of a thin circular disc about its central axis.

Damped Oscillations

Formula	Symbols, units and conditions
$F_d=-bv$	Damping force; $b$ is the damping constant (N s/m), $v$ the speed.
$m\dfrac{d^2x}{dt^2}+b\dfrac{dx}{dt}+kx=0$	Differential equation of a damped harmonic oscillator.
$x=A\,e^{-bt/2m}\cos(\omega' t+\phi)$	Its solution; the amplitude $A\,e^{-bt/2m}$ decays exponentially with time.
$\omega'=\sqrt{\dfrac{k}{m}-\left(\dfrac{b}{2m}\right)^2}\qquad T=\dfrac{2\pi}{\omega'}$	Angular frequency and period of the damped oscillator — damping increases the period and decreases the amplitude.

Free Oscillations, Forced Oscillations and Resonance

Formula	Symbols, units and conditions
$n_{natural}=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$	Natural frequency of a freely oscillating simple pendulum of length $l$ ; for a spring–mass system it is $\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ .
$n_{driver}=n_{natural}$	Condition for resonance — the driven body then absorbs maximum energy and oscillates with maximum amplitude.

Standard Values

$g=9.8$ m/s $^2$ unless the question states otherwise; some board numericals use $g\approx\pi^2\approx10$ m/s $^2$ .
$\pi=3.142$ (the board's own working value); $2\pi=6.284$ .
$1$ dyne/cm $=10^{-3}$ N/m; $1$ dyne $=10^{-5}$ N.
$1$ day $=86400$ s.
SI units: $x$ in m, $v$ in m/s, $a$ in m/s $^2$ , $\omega$ in rad/s, $n$ in Hz, $T$ in s, $k$ in N/m, $E$ in J, $\tau$ in N m, $I$ in kg m $^2$ , $\mu$ in A m $^2$ , $B$ in T (Wb/m $^2$ ).

Short NotesThe chapter subtopic by subtopic, in the textbook's own order.15 subtopics

The whole chapter in brief

5.1–5.2 Periodic and Oscillatory Motion

Periodic motion: any motion that repeats itself after a definite interval of time. The body goes on repeating the same set of movements; the time for one such set is the period.
Oscillatory (vibratory) motion: a periodic motion in which the particle moves to and fro repeatedly along the same path about a mean position.
Every oscillatory motion is periodic, but every periodic motion need not be oscillatory. Uniform circular motion is periodic but not oscillatory.
The simplest oscillatory periodic motion is simple harmonic motion (S.H.M.). If the path is a straight line it is linear S.H.M.; if the path is an arc of a circle it is angular S.H.M.
Period $T$ : smallest time after which the to and fro motion repeats (s). Frequency $n$ : number of oscillations per unit time (Hz). $n=\dfrac{1}{T}$ , $\omega=2\pi n=\dfrac{2\pi}{T}$ .
Harmonic oscillation: expressible by a single harmonic function, $x=a\sin\omega t$ or $x=a\cos\omega t$ . Non-harmonic oscillation: cannot be written as a single harmonic function, e.g. $x=a\sin\omega t+b\sin2\omega t$ .

Periodic motion	Oscillatory motion
Repeats after a definite interval of time	Repeats to and fro about a mean position along the same path
Need not be to and fro	Always to and fro
Example: motion of the earth round the sun, U.C.M.	Example: simple pendulum, spring–mass system, tuning fork prong
Not every periodic motion is oscillatory	Every oscillatory motion is periodic

5.3 Linear S.H.M.

Definition: Linear S.H.M. is the linear periodic motion of a body in which the force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position.
$f=-kx$ , where $k$ is the force constant (restoring force per unit displacement, N/m); $a=-\left(\dfrac{k}{m}\right)x$ .
Conditions required for S.H.M.: (i) oscillation is about a fixed point, (ii) the net force or acceleration is always directed towards that fixed point, (iii) the particle returns to the fixed point because of the restoring force.
Displacement in S.H.M. is always measured from the mean position, and since the motion is along one straight line only $+$ and $-$ signs are needed, not vector notation.
A complete oscillation: the object goes from one extreme to the other and back to the initial position.
Classroom demonstrations of linear S.H.M.: a hydrometer floating in water, depressed and released; mercury in a U-tube, one arm depressed and released; a block attached to a spring on a frictionless horizontal surface.

5.4 Differential Equation of S.H.M.

$\dfrac{d^2x}{dt^2}+\omega^2x=0$ , obtained from $f=-kx$ with $f=ma$ and $\dfrac{k}{m}=\omega^2$ .
Angular form: $I\dfrac{d^2\theta}{dt^2}+c\theta=0$ .
Reading $\omega$ straight off the equation is a standard 2-mark question: in $\dfrac{d^2x}{dt^2}=-36x$ , $\omega=6$ rad/s, so $n=\dfrac{6}{2\pi}=0.9548$ Hz and $T=1.047$ s.

