Oscillations Notes for NEET, Class 11 NCERT

Q: How many questions come from Oscillations in NEET 2027?

You can expect 1 to 2 questions from Oscillations in NEET 2027. The chapter has high PYQ frequency. Time period of pendulum and spring-mass, energy in SHM, velocity and acceleration at given displacement, and series / parallel spring problems are the favourites.

Q: What is Simple Harmonic Motion (SHM)?

SHM is a special kind of oscillatory motion in which the restoring force is directly proportional to the displacement from a fixed point and is always directed toward that point. Mathematically, F equals minus k x, which gives a equals minus omega squared x, with the standard solution x of t equals A sin(omega t plus phi).

Q: What is the time period of a simple pendulum?

For small angles, T equals 2 pi times the square root of L over g, where L is the length of the pendulum and g is the acceleration due to gravity. The period does not depend on the mass of the bob or the (small) amplitude. NEET problems often vary L or g (different planets, lifts) and ask how T changes.

Q: What is the time period of a spring-mass system?

T equals 2 pi times the square root of m over k, where m is the mass and k is the spring constant. This is independent of the amplitude. The period is the same on a horizontal frictionless surface and hanging vertically; gravity only shifts the equilibrium position, it does not change T.

Q: What is the velocity and acceleration in SHM?

For x equals A sin(omega t plus phi), v equals A omega cos(omega t plus phi) and a equals minus A omega squared sin(omega t plus phi). At the mean position (x = 0), v is maximum equal to A omega and a is zero. At the extremes (x = plus minus A), v is zero and a is maximum equal to A omega squared. A useful formula: v equals omega times square root of (A squared minus x squared).

Q: How does the total energy in SHM compare with KE and PE?

The total mechanical energy of SHM is constant: E equals half k A squared equals half m omega squared A squared. KE equals half k (A squared minus x squared) and PE equals half k x squared. KE is maximum at x equals 0; PE is maximum at x equals plus minus A. Their sum is always E.

Q: How do springs combine in series and in parallel?

Series: the same force stretches each spring, so they extend in turn. Effective spring constant is given by 1 over k_eff equals 1 over k_1 plus 1 over k_2. The effective k is smaller than the smallest k. Parallel: both springs share the load and extend by the same amount. k_eff equals k_1 plus k_2. The effective k is larger than the largest k.

Introduction

Oscillation is repetitive motion about a fixed point. A pendulum swinging, a mass bouncing on a spring, a tuning fork, even atoms in a solid: all are oscillators. NEET focuses on the simplest and most useful kind, called simple harmonic motion (SHM).

For NEET 2027 you can expect 1 to 2 questions from this chapter. Repeated favourites: time period of pendulum, time period of spring-mass, energy in SHM, velocity at a given displacement, and series / parallel spring combinations. Lock those down and you have most of the marks.

Periodic vs oscillatory motion

A motion is periodic if it repeats after a fixed time T. Examples: Earth around the Sun (one year), the moon around Earth (one month). Oscillatory motion is a special case of periodic motion in which the body moves back and forth around a fixed point. Every oscillation is periodic; not every periodic motion is oscillatory.

SHM is the cleanest oscillation possible: the motion is sinusoidal in time and the restoring force is linear in displacement.

Simple harmonic motion (SHM)

SHM is defined by a single condition: the restoring force is proportional to the displacement and points back to the mean position.

F = - k x \Leftrightarrow a = - ω^{2} x, ω^{2} = \frac{k}{m} .

The general solution is sinusoidal:

x (t) = A sin (ω t + ϕ) or x (t) = A cos (ω t + ϕ) .

Both forms are equivalent; the choice just shifts what we call the "phase".

Watch a particle execute SHM. The position oscillates as x = A sin(omega t). The dashed lines mark x = +A and x = -A.

Amplitude A: 1.00 m

Time period T: 2.00 s

t = 0.00 s

0.000 m

3.142 m/s

0.000 m/s²

3.142 rad/s

x = A sin (ω t), v = A ω cos (ω t), a = - ω^{2} x

Differential equation of SHM

Newton's second law for a SHM force law gives:

m \frac{d ^{2} x}{d t ^{2}} = - k x \Rightarrow \frac{d ^{2} x}{d t ^{2}} + ω^{2} x = 0.

