JEE Main & Advanced

Oscillations and Waves

Oscillations and Waves for JEE Main & Advanced

1
Module 1

Simple Harmonic Motion

SHM Equations, Energy and PhaseTopic 1

SHM is periodic motion in which restoring force is proportional to displacement and directed toward equilibrium: $F = -kx$, giving $a = -\omega^2 x$ where $\omega = \sqrt{k/m}$.

Solution: $x(t) = A\sin(\omega t + \phi)$, where $A$ = amplitude, $\omega$ = angular frequency, $\phi$ = phase constant.

Key Quantities:

QuantityFormula
Angular frequency$\omega = 2\pi/T = 2\pi f$
Period$T = 2\pi\sqrt{m/k}$
Velocity$v(t) = A\omega\cos(\omega t + \phi)$; $v_{\max} = A\omega$
Acceleration$a(t) = -A\omega^2\sin(\omega t + \phi)$; $a_{\max} = A\omega^2$
$v$ in terms of $x$$v = \omega\sqrt{A^2 - x^2}$

Energy in SHM:

EnergyFormula
Kinetic$K = \dfrac{1}{2}m\omega^2(A^2 - x^2)$
Potential$U = \dfrac{1}{2}m\omega^2 x^2 = \dfrac{1}{2}kx^2$
Total$E = K + U = \dfrac{1}{2}m\omega^2 A^2 = \dfrac{1}{2}kA^2$

Energy is constant; $K$ and $U$ oscillate at twice the SHM frequency.

Phase relation: Velocity leads displacement by $\pi/2$; acceleration leads displacement by $\pi$.

Worked Examples
1

A particle in SHM has amplitude $5$ cm and period $2$ s. Find (a) maximum velocity, (b) maximum acceleration, (c) velocity at $x = 3$ cm.

Show solution

$\omega = 2\pi/T = \pi$ rad/s, $A = 0.05$ m. (a) $v_{\max} = A\omega = 0.05\pi \approx 0.157$ m/s. (b) $a_{\max} = A\omega^2 = 0.05\pi^2 \approx 0.493$ m/s². (c) $v = \omega\sqrt{A^2-x^2} = \pi\sqrt{0.05^2 - 0.03^2} = \pi\sqrt{0.0016} = 0.04\pi \approx 0.126$ m/s.

Final Answer: $v_{\max} \approx 0.157$ m/s; $a_{\max} \approx 0.493$ m/s²; $v(x=3\,\text{cm}) \approx 0.126$ m/s.

2

A particle of mass $0.2$ kg executes SHM with amplitude $0.1$ m and frequency $5$ Hz. Find total energy.

Show solution

$\omega = 2\pi f = 10\pi$ rad/s. $$E = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}(0.2)(10\pi)^2(0.1)^2$$ $$= 0.1 \times 100\pi^2 \times 0.01 = 0.1\pi^2 \approx 0.987 \text{ J}$$

Final Answer: $E \approx 0.987$ J.

✎ Self-Check — 5 questions0 / 5
Q1.

In SHM, the acceleration is proportional to:

Q2.

The total energy of a particle in SHM:

Q3.

A particle in SHM has period $4$ s. The time to reach half amplitude from mean position:

Q4.

The ratio of kinetic to potential energy at $x = A/2$ in SHM:

Q5.

Phase difference between displacement and acceleration in SHM:

Pendulums and Spring SystemsTopic 2

Simple Pendulum: Mass $m$ on light inextensible string of length $L$. For small amplitudes: $$T = 2\pi\sqrt{L/g}$$ Independent of mass and amplitude (isochronous).

Physical Pendulum: Rigid body suspended from pivot at distance $d$ from CM, moment of inertia $I$ about pivot: $$T = 2\pi\sqrt{\frac{I}{mgd}}$$

Torsional Pendulum: $T = 2\pi\sqrt{I/C}$, where $C$ = torsional constant.

Spring-Mass System: $T = 2\pi\sqrt{m/k}$.