5.5 Acceleration, Velocity and Displacement

$a=-\omega^2x$ ; $v=\pm\omega\sqrt{A^2-x^2}$ ; $x=A\sin(\omega t+\phi)$ .
Starting from the mean position: $x=\pm A\sin\omega t$ . Starting from an extreme position: $x=\pm A\cos\omega t$ .
At time $t$ : $v=A\omega\cos(\omega t+\phi)$ and $a=-A\omega^2\sin(\omega t+\phi)$ .
"The particle starts S.H.M." means the observer starts counting time then; the particle was already oscillating. $t=0$ is subjective to the observer.

Quantity	At mean position $(x=0)$	At extreme position $(x=\pm A)$
Displacement	Minimum, $x=0$	Maximum, $x=\pm A$
Velocity	Maximum, $v_{max}=A\omega$	Minimum, $v=0$
Acceleration	Minimum, $a=0$	Maximum, $a_{max}=\mp A\omega^2$
Restoring force	Zero	Maximum
Kinetic energy	Maximum, $\dfrac{1}{2}kA^2$	Zero
Potential energy	Zero	Maximum, $\dfrac{1}{2}kA^2$

5.6 Amplitude, Period and Frequency

Amplitude $A$ : the maximum displacement of the particle from its mean position. Path length $=2A$ .
Period $T$ : the time taken to complete one oscillation. $T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{\sqrt{\text{acceleration per unit displacement}}}=2\pi\sqrt{\dfrac{m}{k}}$ .
Frequency $n$ : number of oscillations per unit time, $n=\dfrac{1}{T}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ .
$\omega^2=\dfrac{k}{m}=\dfrac{\text{force per unit displacement}}{\text{mass}}=$ acceleration per unit displacement.

Springs in series	Springs in parallel
Connected one after the other in a single chain	Connected between the same two points
Each spring carries the same force $f$	Each spring has the same extension $e$
Extensions add: $e=e_1+e_2$	Forces add: $f=f_1+f_2$
$\dfrac{1}{k_s}=\dfrac{1}{k_1}+\dfrac{1}{k_2}$ , so $k_s=\dfrac{k_1k_2}{k_1+k_2}$	$k_p=k_1+k_2$
$n$ identical springs: $k_s=\dfrac{k}{n}$ (softer, lower frequency)	$m$ identical springs: $k_p=mk$ (stiffer, higher frequency)

Two identical springs of constant $k$ : $k_s=\dfrac{k}{2}$ and $k_p=2k$ , so the frequencies are in the ratio $n_s:n_p=1:2$ .

5.7 Reference Circle Method

A linear S.H.M. is the projection of a uniform circular motion (U.C.M.) on any diameter of the reference circle.
Projection of displacement: $y=r\sin(\omega t+\phi)$ — an S.H.M. of amplitude $r$ .
Projection of the tangential velocity $r\omega$ : $v_y=r\omega\cos(\omega t+\phi)$ .
Projection of the centripetal acceleration $r\omega^2$ : $a_y=-r\omega^2\sin(\omega t+\phi)=-\omega^2y$ .
This is why $\omega$ , the angular velocity of the reference circle, is called the angular frequency of the linear S.H.M.

5.8 Phase in S.H.M.

Phase in S.H.M. is the state of oscillation: it fixes the displacement, the direction of velocity and the oscillation number at that instant.
Displacement alone is not enough (two directions of velocity at each position); velocity alone is not enough (two positions for each velocity). Hence a continuously changing quantity is used: the phase angle $\theta=\omega t+\phi$ .
Epoch (initial phase) $\phi$ : the physical quantity that describes the state of oscillation of the particle at the start of the motion, i.e. the phase at $t=0$ .

Phase $\theta$	State of the particle (first oscillation)
$0$	At the mean position, moving to the positive
$\dfrac{\pi}{2}$ or $90^\circ$	At the positive extreme position
$\pi$ or $180^\circ$	At the mean position, moving to the negative
$\dfrac{3\pi}{2}$ or $270^\circ$	At the negative extreme position
$2\pi$ or $360^\circ$	Beginning of the second oscillation

A phase greater than $360^\circ$ tells you the oscillation number: $1110^\circ=3\times360^\circ+30^\circ$ , so during the 4th oscillation the particle is at $A\sin30^\circ=\dfrac{A}{2}$ , moving towards the positive extreme.

5.9 Graphical Representation of S.H.M.

Starting from the mean position: $x=A\sin\omega t$ (sine curve from the origin), $v=A\omega\cos\omega t$ (cosine curve starting at $+A\omega$ ), $a=-A\omega^2\sin\omega t$ (inverted sine curve).
Starting from the positive extreme: $x=A\cos\omega t$ , $v=-A\omega\sin\omega t$ , $a=-A\omega^2\cos\omega t$ .

