The Oscillatory Motion Formula describes the repetitive back-and-forth movement of a body about its mean position, expressed mathematically as \ ( x(t) = A\sin(\omega t + \phi) \). This topic is a core part of NCERT Class 11 Physics, Chapter 14 (Oscillations). It is also heavily tested in CBSE Board Exams and carries significant weightage in JEE Main, JEE Advanced, and NEET. This article covers all key oscillatory motion formulas, derivations, solved examples, exam tips, and FAQs in one place.
Key Oscillatory Motion Formulas at a Glance
Quick reference for the most important oscillatory motion formulas.
- Displacement: \( x(t) = A\sin(\omega t + \phi) \)
- Angular frequency: \( \omega = 2\pi f = \dfrac{2\pi}{T} \)
- Time period of simple pendulum: \( T = 2\pi\sqrt{\dfrac{L}{g}} \)
- Time period of spring-mass system: \( T = 2\pi\sqrt{\dfrac{m}{k}} \)
- Maximum velocity: \( v_{max} = A\omega \)
- Maximum acceleration: \( a_{max} = A\omega^2 \)
- Total energy in SHM: \( E = \dfrac{1}{2}m\omega^2 A^2 \)
What is Oscillatory Motion Formula?
The Oscillatory Motion Formula is the mathematical expression that quantifies the displacement, velocity, acceleration, and energy of a body undergoing periodic to-and-fro motion about a fixed equilibrium position. Oscillatory motion is a special case of periodic motion where the restoring force is always directed toward the mean position and is proportional to the displacement from it.
According to NCERT Class 11 Physics, Chapter 14 — Oscillations, any motion that repeats itself at regular intervals of time is called periodic motion. When that periodic motion is back and forth about a mean position, it is called oscillatory motion. Simple Harmonic Motion (SHM) is the simplest and most important form of oscillatory motion.
The general displacement equation for SHM is \( x(t) = A\sin(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is the angular frequency, \( t \) is the time, and \( \phi \) is the initial phase. This formula is the foundation for understanding pendulums, springs, sound waves, and even molecular vibrations. It is tested in CBSE Class 11 board exams and forms a critical part of JEE and NEET syllabi.
Oscillatory Motion Formula — Expression and Variables
The primary displacement formula for oscillatory motion (SHM) is:
\[ x(t) = A\sin(\omega t + \phi) \]
where \( x(t) \) is the displacement at time \( t \), \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the initial phase angle.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( x(t) \) | Displacement at time t | Metre (m) |
| \( A \) | Amplitude (maximum displacement) | Metre (m) |
| \( \omega \) | Angular frequency | Radian per second (rad/s) |
| \( t \) | Time | Second (s) |
| \( \phi \) | Initial phase angle | Radian (rad) |
| \( T \) | Time period | Second (s) |
| \( f \) | Frequency | Hertz (Hz) |
| \( k \) | Spring constant | Newton per metre (N/m) |
| \( m \) | Mass of oscillating body | Kilogram (kg) |
| \( L \) | Length of pendulum | Metre (m) |
| \( g \) | Acceleration due to gravity | m/s² |
Derivation of Oscillatory Motion Formula
Consider a particle moving in a circle of radius \( A \) with uniform angular velocity \( \omega \). The projection of this circular motion onto a diameter gives SHM. At time \( t = 0 \), the particle makes an angle \( \phi \) with the reference axis. At time \( t \), the angle becomes \( (\omega t + \phi) \). The displacement of the projection is:
\[ x(t) = A\sin(\omega t + \phi) \]
Differentiating once with respect to time gives velocity: \( v(t) = A\omega\cos(\omega t + \phi) \). Differentiating again gives acceleration: \( a(t) = -A\omega^2\sin(\omega t + \phi) = -\omega^2 x \). The negative sign confirms that acceleration is always directed toward the mean position, which is the defining condition of SHM.
