The Speed Of Sound Formula gives the velocity at which sound waves propagate through a medium, expressed as \( v = \sqrt{ rac{B}{ ho}} \) for a general medium and \( v = 331.4\sqrt{1 + rac{T}{273}} \) m/s for air at temperature T. This concept is covered in NCERT Class 9 (Chapter 12: Sound) and revisited in Class 11 Physics (Chapter 15: Waves). It is also an important topic for JEE Main, JEE Advanced, and NEET, where numerical problems on sound velocity in different media appear regularly. This article covers the formula, its derivation, a complete formula sheet, three solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Speed Of Sound Formulas at a Glance
Quick reference for the most important speed of sound formulas used in CBSE and competitive exams.
- Speed of sound in a medium: \( v = \sqrt{\frac{B}{\rho}} \)
- Speed of sound in an ideal gas (Newton-Laplace): \( v = \sqrt{\frac{\gamma P}{\rho}} \)
- Speed of sound in air at temperature T (°C): \( v = 331.4\sqrt{1 + \frac{T}{273}} \) m/s
- Simplified temperature correction: \( v \approx 331 + 0.6T \) m/s
- Speed of sound in a solid rod: \( v = \sqrt{\frac{Y}{\rho}} \)
- Relationship: \( v = f\lambda \)
- Speed of sound in air at 0°C: \( v_0 = 332 \) m/s
What is the Speed Of Sound Formula?
The Speed Of Sound Formula describes how fast a sound wave travels through a given medium. Sound is a mechanical, longitudinal wave. It requires a material medium for propagation — it cannot travel through vacuum. The speed depends on two key properties of the medium: its elasticity (resistance to compression) and its density (mass per unit volume).
In NCERT Class 9, Chapter 12 (Sound), students learn that sound travels at different speeds in solids, liquids, and gases. Solids are the fastest medium and gases are the slowest. In NCERT Class 11, Chapter 15 (Waves), the formula is derived rigorously using bulk modulus and density. The standard value of the speed of sound in air at 0°C is approximately 332 m/s. At room temperature (25°C), it is approximately 346 m/s. The speed of sound increases with temperature and is independent of pressure at constant temperature for an ideal gas.
Understanding this formula is essential for topics like Doppler effect, resonance, standing waves, and echo calculations — all of which appear in CBSE board exams and competitive exams.
Speed Of Sound Formula — Expression and Variables
General Formula for Speed of Sound in a Medium
\[ v = \sqrt{\frac{B}{\rho}} \]
Newton-Laplace Formula for Speed of Sound in an Ideal Gas
\[ v = \sqrt{\frac{\gamma P}{\rho}} \]
Speed of Sound in Air at Temperature T (°C)
\[ v_T = 331.4\sqrt{1 + \frac{T}{273}} \text{ m/s} \]
Simplified Linear Approximation
\[ v_T \approx 331 + 0.6T \text{ m/s} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( v \) | Speed of sound | m/s |
| \( B \) | Bulk modulus of elasticity of the medium | Pascal (Pa) or N/m² |
| \( \rho \) | Density of the medium | kg/m³ |
| \( \gamma \) | Adiabatic index (ratio of specific heats, \( C_p / C_v \)) | Dimensionless |
| \( P \) | Pressure of the gas | Pascal (Pa) |
| \( T \) | Temperature of air in degrees Celsius | °C |
| \( Y \) | Young's modulus (for solid rods) | Pascal (Pa) |
| \( f \) | Frequency of sound wave | Hertz (Hz) |
| \( \lambda \) | Wavelength of sound wave | metre (m) |
Derivation of the Speed Of Sound Formula
Newton originally derived the speed of sound assuming isothermal conditions. He obtained \( v = \sqrt{P/\rho} \), which gave approximately 280 m/s for air — lower than the experimental value of 332 m/s.
Laplace corrected this. He argued that sound propagation is an adiabatic process because compressions and rarefactions occur too rapidly for heat exchange. He replaced \( P \) with \( \gamma P \), giving:
\[ v = \sqrt{\frac{\gamma P}{\rho}} \]
For air, \( \gamma = 1.4 \). At 0°C and 1 atm pressure (\( P = 1.013 \times 10^5 \) Pa), density \( \rho = 1.293 \) kg/m³. Substituting:
\[ v = \sqrt{\frac{1.4 \times 1.013 \times 10^5}{1.293}} \approx 332 \text{ m/s} \]
This matches the experimental value. The temperature dependence arises because, for an ideal gas, \( P/\rho = RT/M \), so \( v = \sqrt{\gamma RT/M} \), where \( R \) is the universal gas constant and \( M \) is the molar mass of the gas.
