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JEE Mains Physics · Formula sheet

Waves formulas

11 formulas, 1 reference table and 29 common traps for JEE Mains Physics Waves, grouped by subtopic.

Full notes with worked examples

Wave Equation and Particle Motion

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Reading speed, frequency and wavelength from a wave equation

Wave speed from the equation

v=ωk=fλω=2πfk=2πλΔϕ=2πλ Δxv = \frac{\omega}{k} = f\lambda \qquad \omega = 2\pi f \qquad k = \frac{2\pi}{\lambda} \qquad \Delta\phi = \frac{2\pi}{\lambda}\,\Delta x

Writing the equation of a travelling wave

A travelling wave

y=f(x∓vt)y=Asin⁡(ωt−kx+ϕ0)k=ωvy = f(x \mp vt) \qquad y = A\sin(\omega t - kx + \phi_0) \qquad k = \frac{\omega}{v}

Particle velocity and intensity of a wave

Particle speed and intensity

vp,max⁡=Aωvp,max⁡v=Ak=2πAλI=P4πr2v_{p,\max} = A\omega \qquad \frac{v_{p,\max}}{v} = Ak = \frac{2\pi A}{\lambda} \qquad I = \frac{P}{4\pi r^{2}}

Common traps

k in cm⁻¹ gives a speed in cm/s

When x is measured in cm, ω/k is in cm/s. Answer options in m/s are then 100 times smaller. Convert k to m⁻¹, or convert the speed at the end.

Expand the common factor before reading

In y = A sin π(300t − x/60) the coefficient of t is 300π, not 300, and that of x is π/60. Here the π cancels in ω/k, but in the frequency f = ω/2π it does not: f is 150 Hz, not 300/2π.

Same signs mean travel along −x

In y = A sin(kx + ωt) the wave moves towards −x, so a velocity asked with its sign is negative. Opposite signs mean +x.

The to-and-fro distance is twice the amplitude

A particle that moves through a total of 6 cm swings 3 cm each side of its mean position. The amplitude is 3 cm, not 6 cm.

A crest at the origin is a cosine

At t = 0 a sine is zero at x = 0, so it cannot describe a crest there. When the stem puts a crest at the origin, the correct option is a cosine.

Particle velocity is not wave velocity

The wave speed ω/k is the same everywhere. The particle velocity ∂y/∂t changes from zero to Aω in every cycle. A question on 'maximum particle velocity' wants Aω.

Scale the radius, not the area or volume

Intensity goes as 1/r². If a sphere's surface area grows 9 times, r grows 3 times and I falls to one ninth, not to one eighty-first.

Wave Speed in Strings, Solids and Gases

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Speed of a transverse wave on a stretched string

Wave speed on a string

v=Tμμ=mLT=YA ΔLLv = \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=Yρv=γPρ=γRTMv2v1=T2T1v = \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}}

Common traps

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.

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.

Superposition and Standing Waves on Strings

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Resultant amplitude of two waves of the same frequency

Resultant amplitude

A2=A12+A22+2A1A2cos⁡ϕϕ=2πλ ΔA^{2} = A_1^{2} + A_2^{2} + 2A_1A_2\cos\phi \qquad \phi = \frac{2\pi}{\lambda}\,\Delta

Standing waves and harmonics of a string fixed at both ends

Harmonics of a string

fn=n2LTμfn+1−fn=v2Ly=2Acos⁡kx sin⁡ωtf_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}} \qquad f_{n+1} - f_n = \frac{v}{2L} \qquad y = 2A\cos kx\,\sin\omega t

Common traps

Amplitudes add only in phase

A₁ + A₂ is the largest possible resultant, reached only when the phase difference is zero or a whole number of cycles. At any other phase use the cosine formula.

A constant inside 2π( ) is not the phase

In sin 2π(x − vt + 0.75) the phase is 2π × 0.75 = 3π/2, not 0.75 rad. Multiply the constant by 2π before putting it into the cosine formula.

Frequency goes as the root of the hanging mass

The tension is mg and f ∝ √T, so f ∝ √m. To raise a sonometer's frequency by a factor of 3, the hanging mass must be 9 times larger, not 3 times.

