Oscillations and Waves Full Chapter for NEET 2026: Important Questions and Quick Revision Notes by PW

Oscillations and Waves combines concepts of force, equilibrium, energy, wave motion and resonance. Many problems become easier when you first identify the equilibrium position, determine the restoring force produced by a small displacement and then compare it with the standard SHM relation.

To strengthen preparation for this chapter, PW’s NEET learning approach focuses on understanding concepts, applying standard equations, solving numerical problems and revising important relationships. Regular practice can help you recognise the physical conditions in oscillation and wave problems and select the appropriate formula.

Simple Harmonic Motion from Restoring Force

A body performs Simple Harmonic Motion (SHM) when its restoring force is directly proportional to displacement and acts opposite to the direction of displacement.

The restoring force is:

F = -mω²x

A useful method for solving small-oscillation problems is:

1. Identify the equilibrium position.

2. Displace the body through a small distance x.

3. Calculate only the additional force produced by this displacement.

4. Equate the restoring force to -mω²x.

5. Find ω and then use f = ω/(2π).

Memory Tip: For small oscillation problems, follow the sequence:

Equilibrium position → Small displacement → Additional force → F = -mω²x

Small Oscillations of a Partially Submerged Sphere

Consider a solid sphere of radius R floating in a liquid with half its volume submerged.

At equilibrium:

Weight = Buoyant Force

If the liquid density is ρ and the sphere density is ρ_b, then:

ρ_b Vg = ρ(V/2)g

Therefore:

ρ_b = ρ/2

When the sphere is pushed downward through a small distance x, the additional submerged volume is:

Additional Volume = πR²x

Therefore, the additional buoyant force is:

Additional Force = ρπR²gx

This additional force acts as the restoring force after a small displacement from equilibrium.

The mass of the sphere is:

m = (ρ/2)(4/3 πR³)

Comparing the restoring force with -mω²x gives:

ω² = 3g/(2R)

Hence:

ω = √[3g/(2R)]

and

f = (1/2π)√[3g/(2R)]

Vertically Oscillating Cylinder in a Liquid

Consider a cylinder of mass m and cross-sectional area A attached to a spring of force constant k and partially submerged in a liquid of density ρ.

After a downward displacement x:

• Additional spring force = -kx

• Additional buoyant force = ρAgx

Therefore, the total restoring force is:

F = -kx + ρAgx

or

F = -(k + ρAg)x

Comparing this with -mω²x:

ω² = (k + ρAg)/m

Therefore:

f = (1/2π)√[(k + ρAg)/m]

The additional spring force and additional buoyant force together provide the restoring force.

SHM of a Mass Attached to an Elastic Wire

For a wire of length L, cross-sectional area A and Young's modulus Y, the extension relation is:

x = FL/(AY)

Therefore, the elastic restoring force is:

F = -(AY/L)x

Comparing this with -mω²x:

mω² = AY/L

Hence:

Y = mω²L/A

Using the numerical values:

Y = 4 × 10⁹ N/m²

Therefore, if:

Y = n × 10⁹ N/m²

then:

n = 4

The block, not the wire, performs SHM. The important step is to identify the restoring force and equate it to ma.

Superposition of Three SHMs

Consider two SHMs:

x₁ = a sin(ωt)

x₂ = a sin(ωt + 2π/3)

A third SHM is added so that the resultant displacement becomes zero.

Equal-amplitude SHMs can be represented using phasor vectors. For the resultant to be zero, the three vectors must form an equilateral arrangement.

Therefore:

• All three amplitudes are equal.

• The phase difference between successive SHMs is 120°.

Hence:

b = a

The phase of the third SHM is:

φ = 4π/3

Memory Tip: Three equal SHM vectors produce zero resultant when they are separated by equal phase angles of 120°.

Shear Oscillations of a Cubical Block

For a cubical block of side l and modulus of rigidity η:

Shear stress = F/l²

Shear strain = x/l

Using:

η = Stress/Strain

we get:

η = F/(lx)

Therefore, the restoring force is:

F = -ηlx

Comparing this with -mω²x:

ω² = ηl/m

Thus, the time period is:

T = 2π√[m/(ηl)]

Dimensional analysis also confirms that m/(ηl) has the dimensions of time squared. Dimensional analysis can therefore help eliminate unsuitable options in numerical questions.

Interpretation of a Wave Equation

Consider a wave equation of the form:

y = a sin(kx - ωt)

The physical meaning of y depends on the type of wave.

Sound travels through compressions and rarefactions, causing pressure and density to vary. Therefore, the physical meaning of y depends on the type of wave being considered.

For a wave equation:

Wave speed = ω/k

For example, if:

ω = 100π and k = 0.5π

then:

Wave speed = 100π/(0.5π) = 200 m/s

Standing Waves, Resonance and Beats

Standing waves are produced when two identical waves of equal frequency and amplitude travel in opposite directions and superpose. Reflection from a fixed or free end can provide the oppositely travelling wave.