$t$	$0$	$T/4$	$T/2$	$3T/4$	$T$	$5T/4$
$\theta$ (from mean)	$0$	$\dfrac{\pi}{2}$	$\pi$	$\dfrac{3\pi}{2}$	$2\pi$	$\dfrac{5\pi}{2}$
$x$	$0$	$A$	$0$	$-A$	$0$	$A$
$v$	$A\omega$	$0$	$-A\omega$	$0$	$A\omega$	$0$
$a$	$0$	$-A\omega^2$	$0$	$A\omega^2$	$0$	$-A\omega^2$

Conclusions from the graphs: all three are periodic functions of time; the displacement–time and acceleration–time curves are sine curves while the velocity–time curve is a cosine curve; the phase difference between displacement and velocity is $\dfrac{\pi}{2}$ , between velocity and acceleration $\dfrac{\pi}{2}$ , and between displacement and acceleration $\pi$ ; all shapes repeat after $2\pi$ radian, i.e. after a time $T$ .

5.10 Composition of Two S.H.M.s (same period, same path)

Two S.H.M.s of the same period along the same path superpose into a single S.H.M. of the same period: $x=R\sin(\omega t+\delta)$ .
$R=\sqrt{A_1^{2}+A_2^{2}+2A_1A_2\cos(\phi_1-\phi_2)}$ and $\delta=\tan^{-1}\left[\dfrac{A_1\sin\phi_1+A_2\sin\phi_2}{A_1\cos\phi_1+A_2\cos\phi_2}\right]$ .

Phase difference $\phi_1-\phi_2$	Resultant amplitude $R$	With $A_1=A_2=A$
$0$ (in phase)	$A_1+A_2$	$2A$
$\dfrac{\pi}{2}$ ( $90^\circ$ )	$\sqrt{A_1^{2}+A_2^{2}}$	$\sqrt2\,A$
$\pi$ ( $180^\circ$ , out of phase)	$\lvert A_1-A_2\rvert$	$0$

$x=a\sin\omega t+b\cos\omega t$ is an S.H.M. of amplitude $\sqrt{a^2+b^2}$ — the $90^\circ$ case.

5.11 Energy of a Particle Performing S.H.M.

$E_k=\dfrac{1}{2}k(A^2-x^2)=\dfrac{1}{2}m\omega^2(A^2-x^2)$ ; with time it varies as $\cos^2\theta$ .
$E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}m\omega^2x^2$ , obtained as the work done against the restoring force; with time it varies as $\sin^2\theta$ .
$E=E_k+E_p=\dfrac{1}{2}m\omega^2A^2=\dfrac{1}{2}kA^2=\dfrac{1}{2}m(v_{max})^2=2\pi^2mn^2A^2$ — independent of $x$ and $t$ , so energy is conserved in S.H.M.
Total energy depends on the mass of the particle, the square of the amplitude, the square of the frequency and the force constant, and is inversely proportional to the square of the period.
At the mean position the energy is entirely kinetic; at the extreme positions it is entirely potential; elsewhere it is partly both.
$E_k=E_p=\dfrac{E}{2}$ at $x=\pm\dfrac{A}{\sqrt2}$ . At $x=\pm\dfrac{A}{2}$ the energy is 25% potential and 75% kinetic.
Graph (energy against displacement $x$ ): $E_p$ is an upward parabola with its minimum (zero) at $x=0$ ; $E_k$ is a downward parabola with its maximum $\dfrac{1}{2}kA^2$ at $x=0$ and zero at $x=\pm A$ ; the total energy is a horizontal straight line at $\dfrac{1}{2}kA^2$ . The two parabolas cross at $x=\pm\dfrac{A}{\sqrt2}$ .

5.12 Simple Pendulum

Ideal simple pendulum: a heavy particle suspended by a massless, inextensible, flexible string from a rigid support.
Practical simple pendulum: a small heavy (dense) sphere, the bob, suspended by a light and inextensible string from a rigid support.
Length of the pendulum: distance between the point of suspension and the centre of gravity of the bob.
In the displaced position, $mg$ resolves into $mg\cos\theta$ along the string (balanced by the tension $T'$ ) and $mg\sin\theta$ perpendicular to the string (the restoring force). For $\theta<10^\circ$ , $\sin\theta\approx\theta=\dfrac{x}{L}$ , so $F=-\dfrac{mg}{L}x\propto-x$ — hence linear S.H.M.
$T=2\pi\sqrt{\dfrac{L}{g}}$ , $n=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}$ .
Laws of the simple pendulum: (a) $T\propto\sqrt{L}$ , (b) $T\propto\dfrac{1}{\sqrt{g}}$ , (c) $T$ is independent of the mass of the bob, (d) $T$ is independent of the amplitude (for small amplitude).
Assumptions in the derivation: the amplitude is very small (at least 20 times smaller than the length), the length of the string is large, and the bob moves in a single vertical plane.
Experimentally $\dfrac{T_\theta}{T_0}$ is $1.02$ at $20^\circ$ , $1.05$ at $50^\circ$ , $1.10$ at $70^\circ$ and $1.18$ at $90^\circ$ — errors of 2%, 5%, 10% and 18%. Hence the recommended maximum angular amplitude is less than $20^\circ$ .
Second's pendulum: a simple pendulum whose period is two seconds. $L_s=\dfrac{g}{\pi^2}$ , about $0.993$ m at $g=9.8$ m/s $^2$ .