Whenever you can put an equation in this form, the motion is SHM with angular frequency $ω = k / m$ . Time period:

T = \frac{2 π}{ω} = 2 π \frac{m}{k} .

Phase, amplitude, frequency, time period

Amplitude A: the maximum displacement from the mean position.
Time period T: time for one complete oscillation. Unit: second.
Frequency f: number of oscillations per second. $f = 1/ T$ . Unit: hertz.
Angular frequency ω: $ω = 2 π f = 2 π / T$ . Unit: rad/s.
Phase: the argument $ω t + ϕ$ . The constant $ϕ$ is the phase at t = 0.

Velocity and acceleration in SHM

Differentiate $x (t) = A sin (ω t + ϕ)$ :

v (t) = A ω cos (ω t + ϕ), a (t) = - A ω^{2} sin (ω t + ϕ) = - ω^{2} x .

Useful results:

Maximum speed (at x = 0): $v_{m a x} = A ω$ .
Maximum acceleration (at x = ±A): $a_{m a x} = A ω^{2}$ .
Speed at a given x: $v = ω A^{2} - x^{2}$ .
Acceleration is always opposite to displacement.

Energy in SHM (KE and PE)

At any instant, the particle has KE and PE. Their sum is the constant total energy E:

K E = \frac{1}{2} m v^{2} = \frac{1}{2} k (A^{2} - x^{2}), P E = \frac{1}{2} k x^{2} .

E = K E + P E = \frac{1}{2} k A^{2} = \frac{1}{2} m ω^{2} A^{2} .

At $x = 0$ : KE is max, PE is zero. At $x = \pm A$ : KE is zero, PE is max equal to E. At $x = A / 2$ : KE = PE = E/2.

Total energy is fixed at half k A squared. As the particle swings, energy moves between KE and PE; their sum always equals E.

Spring constant k: 100 N/m

Amplitude A: 0.100 m

Displacement x: 0.000 m

Energy split (total stays the same)

KE 100%

0.500 J

0.000 J

Total E

0.500 J

K E = \frac{1}{2} k (A^{2} - x^{2}), P E = \frac{1}{2} k x^{2}, E = \frac{1}{2} k A^{2}

Simple pendulum

For a small angle θ, the restoring force on a pendulum of mass m and length L is $F = - m g sin θ \approx - m g θ$ . With $x = L θ$ , this gives $F = - (m g / L) x$ , which is SHM with $k = m g / L$ :

T = 2 π \frac{m}{k} = 2 π \frac{L}{g} .

Two surprising features of this formula:

No mass: the period does not depend on the bob mass.
No amplitude: as long as the angle is small, T does not depend on how big the swing is.

These approximations break down above about 15°. For larger amplitudes the period actually grows slightly.

Watch a simple pendulum swing. For small angles, T is independent of mass and amplitude.

Length L: 1.00 m

g: 9.80 m/s² (Earth = 9.8, Moon = 1.6)

Initial angle θ₀: 15°

Time period

2.007 s

T = 2 π \frac{L}{g}

Spring-mass system

Hooke's law force: $F = - k x$ . With $ω = k / m$ :

T = 2 π \frac{m}{k} .

The period is independent of amplitude. Whether the mass hangs vertically or sits on a horizontal surface, T is the same. Gravity in the vertical case only shifts the equilibrium position; it does not change T.

Two iconic SHM systems with the same kind of formula.

Mass m: 0.50 kg

Spring constant k: 50 N/m

Time period T

0.628 s

T = 2 π \frac{m}{k}

Angular freq ω

10.00 rad/s

Frequency f

1.592 Hz

Series and parallel springs

Parallel (both springs share the same displacement, side by side):

k_{eff} = k_{1} + k_{2} .

Series (springs end-to-end, both stretched by the same force):

\frac{1}{k _{eff}} = \frac{1}{k _{1}} + \frac{1}{k _{2}} \Rightarrow k_{eff} = \frac{k _{1} k _{2}}{k _{1} + k _{2}} .

Memory hook: parallel makes things stiffer, series makes them softer. Same trick as resistors and capacitors.

Same two springs, two arrangements. The effective k changes, and so does the period.

k₁: 50 N/m

k₂: 100 N/m

Mass m: 0.50 kg

Effective spring constant

33.33 N/m

\frac{1}{k _{eff}} = \frac{1}{k _{1}} + \frac{1}{k _{2}}

Time period T

0.770 s

Practice these on the timed test

Try a free 10-question NEET mock test on Oscillations, with instant results and no sign-up needed.

