Formula
Speed of wave on string
v = √(T/μ), where T is tension and μ is linear mass density (mass per unit length). Doubling tension increases speed by √2.
-- NCERT Class 11 Physics, Ch. 14, p. 285Speed Travelling Wave
8 MCQs8 revision cards9-step worked example
Source: NCERT Oscillations and WavesPYQ coverage: NEET 2022Official key: NTA-verifiedLast updated: 26 Sep 2026
Speed Travelling Wave, explained for NEET
The speed of a travelling wave depends on the medium, not on the source. Two formulas govern this topic at the NEET level, and one common error connects them.
Transverse wave on a string. The speed is v = √(T/μ), where T is the tension in the string and μ is the linear mass density (mass per unit length). It can be derived from Newton's laws applied to a small element of the string, but NCERT treats that derivation as outside its scope and obtains the formula by dimensional analysis instead (NCERT Class 11 Physics Chapter 14, page 285). The key relationship: v scales as √T, not as T. Doubling the tension increases speed by a factor of √2, not 2. This is a high-frequency distractor in NEET — students who treat v as directly proportional to T pick the wrong option.
Speed of sound in a gas. The Newton-Laplace formula gives v = √(γP/ρ), where γ is the adiabatic index, P the gas pressure, and ρ the density (NCERT Class 11 Physics Chapter 14, page 287). Newton's original formula used isothermal bulk modulus (B = P), predicting a value about 15% below the measured speed of sound in air. Laplace corrected it by recognising that sound compressions are adiabatic, replacing P with γP.
The connection that trips students: both formulas have the form v = √(elastic property / inertial property). For the string, the elastic property is tension and the inertial property is linear mass density. For the gas, the elastic property is γP (adiabatic bulk modulus) and the inertial property is density ρ.
Watch out: wave speed on a string is independent of frequency and wavelength — those are set by the source. Changing tension changes v, which then changes the wavelength for a given frequency (v = fλ), but does not change the frequency itself.
Can you answer these Speed Travelling Wave MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The speed of a transverse wave on a stretched string depends on:
Show answer and why every option is right or wrong
Answer: C. The wave speed on a string is v = √(T/μ), determined entirely by tension T and linear mass density μ — properties of the medium (NCERT Class 11 Physics Chapter 14, page 285).
Why A is wrong: A is wrong because frequency and wavelength are related to speed by v = fλ, but they do not determine the speed — the medium does. Changing the source frequency changes λ, not v.
Why B is wrong: B is wrong because amplitude affects energy carried by the wave, not its speed. Frequency is set by the source, not the medium.
Why D is wrong: D is wrong because neither wavelength nor amplitude determines wave speed on a string. Speed depends on medium properties (tension, mass density).
MCQ 2Easy RecallPractice
Laplace corrected Newton's formula for the speed of sound in air by replacing the isothermal process with:
Show answer and why every option is right or wrong
Answer: D. Laplace recognised that sound compressions and rarefactions in air are rapid enough to be adiabatic, not isothermal. Replacing bulk modulus B = P (isothermal) with B = γP (adiabatic) corrects the predicted speed (NCERT Class 11 Physics Chapter 14, page 287).
Why A is wrong: A is wrong because isobaric means constant pressure — that would imply no compression at all, which contradicts the nature of sound waves.
Why B is wrong: B is wrong because isochoric means constant volume. Sound transmission involves compressions and rarefactions (volume changes), so a constant-volume assumption is inappropriate.
Why C is wrong: C is wrong because a polytropic process is a generalisation that includes adiabatic as a special case, but Laplace's specific correction identifies the adiabatic process (γ, not a general polytropic index).
MCQ 3Easy RecallPractice
In the Newton-Laplace formula v = √(γP/ρ), what does γ represent?
Show answer and why every option is right or wrong
Answer: B. γ is the adiabatic index, defined as the ratio of specific heats Cp/Cv. For air (diatomic), γ ≈ 1.4 (NCERT Class 11 Physics Chapter 14, page 287).
Why A is wrong: A is wrong because R/M gives the specific gas constant, not the adiabatic index. That quantity appears when rewriting the formula as v = √(γRT/M), but it is not γ itself.
Why C is wrong: C is wrong because the coefficient of thermal expansion describes how volume changes with temperature — it is a thermodynamic property unrelated to the adiabatic index.