Conical pendulum	Simple pendulum
Trajectory and plane of motion of the bob is a horizontal circle	Trajectory and plane of motion is part of a vertical circle
K.E. and gravitational P.E. are constant	K.E. and gravitational P.E. are interconverted; their sum is conserved
The horizontal component of the tension provides the necessary centripetal force	The tangential component of the weight governs the energy conversions
$T=2\pi\sqrt{\dfrac{L\cos\theta}{g}}$	$T=2\pi\sqrt{\dfrac{L}{g}}$
The string makes a fixed angle with the horizontal and can never be horizontal	With a large amplitude the string can be horizontal at some instants

5.13 Angular S.H.M.

Angular S.H.M. is the oscillatory motion of a body in which the torque for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of the angular displacement.
A metallic disc attached centrally to a thin wire, twisted about the axis of the wire and released, performs angular (torsional) oscillations.
$\tau=-c\theta$ , where $c$ is the restoring torque per unit angular displacement; with $\tau=I\alpha$ , $I\dfrac{d^2\theta}{dt^2}+c\theta=0$ and $T=2\pi\sqrt{\dfrac{I}{c}}$ .
$T=\dfrac{2\pi}{\sqrt{\text{angular acceleration per unit angular displacement}}}$ — the angular counterpart of the linear formula.
Magnet in a uniform field: $\tau=\mu B\sin\theta\approx\mu B\theta$ for small $\theta$ ; $\alpha=-\left(\dfrac{\mu B}{I}\right)\theta$ , so the motion is angular S.H.M., with $T=2\pi\sqrt{\dfrac{I}{\mu B}}$ and $n=\dfrac{1}{2\pi}\sqrt{\dfrac{\mu B}{I}}$ .
The symbol $\mu$ is used for the magnetic dipole moment because $m$ is already taken by mass.

5.14 Damped Oscillations

If the amplitude of oscillations of an oscillator is reduced by the application of an external force, the oscillator and its motion are said to be damped.
Damped harmonic oscillations: periodic oscillations of gradually decreasing amplitude; the oscillator is called a damped harmonic oscillator.
Standard arrangement: a block on a spring carries a rod ending in a vane submerged in a liquid; the liquid exerts a drag force, and the mechanical energy of the block–spring system is transferred to thermal energy of the liquid and vane.
$F_d=-bv$ with $b$ the damping constant, so $m\dfrac{d^2x}{dt^2}+b\dfrac{dx}{dt}+kx=0$ , whose solution is $x=A\,e^{-bt/2m}\cos(\omega't+\phi)$ .
$\omega'=\sqrt{\dfrac{k}{m}-\left(\dfrac{b}{2m}\right)^2}$ , so damping increases the period (slows the motion) and decreases the amplitude.
Graph: the displacement–time curve is a cosine curve whose peaks lie on the two exponential envelopes $\pm A\,e^{-bt/2m}$ , closing in on the time axis.

5.15 Free Oscillations, Forced Oscillations and Resonance

Free oscillations: an object allowed to oscillate on its own oscillates with its natural frequency, e.g. a pendulum of length $l$ with $n=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$ .
Forced oscillations: oscillations produced by an external periodic force; the body then oscillates with the driver (forced) frequency, not with its own natural frequency.
Resonance: when the driver frequency equals the natural frequency of the driven body, the body absorbs maximum energy from the source and oscillates with maximum amplitude.
Standard experiment: four pendula A, B, C, D tied to a common string. A (the driver) and C are of equal length, B is shorter and D is longer. When A is set oscillating, B, C and D are forced to oscillate with A's frequency, but C — having the same natural frequency as A — oscillates with the largest amplitude and absorbs the most energy. It is in resonance with A.
Graph: a plot of the square of the amplitude of the driven body against its natural frequency has a sharp peak at the resonant frequency, falling away on either side.

Free oscillations	Forced oscillations
Body oscillates on its own after an initial disturbance	Body is driven by an external periodic force
Frequency is the natural frequency of the body	Frequency is the driver (forced) frequency
Amplitude is decided by the initial disturbance	Amplitude is largest when the driver frequency matches the natural frequency (resonance)

