The Wave Formula gives the mathematical relationship between wave speed, frequency, and wavelength, expressed as \ ( v = f\lambda \). This formula is a core concept in NCERT Class 11 Physics (Chapter 15 — Waves) and appears regularly in CBSE board exams, JEE Main, and NEET. Understanding the Wave Formula helps students solve problems involving sound waves, light waves, and mechanical waves. This article covers the formula expression, derivation, a complete physics formula sheet, three solved examples, CBSE exam tips, common mistakes, JEE/NEET applications, and FAQs.
Key Wave Formulas at a Glance
Quick reference for the most important wave formulas used in CBSE and competitive exams.
- Wave speed: \( v = f\lambda \)
- Frequency: \( f = \frac{v}{\lambda} \)
- Wavelength: \( \lambda = \frac{v}{f} \)
- Time period: \( T = \frac{1}{f} \)
- Angular frequency: \( \omega = 2\pi f \)
- Wave number: \( k = \frac{2\pi}{\lambda} \)
- Speed of sound in a medium: \( v = \sqrt{\frac{B}{\rho}} \)
What is Wave Formula?
The Wave Formula defines the relationship between three fundamental properties of any wave: speed, frequency, and wavelength. A wave is a disturbance that transfers energy from one point to another through a medium, without any net transport of matter. Waves occur in many forms — sound waves, water waves, seismic waves, and electromagnetic waves.
According to NCERT Class 11 Physics, Chapter 15 (Waves), every periodic wave has a characteristic speed at which it travels through a given medium. The Wave Formula states that the speed of a wave equals the product of its frequency and wavelength. This relationship holds true for all types of waves, whether mechanical or electromagnetic.
The Wave Formula is introduced in NCERT Class 11 Physics and is revisited in Class 12 in the context of electromagnetic waves (Chapter 8). It is a foundational concept for understanding topics such as the Doppler effect, resonance, standing waves, and wave interference. Mastering this formula is essential for scoring well in CBSE board exams and competitive entrance tests like JEE and NEET.
Wave Formula — Expression and Variables
The standard Wave Formula is written as:
\[ v = f\lambda \]
This can also be rearranged to find frequency or wavelength:
\[ f = \frac{v}{\lambda} \qquad \lambda = \frac{v}{f} \]
Since frequency and time period are inversely related, the Wave Formula can also be written as:
\[ v = \frac{\lambda}{T} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( v \) | Wave speed (velocity) | metre per second (m/s) |
| \( f \) | Frequency | Hertz (Hz) or s¹ |
| \( \lambda \) | Wavelength | metre (m) |
| \( T \) | Time period | second (s) |
| \( \omega \) | Angular frequency | radian per second (rad/s) |
| \( k \) | Wave number (propagation constant) | radian per metre (rad/m) |
| \( A \) | Amplitude | metre (m) |
Derivation of the Wave Formula
Consider a transverse wave travelling along the positive x-direction. In one complete oscillation, the wave travels a distance equal to one wavelength \( \lambda \) in a time equal to one time period \( T \).
Step 1: Recall the basic speed formula: \( \text{speed} = \frac{\text{distance}}{\text{time}} \)
Step 2: Substitute distance = \( \lambda \) and time = \( T \): \( v = \frac{\lambda}{T} \)
Step 3: Since frequency \( f = \frac{1}{T} \), substitute to get: \( v = f\lambda \)
This derivation is direct and elegant. It shows that the Wave Formula is simply a restatement of the distance-speed-time relationship applied to periodic wave motion.
