JEE Mains Physics · Waves

Wave Speed in Strings, Solids and Gases

Wave speed is √(restoring property ÷ inertia): √(T/μ) on a string, √(Y/ρ) in a rod and √(γP/ρ) = √(γRT/M) in a gas.

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Why this matters

Eleven PYQs, three of them asking for a number, and three from 2026. Four are about transverse waves on a stretched string, two about longitudinal waves in a solid rod, and five about sound in gases: the effect of temperature and molar mass, and why sound is faster in solids. Each is one square root; the marks go on units such as g/cm and on using °C where kelvin is needed.

Concept 1 of 2: Speed of a transverse wave on a stretched string

Tension pulls a displaced bit of string back; its mass per unit length resists. A tighter or lighter string carries a wave faster. Because the speed is a square root, four times the tension only doubles the speed.

Definition

$v = \sqrt{\dfrac{T}{\mu}}$ , where $\mu$ is the mass per unit length in kg/m. So $T = \mu v^{2}$ .
$\mu = \dfrac{\text{mass}}{\text{length}}$ ; for a wire of density $\rho$ and cross-section A, $\mu = \rho A$ .
1 g/m = $10^{-3}$ kg/m and 1 g/cm = 0.1 kg/m.
Time for a pulse to cross a length L: $t = L/v = L\sqrt{\mu/T}$ . Two strings joined under the same tension: $t \propto L\sqrt{\mu}$ .
If the speed comes from an equation, $v = \omega/k$ , then $T = \mu(\omega/k)^{2}$ .
Tension from a stretched wire: $T = \dfrac{YA\,\Delta L}{L}$ , so the extension is $\Delta L = \dfrac{TL}{YA}$ .

Wave speed on a string

v = \sqrt{\frac{T}{\mu}} \qquad \mu = \frac{m}{L} \qquad T = \frac{YA\,\Delta L}{L}

Worked example

A 2 m string of mass 5 g is held at a tension of 100 N. Find the wave speed, the time a pulse takes to cross it, and the tension that would double the speed.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q18Moderate

Example 1 · Waves · Wave Speed in Strings, Solids and Gases

Two strings (A, B) having linear densities

\mu_{A}= 2 \times10^{- 4}\text{ }kg/m

and

\mu_{B}= 4 \times10^{- 4}\text{ }kg/m

and lengths

L_{A}= 2.5\text{ }m

and

L_{B}= 1.5\text{ }m

respectively are joined. Free ends of A and B are tied to two rigid supports C and D, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from C and D ends, take time

t_{1}

and

t_{2}

, respectively, to reach the joint. The ratio

t_{1}/t_{2}

is :

μ is mass per length, not density

The string formula needs kg per metre of string. A density in kg/m³ must first be multiplied by the cross-section area to give μ.

Grams per centimetre is not grams per metre

1 g/cm is 100 g/m, which is 0.1 kg/m. Converting g/cm as if it were g/m makes μ a hundred times too small and the tension a hundred times too small.

Drill 3 more on speed of a transverse wave on a stretched string

Concept 2 of 2: Speed of sound in solids and gases

Sound is a longitudinal wave, and its speed is √(elastic modulus ÷ density). In a rod the modulus is Young's modulus. In a gas the compressions are too fast for heat to flow, so the modulus is γP, and P/ρ = RT/M. The speed in a gas then depends only on temperature and on the gas, not on the pressure.

Definition

Solid rod: $v = \sqrt{\dfrac{Y}{\rho}}$ . Small changes: $\dfrac{\Delta v}{v} = \dfrac{1}{2}\dfrac{\Delta Y}{Y} - \dfrac{1}{2}\dfrac{\Delta\rho}{\rho}$ .
Gas (Newton–Laplace): $v = \sqrt{\dfrac{\gamma P}{\rho}} = \sqrt{\dfrac{\gamma RT}{M}}$ , with T in kelvin and M in kg/mol.
$v \propto \sqrt{T}$ : compare temperatures in kelvin, never in °C.
At the same temperature and the same γ, $v \propto \dfrac{1}{\sqrt{M}}$ .
$\gamma = 1 + \dfrac{2}{f}$ : monatomic 5/3; rigid diatomic 7/5; non-linear polyatomic (f = 6) 4/3.
Changing the pressure at constant temperature changes P and ρ together: v does not change.
Sound is fastest in solids because their elastic modulus is far larger; their higher density does not cancel it.

Speed of sound

v = \sqrt{\frac{Y}{\rho}} \qquad v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma RT}{M}} \qquad \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}

Worked example

Find the speed of sound in nitrogen at 27 °C (

\gamma = 1.4

R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}

M = 0.028\ \text{kg/mol}

). Then find it at 402 °C.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q25Moderate

Example 2 · Waves · Wave Speed in Strings, Solids and Gases

The velocity of sound in air is doubled when the temperature is raised from

0^{\circ}C

\alpha^{\circ}C

. The value of

\alpha

\_\_\_\_

Temperatures go in kelvin

v ∝ √T only with absolute temperature. Doubling the speed from 27 °C means 4 × 300 K = 1200 K, which is 927 °C, not 4 × 27 °C.

Pressure alone does not change the speed

At a fixed temperature, P/ρ stays the same, so √(γP/ρ) does not move. Only temperature, γ and molar mass change the speed of sound in an ideal gas.

Solids are faster because of the modulus

A solid is denser than a gas, which alone would slow sound. Its elastic modulus is larger by a much bigger factor, so sound is faster in solids. Gases have the smaller modulus, not the larger.

Drill 6 more on speed of sound in solids and gases

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Speed of a transverse wave on a stretched string
Wave speed on a string
$v = \sqrt{\frac{T}{\mu}} \qquad \mu = \frac{m}{L} \qquad T = \frac{YA\,\Delta L}{L}$
Speed of sound in solids and gases
Speed of sound
$v = \sqrt{\frac{Y}{\rho}} \qquad v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma RT}{M}} \qquad \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$

Watch out for (5)

μ is mass per length, not density
→ Speed of a transverse wave on a stretched string
Grams per centimetre is not grams per metre
→ Speed of a transverse wave on a stretched string
Temperatures go in kelvin
→ Speed of sound in solids and gases
Pressure alone does not change the speed
→ Speed of sound in solids and gases
Solids are faster because of the modulus
→ Speed of sound in solids and gases

Test yourself on Waves

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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