Traps that cost marks

Displacement is always measured from the mean position. "4 cm from the positive extreme position" with $A=10$ cm means $x=10-4=6$ cm.
"Path length" or "length of path" is $2A$ , not $A$ .
$T=\dfrac{2\pi}{\omega}$ , never $\dfrac{1}{\omega}$ . And $n=\dfrac{\omega}{2\pi}$ , never $\omega$ .
Convert cgs to SI before substituting: $25\times10^3$ dyne/cm is $25$ N/m, not $25\times10^3$ N/m.
The period of a simple pendulum does not depend on the mass of the bob or (for small amplitude) on the amplitude — do not "use" a given mass in the period formula.
A second's pendulum has period $2$ s (one second per swing, two swings per oscillation), not period $1$ s.
$\sin\theta\approx\theta$ needs $\theta$ in radian and $\theta$ small ( $<10^\circ$ ). State that condition — the derivation loses marks without it.
$E_k=E_p$ at $x=\pm\dfrac{A}{\sqrt2}$ , not at $x=\pm\dfrac{A}{2}$ .
Total energy goes as $A^2$ and as $n^2$ : doubling the amplitude quadruples the energy.
In the reference circle, $r$ of the circle is the amplitude $A$ of the S.H.M.
In the composition of two S.H.M.s, $\cos(\phi_1-\phi_2)$ is a cosine of the phase difference — it is the same for $+90^\circ$ and $-90^\circ$ , both giving $0$ .
Keep the calculator in radian mode when $\omega t$ is in radian, and in degree mode when the phase is quoted in degrees.
The mean position is where the net force is zero, not necessarily where the spring is unstretched (for a vertical spring the mean position is the stretched equilibrium position).

Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter. 4 have come up in two or more papers.7