Spring Combinations:

ConfigurationEquivalent $k$
Series ($k_1, k_2$)$k_{\text{eq}} = \dfrac{k_1 k_2}{k_1+k_2}$
Parallel ($k_1, k_2$)$k_{\text{eq}} = k_1 + k_2$
Spring of constant $k$ cut into $n$ equal partseach part has $k' = nk$

Liquid Column in U-Tube: $T = 2\pi\sqrt{L/(2g)}$, where $L$ = total length of liquid column.

Pendulum in accelerated frame: Effective $g$ replaces $g$. E.g., in lift with upward acceleration $a$: $g_{\text{eff}} = g+a$, period decreases.

Damped SHM: $x = A_0 e^{-bt/2m}\cos(\omega' t + \phi)$, $\omega' = \sqrt{\omega_0^2 - (b/2m)^2}$. Energy decays as $e^{-bt/m}$.

Forced/Resonance: At driving frequency = natural frequency, amplitude is maximum (sharp peak for low damping).

Worked Examples
1

A simple pendulum has length $1$ m and is taken to the Moon where $g = 1.6$ m/s². Find new period.

Show solution

$$T = 2\pi\sqrt{L/g} = 2\pi\sqrt{1/1.6} = 2\pi\sqrt{0.625} = 2\pi(0.79) \approx 4.97 \text{ s}$$

Final Answer: $T \approx 4.97$ s.

2

A mass $0.5$ kg is suspended from two springs of $k_1 = 100$ N/m and $k_2 = 200$ N/m connected in parallel. Find the period.

Show solution

$k_{\text{eq}} = 100 + 200 = 300$ N/m. $$T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.5/300} = 2\pi\sqrt{1/600} \approx 0.257 \text{ s}$$

Final Answer: $T \approx 0.26$ s.

✎ Self-Check — 5 questions0 / 5
Q1.

Period of a simple pendulum depends on:

Q2.

A pendulum in a lift falling freely has period:

Q3.

A spring of $k$ is cut into $4$ equal parts. Force constant of each:

Q4.

The period of a pendulum is $2$ s on Earth. At a place where $g = g/4$:

Q5.

At resonance, the amplitude of forced oscillation is:

2
Module 2

Wave Motion

Wave Equation, Travelling Waves and SuperpositionTopic 1

Wave = disturbance propagating through a medium (mechanical waves need a medium; EM waves do not). Two main types:

  • Transverse: Particle motion perpendicular to wave direction (e.g., on a string, EM waves)
  • Longitudinal: Particle motion along wave direction (e.g., sound in fluids)

Wave Quantities:

  • Wavelength $\lambda$, frequency $f$, period $T = 1/f$
  • Wave speed: $v = f\lambda = \omega/k$
  • Angular frequency $\omega = 2\pi f$
  • Wave number $k = 2\pi/\lambda$

Travelling wave: $y(x,t) = A\sin(\omega t - kx + \phi)$ (moving in $+x$).

Wave equation (general): $\dfrac{\partial^2 y}{\partial t^2} = v^2\dfrac{\partial^2 y}{\partial x^2}$.

Wave Speeds:

MediumFormula
Transverse on string$v = \sqrt{T/\mu}$, $\mu$ = mass per unit length
Longitudinal in solid$v = \sqrt{Y/\rho}$
In liquid/gas$v = \sqrt{B/\rho}$
Sound in gas (Newton-Laplace)$v = \sqrt{\gamma P/\rho} = \sqrt{\gamma RT/M}$

Sound speed depends on temperature: $v \propto \sqrt{T}$ (in Kelvin).

Intensity: Power per unit area, $I = \dfrac{1}{2}\rho v\omega^2 A^2$.

For point source: $I \propto 1/r^2$.

Sound Level (decibel): $\beta = 10\log_{10}(I/I_0)$ dB, $I_0 = 10^{-12}$ W/m².

Superposition Principle: If two waves overlap, resultant displacement = vector sum. Leads to interference, beats, and standing waves.

Worked Examples
1

A string of mass per unit length $0.1$ kg/m is stretched with tension $40$ N. Find speed of transverse wave.

Show solution

$$v = \sqrt{T/\mu} = \sqrt{40/0.1} = \sqrt{400} = 20 \text{ m/s}$$

Final Answer: $v = 20$ m/s.

2

The equation of a wave is $y = 0.05 \sin(20t - 4x)$ m. Find amplitude, wavelength, frequency, speed.