Complete Oscillatory Motion Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Displacement in SHM | \( x = A\sin(\omega t + \phi) \) | A = amplitude, ω = angular frequency, φ = phase | m | Class 11, Ch 14 |
| Angular Frequency | \( \omega = 2\pi f = \dfrac{2\pi}{T} \) | f = frequency, T = time period | rad/s | Class 11, Ch 14 |
| Time Period (general) | \( T = \dfrac{1}{f} = \dfrac{2\pi}{\omega} \) | f = frequency, ω = angular frequency | s | Class 11, Ch 14 |
| Time Period (spring-mass) | \( T = 2\pi\sqrt{\dfrac{m}{k}} \) | m = mass, k = spring constant | s | Class 11, Ch 14 |
| Time Period (simple pendulum) | \( T = 2\pi\sqrt{\dfrac{L}{g}} \) | L = length, g = gravitational acceleration | s | Class 11, Ch 14 |
| Velocity in SHM | \( v = \omega\sqrt{A^2 – x^2} \) | A = amplitude, x = displacement | m/s | Class 11, Ch 14 |
| Maximum Velocity | \( v_{max} = A\omega \) | A = amplitude, ω = angular frequency | m/s | Class 11, Ch 14 |
| Acceleration in SHM | \( a = -\omega^2 x \) | ω = angular frequency, x = displacement | m/s² | Class 11, Ch 14 |
| Maximum Acceleration | \( a_{max} = A\omega^2 \) | A = amplitude, ω = angular frequency | m/s² | Class 11, Ch 14 |
| Kinetic Energy in SHM | \( KE = \dfrac{1}{2}m\omega^2(A^2 – x^2) \) | m = mass, A = amplitude, x = displacement | J | Class 11, Ch 14 |
| Potential Energy in SHM | \( PE = \dfrac{1}{2}m\omega^2 x^2 \) | m = mass, ω = angular frequency, x = displacement | J | Class 11, Ch 14 |
| Total Energy in SHM | \( E = \dfrac{1}{2}m\omega^2 A^2 \) | m = mass, ω = angular frequency, A = amplitude | J | Class 11, Ch 14 |
| Restoring Force | \( F = -kx \) | k = spring constant, x = displacement | N | Class 11, Ch 14 |
Oscillatory Motion Formula — Solved Examples
Example 1 (Class 9-10 Level): Finding Time Period of a Simple Pendulum
Problem: A simple pendulum has a length of 1 m. Calculate its time period. Take \( g = 10 \) m/s².
Given: \( L = 1 \) m, \( g = 10 \) m/s²
Step 1: Write the formula for time period of a simple pendulum: \( T = 2\pi\sqrt{\dfrac{L}{g}} \)
Step 2: Substitute the values: \( T = 2\pi\sqrt{\dfrac{1}{10}} \)
Step 3: Simplify: \( T = 2\pi \times \sqrt{0.1} = 2\pi \times 0.3162 \)
Step 4: Calculate: \( T = 2 \times 3.1416 \times 0.3162 \approx 1.99 \) s
Answer
Time period \( T \approx 2 \) s
Example 2 (Class 11-12 Level): Velocity and Acceleration in SHM
Problem: A particle executes SHM with amplitude \( A = 0.05 \) m and angular frequency \( \omega = 20 \) rad/s. Find the velocity and acceleration of the particle when its displacement is \( x = 0.03 \) m.
Given: \( A = 0.05 \) m, \( \omega = 20 \) rad/s, \( x = 0.03 \) m
Step 1: Use the velocity formula in SHM: \( v = \omega\sqrt{A^2 – x^2} \)
Step 2: Substitute values: \( v = 20\sqrt{(0.05)^2 – (0.03)^2} \)
Step 3: Calculate inside the root: \( (0.05)^2 – (0.03)^2 = 0.0025 – 0.0009 = 0.0016 \)
Step 4: Find velocity: \( v = 20 \times \sqrt{0.0016} = 20 \times 0.04 = 0.8 \) m/s
Step 5: Use the acceleration formula: \( a = -\omega^2 x = -(20)^2 \times 0.03 = -400 \times 0.03 = -12 \) m/s²
Answer
Velocity \( v = 0.8 \) m/s; Acceleration \( a = -12 \) m/s² (directed toward mean position)
Example 3 (JEE/NEET Level): Energy Distribution in SHM
Problem: A block of mass 0.5 kg is attached to a spring of spring constant \( k = 200 \) N/m and oscillates on a frictionless surface. The amplitude of oscillation is 0.1 m. Find: (a) the total mechanical energy, (b) the kinetic energy when displacement is half the amplitude, and (c) the velocity at that point.
Given: \( m = 0.5 \) kg, \( k = 200 \) N/m, \( A = 0.1 \) m, \( x = A/2 = 0.05 \) m
Step 1: Find angular frequency: \( \omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{200}{0.5}} = \sqrt{400} = 20 \) rad/s
Step 2: Calculate total energy: \( E = \dfrac{1}{2}m\omega^2 A^2 = \dfrac{1}{2} \times 0.5 \times (20)^2 \times (0.1)^2 \)
Step 3: Simplify: \( E = \dfrac{1}{2} \times 0.5 \times 400 \times 0.01 = 1 \) J
Step 4: Calculate potential energy at \( x = 0.05 \) m: \( PE = \dfrac{1}{2}m\omega^2 x^2 = \dfrac{1}{2} \times 0.5 \times 400 \times (0.05)^2 = 0.25 \) J
Step 5: Find kinetic energy: \( KE = E – PE = 1 – 0.25 = 0.75 \) J
Step 6: Find velocity using \( KE = \dfrac{1}{2}mv^2 \): \( v = \sqrt{\dfrac{2 \times KE}{m}} = \sqrt{\dfrac{2 \times 0.75}{0.5}} = \sqrt{3} \approx 1.73 \) m/s
Answer
(a) Total energy = 1 J; (b) Kinetic energy at \( x = A/2 \) is 0.75 J (which is \( \frac{3}{4} \) of total energy); (c) Velocity \( v \approx 1.73 \) m/s
CBSE Exam Tips 2025-26
- Memorise the displacement equation first. The entire chapter flows from \( x = A\sin(\omega t + \phi) \). Know when to use sine and when to use cosine based on initial conditions.