Complete Physics Wave Formula Sheet
The table below provides a comprehensive reference of all important wave-related formulas covered in NCERT Class 11 and Class 12 Physics.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Wave Speed | \( v = f\lambda \) | v = speed, f = frequency, λ = wavelength | m/s | Class 11, Ch 15 |
| Frequency & Period | \( f = \frac{1}{T} \) | f = frequency, T = time period | Hz | Class 11, Ch 15 |
| Angular Frequency | \( \omega = 2\pi f \) | ω = angular frequency, f = frequency | rad/s | Class 11, Ch 15 |
| Wave Number | \( k = \frac{2\pi}{\lambda} \) | k = wave number, λ = wavelength | rad/m | Class 11, Ch 15 |
| Progressive Wave Equation | \( y = A\sin(\omega t – kx) \) | y = displacement, A = amplitude, ω = angular freq, k = wave number | m | Class 11, Ch 15 |
| Speed of Sound in Solid | \( v = \sqrt{\frac{Y}{\rho}} \) | Y = Young's modulus, ρ = density | m/s | Class 11, Ch 15 |
| Speed of Sound in Fluid | \( v = \sqrt{\frac{B}{\rho}} \) | B = bulk modulus, ρ = density | m/s | Class 11, Ch 15 |
| Newton's Formula (Sound in Air) | \( v = \sqrt{\frac{P}{\rho}} \) | P = pressure, ρ = density | m/s | Class 11, Ch 15 |
| Laplace Correction | \( v = \sqrt{\frac{\gamma P}{\rho}} \) | γ = adiabatic index, P = pressure, ρ = density | m/s | Class 11, Ch 15 |
| Doppler Effect (Source Moving) | \( f’ = f\frac{v \pm v_o}{v \mp v_s} \) | f' = observed freq, v = wave speed, v⊂o; = observer speed, v⊂s; = source speed | Hz | Class 11, Ch 15 |
| Standing Wave Equation | \( y = 2A\sin(kx)\cos(\omega t) \) | A = amplitude, k = wave number, ω = angular freq | m | Class 11, Ch 15 |
| Speed of EM Waves (Vacuum) | \( c = 3 \times 10^8 \) m/s | c = speed of light | m/s | Class 12, Ch 8 |
Wave Formula — Solved Examples
The following three examples progress from basic CBSE level to JEE/NEET application level. Work through each one carefully.
Example 1 (Class 9-10 Level)
Problem: A sound wave has a frequency of 440 Hz and travels through air at a speed of 340 m/s. Calculate the wavelength of this sound wave.
Given: Frequency \( f = 440 \) Hz, Wave speed \( v = 340 \) m/s
Step 1: Write the Wave Formula: \( v = f\lambda \)
Step 2: Rearrange to find wavelength: \( \lambda = \frac{v}{f} \)
Step 3: Substitute the values: \( \lambda = \frac{340}{440} \)
Step 4: Calculate: \( \lambda = 0.773 \) m
Answer
The wavelength of the sound wave is approximately 0.77 m (or 77 cm).
Example 2 (Class 11-12 Level)
Problem: A progressive transverse wave is described by the equation \( y = 0.02\sin(6.28t – 3.14x) \) metres, where t is in seconds and x is in metres. Find the wave speed, frequency, and wavelength of this wave.
Given: Wave equation \( y = 0.02\sin(6.28t – 3.14x) \)
Step 1: Compare with standard form \( y = A\sin(\omega t – kx) \).
Step 2: Identify: \( \omega = 6.28 \) rad/s and \( k = 3.14 \) rad/m.
Step 3: Find frequency: \( f = \frac{\omega}{2\pi} = \frac{6.28}{2\pi} = 1 \) Hz.
Step 4: Find wavelength: \( \lambda = \frac{2\pi}{k} = \frac{2\pi}{3.14} = 2 \) m.
Step 5: Apply the Wave Formula to find speed: \( v = f\lambda = 1 \times 2 = 2 \) m/s.
Alternatively: \( v = \frac{\omega}{k} = \frac{6.28}{3.14} = 2 \) m/s.
Answer
Wave speed = 2 m/s, Frequency = 1 Hz, Wavelength = 2 m.
Example 3 (JEE/NEET Level)
Problem: A source of sound emits waves of frequency 500 Hz. An observer moves towards the source at 20 m/s, while the source moves away from the observer at 30 m/s. If the speed of sound in air is 340 m/s, find the frequency heard by the observer. Use the Doppler effect formula.
Given: Source frequency \( f = 500 \) Hz, Observer speed \( v_o = 20 \) m/s (towards source), Source speed \( v_s = 30 \) m/s (away from observer), Sound speed \( v = 340 \) m/s.
Step 1: Write the Doppler formula: \( f’ = f \cdot \frac{v + v_o}{v + v_s} \)
Step 2: Use the sign convention: observer moving towards source → use \( +v_o \); source moving away from observer → use \( +v_s \) in denominator.