Complete Physics Formula Sheet for Sound Waves
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Speed of sound (general) | \( v = \sqrt{B/\rho} \) | B = bulk modulus, ρ = density | m/s | Class 11, Ch 15 |
| Newton-Laplace formula | \( v = \sqrt{\gamma P / \rho} \) | γ = adiabatic index, P = pressure | m/s | Class 11, Ch 15 |
| Speed of sound in ideal gas | \( v = \sqrt{\gamma RT / M} \) | R = gas constant, T = absolute temp, M = molar mass | m/s | Class 11, Ch 15 |
| Speed of sound in air vs temperature | \( v_T = 331.4\sqrt{1 + T/273} \) | T = temperature in °C | m/s | Class 9, Ch 12 |
| Linear approximation | \( v_T \approx 331 + 0.6T \) | T = temperature in °C | m/s | Class 9, Ch 12 |
| Speed of sound in a solid rod | \( v = \sqrt{Y/\rho} \) | Y = Young's modulus, ρ = density | m/s | Class 11, Ch 15 |
| Wave equation | \( v = f\lambda \) | f = frequency, λ = wavelength | m/s | Class 9, Ch 12 |
| Echo formula (distance) | \( d = \frac{v \times t}{2} \) | v = speed of sound, t = time for echo | m | Class 9, Ch 12 |
| Doppler effect (source moving towards observer) | \( f’ = f\frac{v}{v – v_s} \) | v = speed of sound, v_s = speed of source | Hz | Class 11, Ch 15 |
| Doppler effect (source moving away from observer) | \( f’ = f\frac{v}{v + v_s} \) | v = speed of sound, v_s = speed of source | Hz | Class 11, Ch 15 |
| Speed ratio with temperature (absolute) | \( \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \) | T_1, T_2 = absolute temperatures in Kelvin | Dimensionless | Class 11, Ch 15 |
| Intensity of sound | \( I = \frac{P}{A} \) | P = power, A = area | W/m² | Class 11, Ch 15 |
Solved Examples on Speed Of Sound Formula
Example 1 (Class 9-10 Level)
Problem: The temperature of air is 27°C. Calculate the speed of sound in air at this temperature using the linear approximation formula.
Given: T = 27°C
Step 1: Write the linear approximation formula: \( v_T \approx 331 + 0.6T \)
Step 2: Substitute T = 27: \( v = 331 + 0.6 \times 27 \)
Step 3: Calculate: \( v = 331 + 16.2 = 347.2 \) m/s
Answer
Speed of sound at 27°C ≈ 347.2 m/s
Example 2 (Class 11-12 Level)
Problem: A person fires a gun near a cliff. He hears the echo after 4 seconds. If the temperature of air is 17°C, find the distance of the cliff from the person.
Given: t = 4 s, T = 17°C
Step 1: Find the speed of sound at 17°C using the linear approximation:
\( v = 331 + 0.6 \times 17 = 331 + 10.2 = 341.2 \) m/s
Step 2: Use the echo formula. The sound travels to the cliff and back, so total distance = \( v \times t \).
\( \text{Total distance} = 341.2 \times 4 = 1364.8 \) m
Step 3: Distance of cliff = half the total distance:
\( d = \frac{1364.8}{2} = 682.4 \) m
Answer
Distance of the cliff ≈ 682.4 m
Example 3 (JEE/NEET Level)
Problem: At what temperature will the speed of sound in air be double its value at 0°C? (Use the formula \( v \propto \sqrt{T} \), where T is in Kelvin.)
Given: \( v_1 = v_0 \) at \( T_1 = 273 \) K, \( v_2 = 2v_0 \)
Step 1: Use the relation: \( \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \)
Step 2: Substitute the values: \( \frac{2v_0}{v_0} = \sqrt{\frac{T_2}{273}} \)
Step 3: Square both sides: \( 4 = \frac{T_2}{273} \)
Step 4: Solve for \( T_2 \): \( T_2 = 4 \times 273 = 1092 \) K
Step 5: Convert to Celsius: \( T_2 = 1092 – 273 = 819 \)°C
Answer
The speed of sound doubles at 819°C (1092 K).
CBSE Exam Tips 2025-26
- Memorise the standard value: Speed of sound in air at 0°C is 332 m/s. This value is directly asked in 1-mark questions in CBSE Class 9 and Class 11 exams.