- Learn the energy relationships. CBSE frequently asks 3-mark questions on how KE and PE vary with displacement. Remember that at mean position KE is maximum and PE is zero, while at extreme positions PE is maximum and KE is zero.
- Practice derivations. The derivation of time period for both the spring-mass system and the simple pendulum is a common 5-mark question in board exams. We recommend writing these derivations at least five times before the exam.
- Use dimensional analysis as a check. Always verify your final answer has the correct SI unit. For time period, the answer must be in seconds.
- Note the sign of acceleration. The acceleration formula \( a = -\omega^2 x \) always carries a negative sign. Forgetting this sign costs marks in CBSE 2025-26 exams.
- Distinguish oscillatory from periodic motion. All oscillatory motion is periodic, but not all periodic motion is oscillatory. This distinction is a common 1-mark question in CBSE.
Common Mistakes to Avoid
- Confusing amplitude with displacement. Amplitude \( A \) is the maximum displacement. Displacement \( x \) varies with time. Many students substitute \( x \) where \( A \) is required and lose marks.
- Using wrong initial phase. If the particle starts from the mean position, use \( x = A\sin(\omega t) \). If it starts from the extreme position, use \( x = A\cos(\omega t) \). Mixing these two forms is a very common error.
- Forgetting the negative sign in acceleration. The formula \( a = -\omega^2 x \) must always have a negative sign. This indicates the restoring nature of the force. Writing \( a = +\omega^2 x \) is incorrect.
- Incorrect substitution in pendulum formula. Students sometimes substitute the mass of the bob in \( T = 2\pi\sqrt{L/g} \). The time period of a simple pendulum is independent of mass. Only length and \( g \) matter.
- Unit confusion for angular frequency. Angular frequency \( \omega \) is in rad/s, while frequency \( f \) is in Hz (cycles per second). Always check which one the problem requires before calculating.
JEE/NEET Application of Oscillatory Motion Formula
In our experience, JEE aspirants encounter oscillatory motion formulas in nearly every paper. The topic carries 2–3 questions in JEE Main and at least 1–2 questions in NEET annually. Understanding the Oscillatory Motion Formula deeply can secure 8–12 marks in these competitive exams.
Pattern 1: Spring-Mass System Combinations
JEE Main frequently tests springs in series and parallel. For springs in series, the effective spring constant is \( \dfrac{1}{k_{eff}} = \dfrac{1}{k_1} + \dfrac{1}{k_2} \). For springs in parallel, \( k_{eff} = k_1 + k_2 \). The time period is then calculated using \( T = 2\pi\sqrt{m/k_{eff}} \). This is a standard JEE pattern that appears in almost every session.
Pattern 2: Energy-Based Problems
NEET frequently asks at what displacement the kinetic energy equals the potential energy. Setting \( KE = PE \) gives \( \dfrac{1}{2}m\omega^2(A^2 – x^2) = \dfrac{1}{2}m\omega^2 x^2 \), which simplifies to \( x = \dfrac{A}{\sqrt{2}} \). This result is a must-know for both JEE and NEET.
Pattern 3: Phase and Initial Conditions
JEE Advanced tests the ability to write the correct displacement equation based on initial conditions. If a particle starts at \( x = A \) at \( t = 0 \), the equation is \( x = A\cos(\omega t) \). If it starts at \( x = 0 \) moving in the positive direction, the equation is \( x = A\sin(\omega t) \). Our experts suggest practising at least 10 such initial-condition problems before the exam.
For additional reference on oscillations, students can visit the official NCERT website at ncert.nic.in to access the Class 11 Physics textbook Chapter 14.
FAQs on Oscillatory Motion Formula
We hope this comprehensive guide on the Oscillatory Motion Formula has helped you build a strong foundation. For more Physics formulas, explore our Physics Formulas hub. You can also strengthen your understanding with our detailed articles on the Spring Constant Formula and the Angular Displacement Formula. For wave-related topics, our Flow Rate Formula article provides excellent supplementary reading. Keep practising, and you will master oscillatory motion well before your 2025-26 exams.