Step 3: Substitute values: \( f’ = 500 \times \frac{340 + 20}{340 + 30} \)
Step 4: Calculate: \( f’ = 500 \times \frac{360}{370} \)
Step 5: Simplify: \( f’ = 500 \times 0.9730 \approx 486.5 \) Hz.
Answer
The frequency heard by the observer is approximately 486.5 Hz.
CBSE Exam Tips 2025-26
- Memorise the three forms: Write \( v = f\lambda \), \( f = v/\lambda \), and \( \lambda = v/f \) on your formula sheet. CBSE questions often give two quantities and ask for the third.
- Unit consistency is critical: Always convert frequency to Hz and wavelength to metres before substituting. We recommend double-checking units in every step.
- Link frequency and time period: Many students forget that \( f = 1/T \). CBSE 2025-26 papers often combine both relationships in a single question.
- Wave equation identification: For Class 11 questions on progressive waves, identify \( \omega \) and \( k \) from the equation first. Then compute \( v = \omega/k \) as a shortcut.
- Doppler effect sign convention: Our experts suggest writing a standard rule card: observer approaching → numerator increases; source approaching → denominator decreases. Stick to this consistently.
- Revise NCERT examples: CBSE 2025-26 questions are closely based on NCERT solved examples in Chapter 15. Solve all of them at least twice before the exam.
Common Mistakes to Avoid
- Confusing frequency with angular frequency: Students often substitute \( \omega \) in place of \( f \) in the Wave Formula \( v = f\lambda \). Remember, \( \omega = 2\pi f \), so always use \( f \) (in Hz) with the standard formula.
- Incorrect unit for wave number: The wave number \( k \) has units of rad/m, not m. Mixing this up leads to incorrect dimensional analysis in derivation-based questions.
- Forgetting to convert units: If wavelength is given in centimetres or nanometres, convert to metres before using \( v = f\lambda \). This is a very common error in CBSE practicals and theory papers.
- Wrong sign in Doppler formula: The most frequent JEE/NEET mistake is applying the wrong sign when both source and observer are moving. Always define a positive direction first and apply the convention systematically.
- Assuming wave speed depends on frequency: Wave speed in a given medium depends only on the properties of the medium (density, elasticity), not on frequency. Changing frequency changes wavelength, not speed.
JEE/NEET Application of Wave Formula
In our experience, JEE aspirants encounter the Wave Formula in at least 3–5 questions per paper, spread across mechanics, waves, and modern physics. NEET typically includes 2–3 direct and application-based questions on waves every year.
Application Pattern 1: Wave Equation Analysis
JEE Main frequently gives a wave equation such as \( y = A\sin(\omega t – kx + \phi) \) and asks students to find wave speed, direction of propagation, or phase difference between two points. The key step is always to extract \( \omega \) and \( k \) and apply \( v = \omega/k \), which is a direct consequence of the Wave Formula.
Application Pattern 2: Superposition and Standing Waves
When two waves of equal frequency and amplitude travel in opposite directions, they form a standing wave. The resulting equation \( y = 2A\sin(kx)\cos(\omega t) \) is derived by superimposing \( y_1 = A\sin(\omega t – kx) \) and \( y_2 = A\sin(\omega t + kx) \). JEE Advanced tests this derivation directly. Understanding the Wave Formula is essential to set up these superposition problems correctly.
Application Pattern 3: Doppler Effect in NEET
NEET Biology-Physics integration questions often involve ultrasound frequency shifts in medical imaging. These are direct applications of the Doppler formula, which itself is built on the Wave Formula. Our experts suggest practising at least 10 Doppler problems before the NEET exam. Pay special attention to cases where both source and observer move simultaneously, as these carry higher marks.
For deeper understanding of energy carried by waves, also explore the Wave Power Formula and the Photon Energy Formula, which extend the Wave Formula into quantum and electromagnetic contexts.
FAQs on Wave Formula
We hope this complete guide to the Wave Formula has given you a thorough understanding of wave speed, frequency, wavelength, and their interrelationships. For more related formulas, explore our Physics Formulas hub, and deepen your knowledge with the Wave Power Formula, the Photon Energy Formula, and the Heat Transfer Formula. For official NCERT content, refer to the NCERT official website.