Why D is wrong: D is wrong because P/ρ has dimensions of velocity squared and appears inside the formula, but γ is the dimensionless multiplier that corrects Newton's isothermal prediction to the adiabatic one.
MCQ 4Direct ApplicationPractice
The tension in a stretched string is increased to 4 times its original value while the linear mass density remains unchanged. The wave speed becomes:
Show answer and why every option is right or wrong
Answer: B. v = √(T/μ). If T → 4T and μ is unchanged, v_new = √(4T/μ) = 2√(T/μ) = 2v. The speed doubles (NCERT Class 11 Physics Chapter 14, page 285).
Why A is wrong: A is wrong because this assumes v ∝ T (linear scaling). The actual dependence is v ∝ √T. Quadrupling tension gives √4 = 2 times the speed, not 4 times.
Why C is wrong: C is wrong because √2 would result from doubling the tension (v ∝ √T, so √(2T/μ) = √2 · v). Here the tension is quadrupled, giving factor 2.
Why D is wrong: D is wrong because wave speed on a string directly depends on tension through v = √(T/μ). Changing tension necessarily changes the speed.
MCQ 5Direct ApplicationPractice
A transverse wave travels on a string with speed 40 m/s when the tension is 100 N. If the tension is reduced to 25 N, what is the new wave speed?
Show answer and why every option is right or wrong
Answer: A. v ∝ √T (μ unchanged). v₂/v₁ = √(T₂/T₁) = √(25/100) = √(1/4) = 1/2. So v₂ = 40 × 1/2 = 20 m/s.
Why B is wrong: B is wrong because this assumes v ∝ T (linear). With linear scaling, 25/100 = 1/4, so v = 10. But v ∝ √T, giving √(1/4) = 1/2, so v = 20 m/s.
Why C is wrong: C is wrong because 30 m/s does not correspond to any standard scaling of v with T. The correct ratio is √(25/100) = 1/2, giving 20 m/s.
Why D is wrong: D is wrong because 80 m/s would mean doubling the speed. Reducing tension reduces speed — the factor is √(25/100) = 1/2, not 2.
MCQ 6Direct ApplicationPractice
The speed of sound in oxygen at a given temperature is v. The speed of sound in hydrogen at the same temperature and pressure is (assume both are diatomic):
Show answer and why every option is right or wrong
Answer: A. For an ideal gas, v = √(γRT/M). Both O₂ and H₂ are diatomic (same γ = 1.4), same T. The ratio is v_H₂/v_O₂ = √(M_O₂/M_H₂) = √(32/2) = √16 = 4. So the speed in hydrogen is 4v.
Why B is wrong: B is wrong because this would require M_O₂/M_H₂ = 4, i.e. a molar mass ratio of 4. The actual ratio is 32/2 = 16, giving √16 = 4.
Why C is wrong: C is wrong because this ignores the molar mass dependence. Speed of sound depends on 1/√M at constant γ and T. Hydrogen (M = 2) is much lighter than oxygen (M = 32).
Why D is wrong: D is wrong because this inverts the relationship. Lighter gases have higher sound speed (v ∝ 1/√M). Hydrogen is lighter, so sound travels faster, not slower.
MCQ 7CalculationPractice
A string of length 2.0 m and total mass 80 g is stretched with a tension of 100 N. A transverse wave of frequency 250 Hz is sent along the string. The wavelength of the wave is:
Show answer and why every option is right or wrong
Answer: B. B is correct. The linear mass density is μ = 0.080 kg / 2.0 m = 0.040 kg/m. The wave speed on the string is v = √(T/μ) = √(100/0.040) = √2500 = 50 m/s. The frequency is set by the source, so λ = v/f = 50/250 = 0.20 m.
Why A is wrong: A is wrong because 0.10 m comes from multiplying mass by length instead of dividing: μ = 0.080 × 2.0 = 0.16 kg/m, so v = √(100/0.16) = 25 m/s and λ = 25/250 = 0.10 m. Linear density is mass PER unit length.
Why C is wrong: C is wrong because 0.14 m comes from using the whole mass, 0.080 kg, as if it were μ: v = √(100/0.080) ≈ 35 m/s and λ ≈ 0.14 m. The wave speed depends on mass per metre, not on the total.
Why D is wrong: D is wrong because 10 m comes from leaving out the square root: v = T/μ = 2500 gives λ = 2500/250 = 10 m. The units flag it — T/μ is in m²/s², and only its square root is a speed.