1
Length of a second's pendulum
Define second's pendulum. Derive a formula for the length of second's pendulum.
Asked in: March 2022 (Q.30, 4 marks)
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Step 1
A simple pendulum whose period is two seconds is called second's pendulum.
Step 2
Time period of simple pendulum, $T=2\pi\sqrt{\dfrac{L}{g}}$
For a second's pendulum, $2=2\pi\sqrt{\dfrac{L_s}{g}}$
Where, $L_s$ is the length of second's pendulum, having period T = 2 s.
$\therefore L_s=\dfrac{g}{\pi^2}$
2
Differential equation of linear S.H.M. -- acceleration and velocity Repeats
State the differential equation of linear S.H.M. Hence, obtain the expression for: (i) acceleration (ii) velocity
Asked in: March 2018; March 2019; February 2023 (Q.21, 3 marks)
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Differential equation of linear S.H.M.: The differential equation of linear S.H.M. is given as,
$\dfrac{d^2x}{dt^2}+\omega^2x=0$
Step 1
Expression for acceleration in linear S.H.M:
a. From differential equation,
$\dfrac{d^2x}{dt^2}+\omega^2x=0$
$\dfrac{d^2x}{dt^2}=-\omega^2x\qquad....(1)$
b. But, linear acceleration is given by,
$\dfrac{d^2x}{dt^2}=a\qquad....(2)$
From equations (1) and (2),
$a=-\omega^2x\qquad....(3)$
Equation (3) gives acceleration in linear S.H.M.
Step 2
Expression for velocity in linear S.H.M:
a. From differential equation of linear S.H.M,
$\dfrac{d^2x}{dt^2}=-\omega^2x$
$\therefore \dfrac{d}{dt}\left(\dfrac{dx}{dt}\right)=-\omega^2x$
$\therefore \dfrac{dv}{dt}=-\omega^2x\qquad\left(\because\dfrac{dx}{dt}=v\right)$
$\therefore \dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=-\omega^2x$
$\therefore v\dfrac{dv}{dx}=-\omega^2x\qquad\left(\because\dfrac{dx}{dt}=v\right)$
$\therefore v\,dv=-\omega^2x\,dx\qquad....(4)$
b. Integrating both sides of equation (4),
$\displaystyle\int v\,dv=\int -\omega^2x\,dx$
$\dfrac{v^2}{2}=-\dfrac{\omega^2x^2}{2}+C\qquad....(5)$
where, C is the constant of integration.
c. At extreme position, x = ± A and v = 0.
Substituting these values in equation (5),
$0=-\dfrac{\omega^2A^2}{2}+C$
$\therefore C=\dfrac{\omega^2A^2}{2}\qquad....(6)$
d. Substituting equation (6) in equation (5),
$\dfrac{v^2}{2}=-\dfrac{\omega^2x^2}{2}+\dfrac{\omega^2A^2}{2}$
$\therefore v^2=\omega^2A^2-\omega^2x^2$
$\therefore v^2=\omega^2(A^2-x^2)$
$\therefore v=\pm\omega\sqrt{A^2-x^2}$
This is the required expression for velocity in linear S.H.M.
Step 3
Expression for displacement in linear S.H.M:
a. From differential equation of linear S.H.M, velocity is given by,
$v=\omega\sqrt{A^2-x^2}\qquad....(1)$
But, in linear motion, $v=\dfrac{dx}{dt}\qquad....(2)$
From equation (1) and (2),
$\dfrac{dx}{dt}=\omega\sqrt{A^2-x^2}$
$\therefore \dfrac{dx}{\sqrt{A^2-x^2}}=\omega\,dt\qquad....(3)$
b. Integrating both sides of equation (3),
$\displaystyle\int\dfrac{dx}{\sqrt{A^2-x^2}}=\int\omega\,dt$
$\therefore \sin^{-1}\left(\dfrac{x}{A}\right)=\omega t+\phi$
where, φ is constant of integration which depends upon initial condition (phase angle)
$\therefore \dfrac{x}{A}=\sin(\omega t+\phi)$
$\therefore x=A\sin(\omega t+\phi)$
This is required expression for displacement of a particle performing linear S.H.M. at time t.
3
Period of a simple pendulum performing S.H.M. Repeats
Obtain the expression for the period of a simple pendulum performing S.H.M.
Asked in: March 2013; October 2013; July 2016; July 2023 (Q.18, 3 marks)
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Definition: An ideal simple pendulum is a heavy particle suspended by a massless, inextensible, flexible string from a rigid support.
Step 1
Let 'm' be the mass of the bob and 'T' be the tension in the string. The pendulum remains in equilibrium in the position OA, with the centre of gravity of the bob, vertically below the point of suspension O.
Step 2
If now the pendulum is displaced through a small angle θ, called angular amplitude, and released, it begins to oscillate on either side of the mean (equilibrium) position in a single vertical plane.
Step 3
In the displaced position (extreme position), two forces are acting on the bob.
a. Force T′ due to tension in the string, directed along the string, towards the support.
b. Weight mg, in the vertically downward direction.
Step 4
At the extreme positions, there should not be any net force along the string.
Step 5
The component of mg can only balance the force due to tension.
Thus, weight mg is resolved into two components;
a. The component mg cosθ along the string, which is balanced by the tension T′.
b. The component mg sinθ perpendicular to the string is the restoring force acting on mass m tending to return it to the equilibrium position.
$\therefore$ Restoring force, $F=-mg\sin\theta$
Step 6
As θ is very small (θ < 10°),
$\sin\theta\approx\theta^c$
$\therefore F\approx -mg\theta$
From the figure,
For small angle, $\theta=\dfrac{x}{L}$
$\therefore F=-mg\dfrac{x}{L}\qquad....(1)$
As m, g and L are constant, $F\propto -x$
Step 7