- Use the correct formula for the level: For Class 9 problems, use \( v = 331 + 0.6T \). For Class 11 and competitive exams, use \( v = \sqrt{\gamma RT/M} \) or the ratio method \( v \propto \sqrt{T} \).
- Echo problems are common: In CBSE Class 9, echo-based numerical questions are very frequent. Always remember that sound travels twice the distance (to the reflector and back). Use \( d = vt/2 \).
- Temperature must be in Kelvin when using the ratio formula \( v_1/v_2 = \sqrt{T_1/T_2} \). We recommend converting °C to K first by adding 273.
- Know the order of speed: Speed of sound is greatest in solids, then liquids, then gases. This fact appears in MCQ and short-answer questions in 2025-26 board exams.
- Laplace correction: For Class 11 boards, you must explain why Laplace replaced \( P \) with \( \gamma P \) — because sound propagation is adiabatic, not isothermal.
Common Mistakes to Avoid
- Using °C instead of Kelvin in the ratio formula: The formula \( v \propto \sqrt{T} \) requires T in Kelvin. Using Celsius directly gives a completely wrong answer. Always add 273 to convert.
- Forgetting the factor of 2 in echo problems: Sound travels to the wall and back. Many students calculate \( d = v \times t \) and forget to divide by 2. The correct formula is \( d = vt/2 \).
- Confusing Newton's and Laplace's formulas: Newton used \( v = \sqrt{P/\rho} \) (isothermal). Laplace corrected it to \( v = \sqrt{\gamma P/\rho} \) (adiabatic). Use Laplace's formula in all calculations.
- Assuming speed of sound is independent of temperature: Speed of sound in a gas increases with temperature. Only pressure has no effect (at constant temperature and density).
- Mixing up frequency, wavelength, and speed: Speed \( v = f\lambda \). If frequency increases, wavelength decreases (speed stays constant in the same medium). Students often incorrectly assume speed changes when frequency changes.
JEE/NEET Application of Speed Of Sound Formula
In our experience, JEE aspirants encounter the Speed Of Sound Formula in several important contexts. Understanding these patterns will save valuable time in the exam hall.
Pattern 1: Temperature Ratio Problems
JEE Main frequently asks: “At what temperature does the speed of sound become X times its value at Y°C?” Use the relation \( v \propto \sqrt{T} \) (T in Kelvin). Square the velocity ratio to get the temperature ratio. This is a high-frequency question type in JEE Main Paper 1.
Pattern 2: Doppler Effect Combined with Speed of Sound
Both JEE and NEET combine the Doppler effect formula with the speed of sound. The Doppler formula \( f’ = f(v \pm v_o)/(v \mp v_s) \) uses the speed of sound \( v \) as the reference. Problems often ask for the apparent frequency when a source moves towards or away from a stationary observer. Our experts suggest practising at least 10 Doppler problems before your exam.
Pattern 3: Speed of Sound in Different Media
NEET Biology-Physics interface questions sometimes involve ultrasound in human tissue (used in medical imaging). The speed of sound in human soft tissue is approximately 1540 m/s. For JEE Advanced, questions may involve deriving the speed of sound in a gas mixture using the effective \( \gamma \) and molar mass. Know that for a diatomic gas (like air), \( \gamma = 1.4 \), and for a monatomic gas (like Helium), \( \gamma = 1.67 \).
Key Competitive Exam Values to Memorise
| Medium | Speed of Sound (approx.) |
|---|---|
| Air at 0°C | 332 m/s |
| Air at 25°C | 346 m/s |
| Water (at 25°C) | 1498 m/s |
| Steel | 5100 m/s |
| Aluminium | 6420 m/s |
| Human soft tissue | 1540 m/s |
FAQs on Speed Of Sound Formula
Explore More Physics Formulas
Strengthen your understanding of waves and fluid physics with these related formula guides on ncertbooks.net:
- Learn how pressure changes in fluids using the Bernoulli's Equation Formula — a key topic in Class 11 fluid mechanics and JEE.
- Understand upward forces in fluids with the Buoyancy Formula, which is closely related to density concepts used in the speed of sound derivation.
- Explore another wave-based phenomenon with the Lightning Formula, which uses the speed of sound to calculate the distance of a lightning strike.
- For the complete collection of Physics Formulas for CBSE and JEE/NEET, visit our Physics Formulas Hub.
- For official NCERT resources and syllabus details, refer to the NCERT official website.