Memory Tip: Standing wave = incident wave + reflected wave travelling in the opposite direction.

At a fixed wall:

• A node forms at the wall.

• The nearest antinode is at a distance of λ/4 from the wall.

For sound with frequency 660 Hz and speed 330 m/s:

λ = v/f

λ = 330/660 = 0.5 m

Therefore, the distance of the nearest antinode from the fixed wall is:

d = λ/4 = 0.125 m

Beat Frequency

The beat frequency is:

f_beat = |f₁ - f₂|

If the angular frequencies are 100π rad/s and 92π rad/s, the corresponding frequencies are:

f₁ = 50 Hz

f₂ = 46 Hz

Therefore:

f_beat = |50 - 46| = 4 Hz

The observer hears maximum intensity four times per second.

Resonance-Tube Experiment

For the first and second resonance lengths:

L₁ + e = λ/4

L₂ + e = 3λ/4

Subtracting the two equations eliminates the end correction:

λ = 2(L₂ - L₁)

For:

L₁ = 30.7 cm

L₂ = 63.2 cm

we get:

λ = 2(63.2 - 30.7)

λ = 65 cm = 0.65 m

Using:

v = fλ

with f = 512 Hz:

v = 512 × 0.65 = 332.8 m/s

Compared with 330 m/s, the difference is:

2.8 m/s = 280 cm/s

At first resonance, the effective length is λ/4. The actual air-column length is slightly shorter because of end correction.

Fundamental Modes of Standing-Wave Systems

The fundamental wavelength differs according to the type of standing-wave system.

An organ pipe supports longitudinal waves, while a stretched string supports transverse waves.

Important Frequency and Wave-Speed Relations

For a stretched string:

Wave speed = √(Tension/Linear mass density)

For the same string, wave speed is proportional to the square root of tension.

If the extension increases from x to 3x/2, the tension also becomes 3T/2. Therefore:

New speed = Old speed × √(3/2)

For a body suspended partly in water, buoyancy reduces the tension. Since frequency is proportional to the square root of tension, the new frequency decreases.

Charged Spring-Mass System

For a spring-mass system placed in a constant electric field, the electric force shifts the equilibrium position.

The new equilibrium displacement is:

x₀ = qE/k

The constant electric force does not change the frequency.

Therefore, the system continues to perform SHM with the same frequency but with a shifted mean position.

Oscillators with Springs, Pendulums and Gas

Different oscillating systems have different expressions for their angular frequency and time period.

Torsional Pendulum

For a torsional pendulum:

ω = √(C/I)

where:

• C = torsional constant

• I = moment of inertia

Springs in Series

For two springs in series:

Equivalent spring constant = k₁k₂/(k₁ + k₂)

Elastic Wire as a Spring

A wire behaves like a spring with an equivalent spring constant:

Equivalent wire constant = AY/L

where A is the cross-sectional area, Y is Young's modulus and L is the length of the wire.

Pendulum with an Accelerating Support

If the support of a pendulum accelerates upward:

Effective gravity = g + a

Therefore, upward acceleration decreases the time period.

Gas-Piston Oscillator

For an isolated gas-piston system, the process is adiabatic.

The equilibrium gas pressure is:

p = p₀ + mg/A

Using the adiabatic relation and the volume change:

dV = Ax

the resulting pressure change produces a restoring force proportional to displacement. Therefore, the system performs SHM.

Important Questions-

Important Questions 

Q1.A tuning fork of frequency 512 Hz is used in a resonance tube experiment. The water level is 30.7 cm at the first resonance and 63.2 cm at the second. What is the error in calculating the velocity of sound?  (IIT JEE 2005, Resonance Column) 

Answer: 280 cm/s
Explanation: λ = 2(L₂ − L₁) = 2 × 32.5 = 65 cm. So v = fλ = 512 × 0.65 = 332.8 m/s. Taking 330 m/s as the standard value, the error is 2.8 m/s = 280 cm/s.

Q2.  A solid sphere of radius R floats in a liquid of density ρ with half its volume submerged. If it is slightly pushed down and released, it performs SHM. Find the frequency of oscillation.

Answer: f = (1/2π)√(3g/2R)
Explanation: In equilibrium, the body's density is ρ/2. When pushed down by x, the extra buoyant force ρ(πR²x)g acts as the restoring force. Equating this to mω²x with m = (ρ/2)(4/3)πR³ gives ω² = 3g/2R.

Q3. A point mass is subjected to two simultaneous displacements along x: x₁ = a sin ωt and x₂ = a sin(ωt + 2π/3). Adding a third displacement x₃ = b sin(ωt + φ) brings the mass to rest. Find b and φ.(IIT JEE 2011, Superposition of SHMs)

Answer: b = a, φ = 4π/3
Explanation: Treat the displacements as vectors. Three vectors sum to zero only if they have equal magnitude and are 120° apart. The vectors are at 0° and 120°, so the third is at 240° = 4π/3, with magnitude a.