Show solution

Compare with $y = A\sin(\omega t - kx)$:

  • $A = 0.05$ m
  • $\omega = 20$ rad/s, so $f = 20/2\pi \approx 3.18$ Hz
  • $k = 4$ rad/m, so $\lambda = 2\pi/4 = \pi/2 \approx 1.57$ m
  • $v = \omega/k = 20/4 = 5$ m/s

Final Answer: $A = 0.05$ m, $\lambda \approx 1.57$ m, $f \approx 3.18$ Hz, $v = 5$ m/s.

✎ Self-Check — 5 questions0 / 5
Q1.

The velocity of sound in air increases with:

Q2.

In a transverse wave, particle motion is:

Q3.

Wave speed on string $= \sqrt{T/\mu}$. If $T$ doubles, speed becomes:

Q4.

The unit of intensity of sound is:

Q5.

The sound level for $I = 10^{-6}$ W/m² ($I_0 = 10^{-12}$):

Standing Waves, Beats and Doppler EffectTopic 2

Standing Wave: Superposition of two identical waves travelling in opposite directions: $$y = 2A\sin(kx)\cos(\omega t)$$ Nodes at $kx = n\pi$ (zero amplitude); antinodes at $kx = (2n+1)\pi/2$ (max). Distance between consecutive nodes (or antinodes) = $\lambda/2$.

Vibrating String (both ends fixed): Allowed wavelengths $\lambda_n = 2L/n$, frequencies $f_n = nv/(2L)$, $n = 1, 2, 3, \ldots$ Fundamental: $f_1 = v/(2L)$.

Air Column:

ConfigurationFundamental $f_1$Harmonics present
Closed pipe (one end closed)$v/(4L)$Only odd: $f_1, 3f_1, 5f_1, \ldots$
Open pipe (both ends open)$v/(2L)$All: $f_1, 2f_1, 3f_1, \ldots$

End correction for resonance tube: $L_{\text{eff}} = L + 0.6r$ (single end open) or $L + 1.2r$ (both ends open).

Beats: Two slightly different frequency waves give resultant whose amplitude varies. Beat frequency: $$f_{\text{beat}} = |f_1 - f_2|$$ Only audible if difference $\leq 10$ Hz.

Doppler Effect (sound, source frequency $f_0$, speed of sound $v$, source velocity $v_s$ toward observer, observer velocity $v_o$ toward source): $$f' = f_0 \cdot \frac{v + v_o}{v - v_s}$$

Sign convention: velocities are positive toward each other. Examples:

  • Source toward stationary observer: $f' = f_0 \cdot v/(v - v_s)$ — higher frequency
  • Observer toward stationary source: $f' = f_0 \cdot (v + v_o)/v$ — higher frequency
  • Both moving apart: $f' = f_0 \cdot (v - v_o)/(v + v_s)$ — lower frequency

For light (EM): relativistic Doppler $f' = f_0\sqrt{(c+v)/(c-v)}$ approaching.

Worked Examples
1

A string of length $1$ m fixed at both ends has linear mass density $10^{-3}$ kg/m and tension $40$ N. Find the fundamental frequency.

Show solution

$v = \sqrt{T/\mu} = \sqrt{40/10^{-3}} = \sqrt{40000} = 200$ m/s. $$f_1 = v/(2L) = 200/2 = 100 \text{ Hz}$$

Final Answer: $f_1 = 100$ Hz.

2

A car moves at $20$ m/s toward a stationary observer, sounding a horn of $400$ Hz. Find the frequency heard. Speed of sound $= 340$ m/s.

Show solution

$$f' = f_0 \cdot \frac{v}{v - v_s} = 400 \cdot \frac{340}{340 - 20} = 400 \cdot \frac{340}{320} = 425 \text{ Hz}$$

Final Answer: $f' = 425$ Hz.

✎ Self-Check — 5 questions0 / 5
Q1.

Distance between consecutive nodes in standing wave:

Q2.

Fundamental frequency of closed pipe of length $L$:

Q3.

Two waves of frequencies $200$ Hz and $206$ Hz produce beats of frequency:

Q4.

Doppler effect is observed when:

Q5.

A closed pipe and open pipe of same length have fundamental frequency ratio:

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