Thus, for small displacement, the restoring force is directly proportional to the displacement and is oppositely directed. Hence the bob of a simple pendulum performs linear S.H.M. for small amplitudes.
Step 8
The period T of oscillation of a pendulum is given by,
$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{\sqrt{\text{acceleration per unit displacement}}}$
Using equation (1),
$F=-mg\dfrac{x}{L}$
$\therefore ma=-mg\dfrac{x}{L}\qquad(\because F=ma)$
$\therefore a=-g\dfrac{x}{L}$
$\therefore \dfrac{a}{x}=-\dfrac{g}{L}=\dfrac{g}{L}$ (in magnitude)
Substituting in the expression for T,
$T=2\pi\sqrt{\dfrac{L}{g}}$
This gives the expression for the time period of a simple pendulum.
Step 9
Period of simple pendulum depends on the length of the pendulum and acceleration due to gravity.
4
Composition of two SHMs of the same period along the same path (resultant amplitude)
Discuss analytically the composition of two linear SHMs having the same period and along the same path. Obtain the expression for resultant amplitude.
Asked in: July 2022 (Q.27, 4 marks)
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[From 5.10, Q1 (except Initial Phase)]
Step 1
Consider a particle simultaneously subjected to two S.H.M.s having the same period and along same path (let it be along the x-axis), but of different amplitudes and initial phases. The resultant displacement at any instant is equal to the vector sum of its displacements due to both the S.H.M.s at that instant.
Step 2
Let the two linear S.H.M's be given by equations,
$x_1=A_1\sin(\omega t+\phi_1)\qquad....(1)$
$x_2=A_2\sin(\omega t+\phi_2)\qquad....(2)$
where $A_1$ , $A_2$ are amplitudes; $\phi_1$ , $\phi_2$ are initial phase angles and $x_1$ , $x_2$ are the displacement of two S.H.M's in time 't'. ω is same for both S.H.M's.
Step 3
The resultant displacement of the two S.H.M's is given by, $x=x_1+x_2\qquad....(3)$
Step 4
Using equations (1) and (2), equation (3) can be written as,
$x=A_1\sin(\omega t+\phi_1)+A_2\sin(\omega t+\phi_2)$
$=A_1[\sin\omega t\cos\phi_1+\cos\omega t\sin\phi_1]+A_2[\sin\omega t\cos\phi_2+\cos\omega t\sin\phi_2]$
$=A_1\sin\omega t\cos\phi_1+A_1\cos\omega t\sin\phi_1+A_2\sin\omega t\cos\phi_2+A_2\cos\omega t\sin\phi_2$
$=[A_1\sin\omega t\cos\phi_1+A_2\sin\omega t\cos\phi_2]+[A_1\cos\omega t\sin\phi_1+A_2\cos\omega t\sin\phi_2]$
$\therefore x=\sin\omega t[A_1\cos\phi_1+A_2\cos\phi_2]+\cos\omega t[A_1\sin\phi_1+A_2\sin\phi_2]\qquad....(4)$
Step 5
As $A_1$ , $A_2$ , $\phi_1$ and $\phi_2$ are constants, we can combine them in terms of another convenient constants R and δ as
$A_1\cos\phi_1+A_2\cos\phi_2=R\cos\delta\qquad....(5)$
and $A_1\sin\phi_1+A_2\sin\phi_2=R\sin\delta\qquad....(6)$
Step 6
Using equations (5) and (6), equation (4) can be written as,
$x=\sin\omega t.\,R\cos\delta+\cos\omega t.\,R\sin\delta=R[\sin\omega t\cos\delta+\cos\omega t\sin\delta]$
$\therefore x=R\sin(\omega t+\delta)\qquad....(7)$
Equation (7) is the equation of an S.H.M. of the same angular frequency (hence, the same period) but of amplitude R and initial phase δ. It shows that the combination (superposition) of two linear S.H.M.s of the same period and occurring along the same path is also an S.H.M.
Resultant Amplitude:
Resultant amplitude is,
$R=\sqrt{(R\sin\delta)^2+(R\cos\delta)^2}$
Squaring and adding equations (5) and (6) we get,
$(A_1\cos\phi_1+A_2\cos\phi_2)^2+(A_1\sin\phi_1+A_2\sin\phi_2)^2=R^2\cos^2\delta+R^2\sin^2\delta$
$\therefore A_1^2\cos^2\phi_1+A_2^2\cos^2\phi_2+2A_1A_2\cos\phi_1\cos\phi_2+A_1^2\sin^2\phi_1+A_2^2\sin^2\phi_2+2A_1A_2\sin\phi_1\sin\phi_2=R^2(\cos^2\delta+\sin^2\delta)$
$\therefore A_1^2(\cos^2\phi_1+\sin^2\phi_1)+A_2^2(\cos^2\phi_2+\sin^2\phi_2)+2A_1A_2(\cos\phi_1\cos\phi_2+\sin\phi_1\sin\phi_2)=R^2$
$\therefore A_1^2+A_2^2+2A_1A_2\cos(\phi_1-\phi_2)=R^2$
$\therefore R=\pm\sqrt{A_1^2+A_2^2+2A_1A_2\cos(\phi_1-\phi_2)}\qquad....(8)$
Equation (8) represents resultant amplitude of two S.H.M's.
Initial Phase:
Dividing equation (6) by (5), we get,
$\dfrac{A_1\sin\phi_1+A_2\sin\phi_2}{A_1\cos\phi_1+A_2\cos\phi_2}=\dfrac{R\sin\delta}{R\cos\delta}=\tan\delta$
$\therefore \delta=\tan^{-1}\left[\dfrac{A_1\sin\phi_1+A_2\sin\phi_2}{A_1\cos\phi_1+A_2\cos\phi_2}\right]\qquad....(9)$
Equation (9) represents resultant or initial phase of two S.H.M's.
Step 7
a. When phase difference is zero radians:
Two S.H.M.s are in phase, i.e., $\phi_1-\phi_2=0$
$\therefore \cos(\phi_1-\phi_2)=1$
$\therefore$ From equation (8),
$R=\sqrt{A_1^2+A_2^2+2A_1A_2}=\sqrt{(A_1+A_2)^2}$
$\therefore R=\pm(A_1+A_2)$
b. When phase difference is $\dfrac{\pi}{2}$ radians:
Two S.H.M.s are out of phase, i.e., $\phi_1-\phi_2=\dfrac{\pi}{2}$ ,
$\therefore \cos(\phi_1-\phi_2)=0$
From equation (8),
$R=\sqrt{A_1^2+A_2^2}$
5
Differential equation of linear S.H.M. (general form) Repeats
Obtain the differential equation of linear simple harmonic motion.
Asked in: March 2017; February 2024 (Q.6, 2 marks)
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Step 1
In a linear S.H.M., the force is directed towards the mean position and its magnitude is directly proportional to the displacement of the body from mean position.
