NEET Physics — Oscillations Complete Chapter Guide

Chapter Overview & Weightage

Oscillations covers simple harmonic motion (SHM), spring-mass systems, the simple pendulum, energy in SHM, and damped/forced oscillations. NEET focuses on SHM equations and time period calculations.

This chapter carries 3-4% weightage in NEET with 1-2 questions. Time period formulas for spring-mass and pendulum systems, and energy in SHM are the most tested.

Key Concepts You Must Know

Tier 1 (Core)

SHM condition: restoring force $F \propto -x$ (proportional to displacement, opposite direction)
Displacement: $x = A\sin(\omega t + \phi)$ or $x = A\cos(\omega t + \phi)$
Velocity: $v = A\omega\cos(\omega t + \phi)$ , max at mean position ( $v_{max} = A\omega$ )
Acceleration: $a = -\omega^2 x$ , max at extreme positions ( $a_{max} = A\omega^2$ )
Time period: spring-mass $T = 2\pi\sqrt{m/k}$ , pendulum $T = 2\pi\sqrt{l/g}$

Tier 2 (Frequently tested)

Energy in SHM: $KE = \frac{1}{2}m\omega^2(A^2 - x^2)$ , $PE = \frac{1}{2}m\omega^2 x^2$ , Total $E = \frac{1}{2}m\omega^2 A^2$ = constant
At mean position: KE = max, PE = 0. At extreme: KE = 0, PE = max.
Springs in series: $1/k_{eq} = 1/k_1 + 1/k_2$ . In parallel: $k_{eq} = k_1 + k_2$ .
Effective length changes in pendulum (with temperature, in accelerating systems)

Important Formulas

x = A\sin(\omega t + \phi)

v = \omega\sqrt{A^2 - x^2}

a = -\omega^2 x

T = \frac{2\pi}{\omega}, \quad f = \frac{\omega}{2\pi}

Spring-mass: $T = 2\pi\sqrt{\frac{m}{k}}$ (independent of amplitude and $g$ )

Simple pendulum: $T = 2\pi\sqrt{\frac{l}{g}}$ (independent of mass and amplitude for small angles)

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

PE = \frac{1}{2}m\omega^2 x^2

E_{total} = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}kA^2 = \text{constant}

At $x = A/\sqrt{2}$ : $KE = PE = E/2$ (equal split)

NEET loves asking: “At what displacement are KE and PE equal?” Answer: $x = A/\sqrt{2}$ . At this point, each is half the total energy. Also, KE and PE each average to $E/2$ over a complete cycle.

Solved Previous Year Questions

PYQ 1 — NEET 2024

Problem: The time period of a spring-mass system is 2 s. If the mass is doubled, the new time period is:

Solution:

$T = 2\pi\sqrt{m/k}$

$T' = 2\pi\sqrt{2m/k} = \sqrt{2} \times 2\pi\sqrt{m/k} = \sqrt{2} \times T = \sqrt{2} \times 2 = \mathbf{2\sqrt{2} \text{ s}}$

PYQ 2 — NEET 2023

Problem: The maximum velocity of a particle in SHM is 4 m/s and maximum acceleration is 8 m/s $^2$ . Find the time period.

Solution:

$v_{max} = A\omega = 4$ , $a_{max} = A\omega^2 = 8$

Dividing: $\omega = a_{max}/v_{max} = 8/4 = 2$ rad/s

$T = 2\pi/\omega = 2\pi/2 = \mathbf{\pi \text{ s}}$

PYQ 3 — NEET 2022

Problem: A simple pendulum has time period $T$ on Earth. On a planet where $g$ is 4 times that of Earth, the time period is:

Solution:

$T = 2\pi\sqrt{l/g}$

$T' = 2\pi\sqrt{l/4g} = T/2 = \mathbf{T/2}$

Expert Strategy

Day 1: SHM equations — displacement, velocity, acceleration. Understand the phase relationships: velocity leads displacement by $\pi/2$ , acceleration is opposite to displacement.

Day 2: Time period formulas and energy. Practice problems where mass, spring constant, or length changes. Know the energy split at different positions.

Common Traps

Trap 1 — Time period of a spring-mass system doesn’t depend on $g$ . $T = 2\pi\sqrt{m/k}$ has no $g$ . The same spring-mass system has the same time period on Earth and on the Moon. Only the pendulum’s time period depends on $g$ .

Trap 2 — Time period of SHM is independent of amplitude. Doubling the amplitude does NOT change the time period. This is a defining property of SHM. (For large-angle pendulums, this breaks down, but NEET assumes small angles.)

Trap 3 — Velocity is maximum at the MEAN position, not at the extreme. At extremes, velocity = 0 and acceleration = maximum. At mean position, velocity = maximum and acceleration = 0. Students often swap these.

Trap 4 — Springs in series give LOWER effective $k$ , not higher. Series: $1/k_{eq} = 1/k_1 + 1/k_2$ (like resistors in parallel). Parallel: $k_{eq} = k_1 + k_2$ (like resistors in series). The analogy is opposite to what you’d expect.