Damped oscillations

Real oscillators slow down over time because of friction or air drag. Adding a velocity-proportional damping force $- b v$ changes the equation to:

m \overset{x}{¨} + b \overset{x}{˙} + k x = 0.

For weak damping ( $b < ω_{0}$ where $ω_{0} = k / m$ ):

x (t) = A_{0} e^{- b t / (2 m)} cos (ω_{d} t + ϕ), ω_{d} = ω_{0}^{2} - (b /2 m)^{2} .

Amplitude decays exponentially. The damped frequency is slightly less than the natural frequency. NEET only asks for the qualitative picture: weakly damped, critically damped, or overdamped.

Real oscillators slow down due to friction. Amplitude decays as e^(-b t), and the actual frequency is slightly less than omega_0 (the natural frequency).

Natural ω₀: 6.00 rad/s

Damping b: 0.40 1/s

Damped angular freq ω_d

5.99 rad/s

T_d = 1.05 s

ω_{d} = ω_{0}^{2} - b^{2}

Solid: x(t) = e^(-bt) cos(ω_d t). Dashed: amplitude envelope.

Forced oscillations and resonance

Apply a periodic external force $F_{0} cos (ω t)$ . The system eventually oscillates at the driving frequency omega, not at its natural frequency omega_0. The amplitude depends on how close omega is to omega_0.

When the driving frequency matches the natural frequency, the amplitude is enormous. This is resonance:

ω = ω_{0} \Rightarrow maximum amplitude .

Real-world examples of resonance: pushing a swing at just the right rhythm, the destruction of the Tacoma Narrows Bridge, tuning a radio to a specific station.

SHM as projection of circular motion

Imagine a particle moving with uniform speed around a circle of radius A. Its projection on a diameter oscillates as $A cos (ω t)$ . So SHM is exactly the "shadow" of uniform circular motion. This phasor picture is useful in waves, AC and optics later on.

SHM is the projection of uniform circular motion. The arrow rotates around the circle; its vertical projection traces a sine wave.

Time period T: 2.00 s

x (t) = A sin (ω t), ω = \frac{2 π}{T} ∣ angle = 0.00 rad, x = 0.000 A

Worked NEET problems

NEET-style problem · Spring-mass T

Question

A 200 g block on a frictionless horizontal surface is attached to a spring with

k = 50 N/m

. Find the time period.

Solution

T = 2 π m / k = 2 π 0.2/50 = 2 π 0.004 = 2 π \times 0.0632 \approx 0.397 s .

About 0.4 s.

NEET-style problem · Pendulum

Question

A pendulum has a time period of 2 s on Earth (

g = 9.8 m/s^{2}

). Find its time period on the Moon (

g = 1.6 m/s^{2}

Solution

T \propto 1/ g \Rightarrow \frac{T _{moon}}{T _{earth}} = \frac{g _{earth}}{g _{moon}} = 9.8/1.6 \approx 2.47.

So $T_{moon} \approx 2 \times 2.47 \approx 4.95 s$ .

NEET-style problem · Velocity in SHM

Question

A particle in SHM has amplitude

A = 5 cm

and angular frequency

ω = 4 rad/s

. Find the speed at

x = 3 cm

Solution

v = ω A^{2} - x^{2} = 4 5^{2} - 3^{2} = 416 = 16 cm/s = 0.16 m/s .

NEET-style problem · Energy in SHM

Question

A particle of mass

m = 0.5 kg

in SHM has amplitude

A = 0.1 m

and angular frequency

ω = 10 rad/s

. Find the total energy.

Solution

E = \frac{1}{2} m ω^{2} A^{2} = \frac{1}{2} (0.5) (10)^{2} (0.1)^{2} = 0.25 J .

NEET-style problem · Springs in series

Question

Two springs of constants

k_{1} = 100 N/m

and

k_{2} = 200 N/m

are connected in series. A 0.5 kg mass is attached. Find the time period.

Solution

k_{eff} = \frac{k _{1} k _{2}}{k _{1} + k _{2}} = \frac{100 \times 200}{300} = 66.67 N/m .

T = 2 π m / k_{eff} = 2 π 0.5/66.67 = 2 π 0.0075 \approx 0.544 s .