$\therefore f\propto -x$
$\therefore f=-kx\qquad....(1)$
where, k is force constant and x is displacement from the mean position.
Step 2
According to Newton's second law of motion,
$f=ma\qquad....(2)$
From equations (1) and (2),
$ma=-kx\qquad....(3)$
Step 3
The velocity of the particle is given by,
$v=\dfrac{dx}{dt}$ ,
$\therefore$ Acceleration, $a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}\qquad....(4)$
Step 4
Substituting equation (4) in equation (3),
$m\dfrac{d^2x}{dt^2}=-kx$
$\therefore \dfrac{d^2x}{dt^2}+\dfrac{k}{m}x=0$
Step 5
Substituting $\dfrac{k}{m}=\omega^2$ , where ω is the angular frequency,
$\therefore \dfrac{d^2x}{dt^2}+\omega^2x=0$
This is the differential equation of linear S.H.M.
6
Potential energy of a particle performing S.H.M. Repeats
Obtain an expression for the potential energy of a particle performing S.H.M. Hence evaluate the potential energy (i) at the mean position and (ii) at the extreme position.
Asked in: March 2015; March 2019
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Expression for potential energy:
Step 1
The restoring force acting on the particle at point P is given by, $f=-kx$
where k is the force constant.
Step 2
Suppose that the particle is displaced further by an infinitesimal displacement 'dx' against the restoring force 'f'.
Step 3
The external work done (dW) during this displacement is
$dW=f(-dx)=-kx(-dx)=kx\,dx$
Step 4
The total work done on the particle to displace it from O to P is given by,
$W=\int_0^x dW=\int_0^x kx\,dx=\dfrac{1}{2}kx^2$
Step 5
This work done is stored as the potential energy (P.E.) $E_p$ of the particle at displacement x.
$\therefore E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}m\omega^2x^2$
Step 6
At time t,
$\therefore E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}kA^2\sin^2(\omega t+\phi)$
$=\dfrac{1}{2}mA^2\omega^2\cos^2(\omega t+\phi)$
Thus, with time, it varies as $\sin^2\theta$ .
Step 7
At the mean position, x = 0
$\therefore$ Potential energy, $(E_p)_{min}=0$
Step 8
At the extreme positions, x = ± A
$\therefore$ Potential energy, $(E_p)_{max}=\dfrac{1}{2}m\omega^2A^2$
7
Law of conservation of energy for a particle performing S.H.M.
Prove the law of conservation of energy for a particle performing simple harmonic motion. Hence graphically show the variation of kinetic energy and potential energy with instantaneous displacement.
Asked in: March 2017
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Expression for potential energy:
Step 1
The restoring force acting on the particle at point P is given by, $f=-kx$
where k is the force constant.
Step 2
Suppose that the particle is displaced further by an infinitesimal displacement 'dx' against the restoring force 'f'.
Step 3
The external work done (dW) during this displacement is
$dW=f(-dx)=-kx(-dx)=kx\,dx$
Step 4
The total work done on the particle to displace it from O to P is given by,
$W=\int_0^x dW=\int_0^x kx\,dx=\dfrac{1}{2}kx^2$
Step 5
This work done is stored as the potential energy (P.E.) $E_p$ of the particle at displacement x.
$\therefore E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}m\omega^2x^2$
Step 6
At time t,
$\therefore E_p=\dfrac{1}{2}kx^2=\dfrac{1}{2}kA^2\sin^2(\omega t+\phi)$
$=\dfrac{1}{2}mA^2\omega^2\cos^2(\omega t+\phi)$
Thus, with time, it varies as $\sin^2\theta$ .
Step 7
At the mean position, x = 0
$\therefore$ Potential energy, $(E_p)_{min}=0$
Step 8
At the extreme positions, x = ± A
$\therefore$ Potential energy, $(E_p)_{max}=\dfrac{1}{2}m\omega^2A^2$
Step 1
Expression for kinetic energy:
a. Consider a particle of mass m, performing a linear S.H.M. along the path MN about the mean position O as shown in figure.
b. At a given instant, let the particle be at P, at a distance x from O.
c. Velocity of the particle in S.H.M. is given as $v=\omega\sqrt{A^2-x^2}=A\omega\cos(\omega t+\phi)$
where x is the displacement of the particle performing S.H.M. and A is the amplitude of S.H.M.
d. Thus, the kinetic energy,
$E_k=\dfrac{1}{2}mv^2$
$=\dfrac{1}{2}m\omega^2(A^2-x^2)\qquad....(1)$
$=\dfrac{1}{2}k(A^2-x^2)$
This is the kinetic energy at displacement x.
e. Also, at time t, kinetic energy is,
$E_k=\dfrac{1}{2}mv^2=\dfrac{1}{2}mA^2\omega^2\cos^2(\omega t+\phi)$
$=\dfrac{1}{2}kA^2\cos^2(\omega t+\phi)$
Thus, with time, it varies as $\cos^2\theta$ .
iii. Expression for total energy:
a. The total energy of the particle is the sum of its kinetic energy and potential energy.
$\therefore E=E_k+E_p$
b. Using equation (1) and equation (2), we get
$E=\dfrac{1}{2}m\omega^2(A^2-x^2)+\dfrac{1}{2}m\omega^2x^2$
$E=\dfrac{1}{2}m\omega^2A^2=\dfrac{1}{2}kA^2=\dfrac{1}{2}m(v_{max})^2\qquad....(3)$
This expression gives the total energy of the particle at point P.
iv. As m, ω and A are constant, the total energy of the particle at any point P is constant (independent of x and t). In other words, the energy is conserved in S.H.M.

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