Q4. A uniform cylinder of length l, mass m and cross-sectional area A hangs vertically from a massless spring of constant k. At equilibrium it is half-submerged in a liquid of density ρ. It is given a small downward push and released. Find the frequency of oscillation.(IIT JEE 1990, Cylinder on Spring in Liquid)

Answer: f = (1/2π)√((k + Aρg)/m)
Explanation: For a downward displacement x, the extra spring force is kx and the extra buoyant force is ρAxg. Together they act as the restoring force, so mω² = k + Aρg.

Q5. A mass is suspended from a wire of negligible mass and length 1 m. When pulled slightly down and released, it performs SHM with angular frequency 140 rad/s. If Young's modulus of the wire is N × 10⁹ N/m², find N.(IIT JEE 2010, Mass on a Wire)

Answer: N = 4
Explanation: The restoring force is F = (AY/L)x, so mω² = AY/L. The teacher's working uses m = 0.1 kg and A = 4.9 × 10⁻⁷ m² (these values were read out somewhat unclearly). Then Y = mω²L/A = (0.1 × 140² × 1)/(4.9 × 10⁻⁷) = 4 × 10⁹ N/m².

Q6. The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment the displacement is y = A sin(πx/L) cos ωt, with energy E₁. In another it is y = A sin(2πx/L) cos 2ωt, with energy E₂. How are E₁ and E₂ related? (IIT JEE 2001, Stretched Wire)

Answer: E₂ = 4E₁
Explanation: Energy ∝ f²A²v for the same wire (same μ). The amplitude A is the same. Frequency doubles, and wave speed v = ω/k stays the same because k also doubles. So E ∝ f² gives 4 times the energy.

Q7. Sound waves of frequency 660 Hz fall normally on a perfectly reflecting wall. What is the shortest distance from the wall at which the air particles have maximum amplitude of vibration? (Take v = 330 m/s.) (IIT JEE 1984, Reflection from a Wall)

Answer: 0.125 m
Explanation: Incident and reflected waves form a standing wave with a displacement node at the wall. The nearest antinode (maximum amplitude) is λ/4 away. λ = v/f = 330/660 = 0.5 m, so the distance is 0.5/4 = 0.125 m.

Q8.Two plane harmonic sound waves have angular frequencies ω₁ = 100π and ω₂ = 92π (all parameters in MKS units). How many times does an observer hear maximum intensity in one second?  (IIT JEE 2006, Beats)

Answer: 4
Explanation: This is beat frequency. f₁ = ω₁/2π = 50 Hz and f₂ = ω₂/2π = 46 Hz. Beats per second = 50 − 46 = 4.

Q9. Standing waves can be produced: (IIT JEE 2002, More Than One Correct)
(a) on a string clamped at both ends
(b) on a string clamped at one end and free at the other
(c) when the incident wave gets reflected from a wall
(d) when two identical waves with a phase difference move in the same direction

Answer: (a), (b) and (c)
Explanation: A standing wave needs two waves travelling in opposite directions to superpose. Reflection at a fixed or free end, or from a wall, provides this. In (d) the waves move in the same direction, so no standing wave forms.

Q10.A mass M oscillates in SHM with amplitude A, attached to two massless springs of constants k₁ and k₂ connected in series, with the outer end of the first spring fixed. What is the amplitude of point P, the junction between the two springs? (IIT JEE 2009, Springs in Series)

Answer: k₂A/(k₁ + k₂)
Explanation: Let the extensions be x₁ and x₂. Then x₁ + x₂ = A, and in series the spring forces are equal, so k₁x₁ = k₂x₂. Solving gives x₁ = k₂A/(k₁ + k₂). Point P moves as much as the first spring stretches, so its amplitude is x₁.

Oscillations and Waves: Key Takeaways

Before moving on from this chapter, make sure you can recall:

• The condition for Simple Harmonic Motion

• How to identify the restoring force after a small displacement

• SHM involving buoyancy, springs and elastic wires

• Superposition of multiple SHMs using phasor representation

• Shear oscillations and the role of modulus of rigidity

• Interpretation of wave equations

• The relationship v = ω/k

• Formation of standing waves

• Position of nodes and antinodes

• Beat frequency

• Resonance-tube relations and end correction

• Fundamental modes of pipes and strings

• Frequency and wave-speed relations for stretched strings

• Torsional pendulum and spring combinations

• Effect of acceleration on a pendulum

• Oscillation of a gas-piston system

For NEET 2026 preparation, focus on identifying the physical situation before applying an equation. In SHM problems, start with the equilibrium position and additional restoring force; in wave problems, identify the wave type and relevant parameters before using the standard relation.

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