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Summary cheat sheet

SHM: $a = - ω^{2} x$ , $x = A sin (ω t + ϕ)$ .
Time period: $T = 2 π / ω$ .
v(x): $v = ω A^{2} - x^{2}$ , $v_{m a x} = A ω$ .
a(x): $a_{m a x} = A ω^{2}$ .
Energy: $E = \frac{1}{2} k A^{2} = \frac{1}{2} m ω^{2} A^{2}$ , KE/PE share at x = A/√2 is half each.
Spring-mass: $T = 2 π m / k$ .
Pendulum: $T = 2 π L / g$ (small angle).
Springs parallel: $k_{eff} = k_{1} + k_{2}$ .
Springs series: $k_{eff} = k_{1} k_{2} / (k_{1} + k_{2})$ .
Damped: amplitude decays as $e^{- b t / (2 m)}$ , $ω_{d} < ω_{0}$ .
Resonance: driving frequency = natural frequency.

Next: try the interactive widgets for SHM visualiser, pendulum, energy bar and springs, or work through the 32 NEET PYQs with full solutions. To time yourself, take the free 10-question mock test.

Frequently asked questions

How many questions come from Oscillations in NEET 2027?

You can expect 1 to 2 questions from Oscillations in NEET 2027. The chapter has high PYQ frequency. Time period of pendulum and spring-mass, energy in SHM, velocity and acceleration at given displacement, and series / parallel spring problems are the favourites.

What is Simple Harmonic Motion (SHM)?

SHM is a special kind of oscillatory motion in which the restoring force is directly proportional to the displacement from a fixed point and is always directed toward that point. Mathematically, F equals minus k x, which gives a equals minus omega squared x, with the standard solution x of t equals A sin(omega t plus phi).

What is the time period of a simple pendulum?

For small angles, T equals 2 pi times the square root of L over g, where L is the length of the pendulum and g is the acceleration due to gravity. The period does not depend on the mass of the bob or the (small) amplitude. NEET problems often vary L or g (different planets, lifts) and ask how T changes.

What is the time period of a spring-mass system?

T equals 2 pi times the square root of m over k, where m is the mass and k is the spring constant. This is independent of the amplitude. The period is the same on a horizontal frictionless surface and hanging vertically; gravity only shifts the equilibrium position, it does not change T.

What is the velocity and acceleration in SHM?

For x equals A sin(omega t plus phi), v equals A omega cos(omega t plus phi) and a equals minus A omega squared sin(omega t plus phi). At the mean position (x = 0), v is maximum equal to A omega and a is zero. At the extremes (x = plus minus A), v is zero and a is maximum equal to A omega squared. A useful formula: v equals omega times square root of (A squared minus x squared).

How does the total energy in SHM compare with KE and PE?

The total mechanical energy of SHM is constant: E equals half k A squared equals half m omega squared A squared. KE equals half k (A squared minus x squared) and PE equals half k x squared. KE is maximum at x equals 0; PE is maximum at x equals plus minus A. Their sum is always E.

How do springs combine in series and in parallel?

Series: the same force stretches each spring, so they extend in turn. Effective spring constant is given by 1 over k_eff equals 1 over k_1 plus 1 over k_2. The effective k is smaller than the smallest k. Parallel: both springs share the load and extend by the same amount. k_eff equals k_1 plus k_2. The effective k is larger than the largest k.

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OscillationsNEET Physics · Class 11 · NCERT Chapter 13

Introduction

Periodic vs oscillatory motion

Simple harmonic motion (SHM)

Differential equation of SHM

Phase, amplitude, frequency, time period

Velocity and acceleration in SHM

Energy in SHM (KE and PE)

Simple pendulum

Spring-mass system

Series and parallel springs

Damped oscillations

Forced oscillations and resonance

SHM as projection of circular motion

Worked NEET problems

Summary cheat sheet

Frequently asked questions

How many questions come from Oscillations in NEET 2027?

What is Simple Harmonic Motion (SHM)?

What is the time period of a simple pendulum?

What is the time period of a spring-mass system?

What is the velocity and acceleration in SHM?

How does the total energy in SHM compare with KE and PE?

How do springs combine in series and in parallel?

Continue with the next chapter notes

Track Your NEET Score Across All 90 Chapters