MCQ 8CalculationPractice
The speed of sound in air at 0 °C is 330 m/s. If the temperature is raised to 819 °C, the speed of sound becomes approximately:
Show answer and why every option is right or wrong
Answer: D. D is correct. In an ideal gas the speed of sound goes as the square root of the absolute temperature. T₁ = 0 + 273 = 273 K and T₂ = 819 + 273 = 1092 K, so T₂/T₁ = 4 and v₂ = 330 × √4 = 660 m/s.
Why A is wrong: A is wrong because 330 m/s assumes the speed of sound does not depend on temperature. In an ideal gas v ∝ √T on the absolute scale, so heating the air raises it.
Why B is wrong: B is wrong because 467 m/s ≈ 330 × √2 is what doubling the ABSOLUTE temperature gives (T₂ = 546 K, i.e. 273 °C). Raising it to 819 °C makes T₂ = 1092 K, four times 273 K.
Why C is wrong: C is wrong because 572 m/s ≈ 330 × √3 corresponds to T₂/T₁ = 3, i.e. 819 K. That is the error of reading 819 °C as 819 K — forgetting to add 273.
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Speed Travelling Wave: quick recall before you leave
How do you solve a Speed Travelling Wave question? A worked example
Pattern: Wave speed on a string changes when tension changes (based on PYQ pattern NEET pattern: wave speed tension — observed NEET 2022).
- 1
Given
• v₁ = 60 m/s• T₁ = 36 N• T₂ = 144 N• μ unchanged
- 2
Required
• v₂ = ?
- 3
Concept
Wave speed on a string depends on tension and linear mass density: v = √(T/μ). When μ is constant, speed scales as the square root of tension.
- 4
Formula
v₂/v₁ = √(T₂/T₁)
- 5
Substitution
v₂/v₁ = √(144/36) = √4
- 6
Calculation
v₂/v₁ = 2
v₂ = 2 × 60 = 120 m/s
Note on exact values: 36 N, 144 N, and 60 m/s are problem-defined exact values. The integers 2 and 4 are exact. These do not limit significant figures. - 7
Final answer
v₂ = 120 m/s
- 8
Common trap
Treating v as directly proportional to T (linear scaling) would give v₂ = 60 × (144/36) = 60 × 4 = 240 m/s — exactly double the correct answer. The square-root dependence is the key: v ∝ √T.
- 9
Similar NEET-style question
A wave travels at speed v on a wire under tension T. The wire is replaced by one with the same material but double the diameter (hence 4× the cross-sectional area and 4× the linear mass density). If tension remains the same, what is the new speed? (Answer: v/2, since v = √(T/μ) and μ → 4μ gives v → v/√4 = v/2.)
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What to remember before solving Speed Travelling Wave questions
Formula
Speed of sound in gas
v = √(γ P/ρ) (Newton-Laplace formula). For air at STP: v ≈ 343 m/s. Increases with temperature: v ∝ √T.
-- NCERT Class 11 Physics, Ch. 14, p. 287Which Speed Travelling Wave formulas do you need for NEET?
2 formulas — click to collapse
Speed of sound in gas (Newton-Laplace)
Speed of sound in gas. Adiabatic index gamma, pressure P, density rho. Increases with sqrt(T).
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | speed of sound | m/s |
| gamma | adiabatic index | - |
| P | pressure | Pa |
| rho | density | kg/m^3 |
Valid when
- Ideal gas
- Adiabatic compression/expansion of sound waves
Wave speed on string
Speed of transverse wave on string under tension T, linear mass density mu.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | wave speed | m/s |
| T | tension | N |
| mu | linear mass density | kg/m |
Valid when
- Stretched uniform string
- Small amplitude
More in Oscillations and Waves: 6 exam traps and mistakes · 8 formulas · 5 question patterns from its other lessons.
Speed Travelling Wave questions from past NEET papers
1 question from NEET 2022. Answers verified against NTA official keys. — click to collapse
NEET 2022
11 : 2
21 : 1
3√2 : 1
41 : √2
NTA Answer: Option 4(final)
How does NEET ask about Speed Travelling Wave?
1 recurring pattern from past papers — click to collapse
Given tension change, find new wave speed on string. v ∝ sqrt(T).
Direct ApplicationEasy
Common distractors
uses linear tension scaling
Treats v ∝ T not sqrt(T)
Sources
NCERT refs: Class 11 Physics Chapter 14, p.285 | Class 11 Physics Chapter 14, p.287
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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