The difference of two resonances is the fundamental

Two neighbouring resonances of a string differ by v/2L. That difference is the first harmonic itself; it does not tell you n until you divide either frequency by it.

A standing wave's amplitude depends on position

In y = 2A cos kx sin ωt, the factor 2A is only the amplitude at an antinode. At any other point the amplitude is |2A cos kx|, and at a node it is zero.

Organ Pipes and the Resonance Tube

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Resonance tube and end correction

Resonance tube

ln+e=(2n−1)λ4l2−l1=λ2e=0.3dl_n + e = \frac{(2n - 1)\lambda}{4} \qquad l_2 - l_1 = \frac{\lambda}{2} \qquad e = 0.3d

Harmonics and overtones of open and closed pipes

Vibrating systemFundamentalHarmonics presentFirst overtonekth overtone
String fixed at both endsv/2Lv/2L (λ=2L\lambda = 2L)All: 1, 2, 3, …2nd harmonic, 2f12f_1(k + 1)th harmonic
Pipe open at both endsv/2Lv/2L (λ=2L\lambda = 2L)All: 1, 2, 3, …2nd harmonic, 2f12f_1(k + 1)th harmonic
Pipe closed at one endv/4Lv/4L (λ=4L\lambda = 4L)Odd only: 1, 3, 5, …3rd harmonic, 3f13f_1(2k + 1)th harmonic
A closed pipe has no 2nd harmonic, so its first overtone is three times the fundamental.
A closed pipe of length L has the same fundamental as an open pipe of length 2L.

Common traps

Overtone number is not harmonic number

The first overtone is the first frequency above the fundamental. In an open pipe that is the 2nd harmonic; in a closed pipe it is the 3rd, because the 2nd does not exist.

Water makes the closed pipe shorter

Water poured into a closed pipe takes the place of air, so the vibrating column is shorter and the note is higher. Use the new air-column length, not the full pipe.

The end correction is added, not subtracted, to the column

The antinode is just outside the open end, so the effective length is l + e. The measured column is therefore shorter than λ/4 by e.

Use the difference to get the speed

v = 2f(l₂ − l₁) needs no end correction. Using v = 4f l₁ instead ignores e and gives a speed that is too low.

0.3 times the diameter, not the radius

The end correction is 0.3d, which is 0.6r. Using the radius in 0.3d halves the correction.

Beats and the Doppler Effect

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Beat frequency of two close frequencies

Beats

fbeat=∣f1−f2∣fbeat=v∣1λ1−1λ2∣f_{\text{beat}} = |f_1 - f_2| \qquad f_{\text{beat}} = v\left|\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right|

Doppler effect for a moving source and observer

Doppler effect

f′=f v±vov∓vsΔλλ=vcf' = f\,\frac{v \pm v_o}{v \mp v_s} \qquad \frac{\Delta\lambda}{\lambda} = \frac{v}{c}

Doppler effect for an echo from a wall

Echo from a wall

fecho=f v+uv−uf_{\text{echo}} = f\,\frac{v + u}{v - u}

Common traps

Beats in a time are not beats per second

Ten beats in 2 s is a beat frequency of 5 Hz. Using 10 as the frequency difference doubles every answer that follows.

Test both candidate frequencies

A beat count gives two possible frequencies, one above and one below. Only the change after loading or filing tells which one is right; never take the higher one by default.

The envelope frequency is half the beat frequency

In a cos(Δt) cos(ω̄t) the envelope cos(Δt) has frequency Δ/2π, but the loudness peaks twice in each of its cycles. The beat frequency is Δ/π.

Set each sign by 'towards raises'

Do not memorise one sign pattern. For the observer term, motion towards the source adds to v on top; for the source term, motion towards the observer subtracts from v below. Check that the answer rises when they close in.

A chase is not an approach

When the observer follows the source in the same direction, the observer moves towards the source but the source moves away from the observer. Both signs are +, and the two effects partly cancel.

An echo is shifted twice

Applying the Doppler formula once, for the source only, misses the second shift as the driver moves into the reflected sound. The echo heard by an approaching driver is f(v + u)/(v − u).

The driver hears his own horn unshifted

The driver moves with the horn, so the direct sound reaches him at its true frequency. A beat or a 'change' between horn and echo is measured from f, not from a shifted value.

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