JEE Main chapters · Class 11 · Physics

Oscillations and Waves: JEE Main Physics formulas and notes

Oscillations and Waves is two chapters in one and together they give about 3 questions in JEE Mains, with SHM and springs, pendulums, standing waves in pipes, beats and the Doppler effect being the favourites. Most of the questions use a few formulas with a clear condition, so learning when each holds is half the work. About 12 hours is enough.

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Quick revision

• SHM: a = −ω²x, v = ω√(A² − x²), T = 2π/ω
• E = ½kA² = ½mω²A², KE = ½k(A² − x²), KE = PE at x = A/√2
• Spring: T = 2π√(m)/(k); parallel k₁ + k₂; series k₁k₂ ÷ (k₁ + k₂); k ∝ 1/L
• Simple pendulum T = 2π√(L)/(g(eff)); lift g ± a; liquid g(1 − σ/ρ)
• Physical pendulum 2π√(I)/(mgd); torsion 2π√(I)/(C)
• Float 2π√(h)/(g); U-tube 2π√(L)/(2g); tunnel 2π√(R)/(g)
• Same-direction SHMs: A² = A₁² + A₂² + 2A₁A₂ cos δ
• Damped: A = A₀e^−bt/2m, E = E₀e^−bt/m; resonance at ω(d) = ω₀
• Wave: y = A sin(kx − ωt), v = fλ = ω/k, Δφ = 2πΔx/λ
• String v = √(T)/(μ); gas v = √(γP)/(ρ) = √(γRT)/(M); rod v = √(Y)/(ρ)
• β = 10 log₁₀(I/I₀), I ∝ 1/r², I ∝ A²
• Beats f(b) = |f₁ − f₂|; I(max)/I(min) = ((A₁ + A₂)/(A₁ − A₂))²
• String and open pipe: f = nv/2L; closed pipe: (2n − 1)v/4L
• End correction 0.6r; resonance tube v = 2f(L₂ − L₁)
• Doppler: f′ = f(v − v(o))/(v − v(s)), + from S to O
• Echo from a reflector: f″ = f(v + u)/(v − u)

Formulas with solved examples

SHM and energy

Definition and basic quantities of SHM

• F = −kx, so a = −ω²x
• the restoring force (or acceleration) is proportional to the displacement from the mean position and directed towards it
• ω = √(k)/(m), T = (2π)/(ω), f = (1)/(T)
• a graph of a against x is a straight line of negative slope −ω²
• the motion is periodic and the equation is linear in x

How to apply. To prove a motion is SHM, write the net force (or torque) about the equilibrium position and show that it equals −(constant) × displacement; the constant gives ω². This holds for small displacements only in a pendulum.

Example. x = 0.02 m and ω = 10 rad/s: a = −ω²x = −100 × 0.02 = −2 m/s², directed towards the mean position.

Displacement equation and initial phase

• x = A sin(ωt + φ) or x = A cos(ωt + φ′), with φ′ = φ − π/2
• A = amplitude, (ωt + φ) = phase, φ = initial phase (epoch)
• from the starting condition: x₀ = A sin φ, v₀ = Aω cos φ, so tan φ = (ωx₀)/(v₀) and A = √(x₀² + (v₀²)/(ω²))
• x = a sin ωt + b cos ωt is also SHM with A = √(a² + b²)

How to apply. Compare the given equation with the standard form to read A, ω and φ. When the position and velocity at t = 0 are given, find φ from tan φ and put A from the second formula; take care of the quadrant.

Example. x = 0.1 sin(20πt + π/6) m: A = 0.1 m, f = 10 Hz, T = 0.1 s; at t = 0, x = 0.05 m and v = 0.1 × 20π × cos 30° = 5.44 m/s. Starting with x₀ = 0.03 m, v₀ = 0.4 m/s and ω = 10: A = 0.05 m, tan φ = 0.75, φ = 36.9°.

Velocity and acceleration, and their phases

• v = Aω cos(ωt + φ) = ±ω√(A² − x²)
• v(max) = Aω at x = 0
• a = −Aω² sin(ωt + φ) = −ω²x
• a(max) = Aω² at the extremes, where v = 0
• velocity leads displacement by π/2, acceleration leads displacement by π (acceleration is in antiphase with x)
• the v–x graph is an ellipse x²/A² + v²/(Aω)² = 1

How to apply. Use v = ω√(A² − x²) when the position is given; use the maximum values for the extreme questions. At the mean position the speed is largest and the acceleration is zero, and at the extreme the opposite holds.

Example. A = 0.1 m, ω = 10 rad/s, x = 0.06 m: v = 10 × √(0.01 − 0.0036) = 10 × 0.08 = 0.8 m/s, and a = −ω²x = −6 m/s². The maximum values are 1 m/s and 10 m/s².

Time between two positions and the circular-motion model

• SHM is the projection of uniform circular motion of radius A and angular speed ω on a diameter
• t = (1)/(ω) (θ₂ − θ₁) with θ = sin⁻¹(x)/(A) measured from the mean position
• mean → A : (T)/(4)
• mean → (A)/(2) : (T)/(12)
• mean → (A)/(√2) : (T)/(8)
• (A)/(2) → A : (T)/(6)
• extreme → extreme : (T)/(2)
• x = A cos ωt starting from an extreme

How to apply. Convert the positions to angles on the reference circle, find the angle swept and divide by ω. The time is longer where the particle moves slowly (near the extremes) than near the mean position for the same distance.

Example. T = 2.4 s: x = 0 to x = A/2 takes 2.4 ÷ 12 = 0.2 s, and x = A/2 to x = A takes 2.4 ÷ 6 = 0.4 s. The particle takes twice as long over the second half-amplitude.

Energy of an oscillator

• KE = ½mv² = ½mω²(A² − x²) = ½k(A² − x²)
• PE = ½kx² = ½mω²x²
• E = KE + PE = ½kA² = ½mω²A² = constant
• KE is greatest (E) at x = 0 and zero at x = ±A; PE is the opposite
• KE = PE at x = ± (A)/(√2)
• holds only when there is no damping (energy conserved)

How to apply. Use the energy form when a question gives positions and not times. To find where KE is a fraction of E, use KE ÷ E = 1 − x²/A² . The total energy depends on k and A only, not on the phase.

Example. m = 0.2 kg, ω = 10 rad/s, A = 0.1 m: E = ½ × 0.2 × 100 × 0.01 = 0.1 J. At x = 0.06 m: PE = 0.036 J, KE = 0.064 J; at x = A/√2 = 0.0707 m KE = PE = 0.05 J.

KE and PE curves and their averages

• KE and PE each change with a period (T)/(2) (frequency 2f) while x changes with period T
• the KE–x and PE–x graphs are parabolas, KE an inverted one, crossing at x = ± (A)/(√2) at the height (E)/(2)
• average over time: ⟨KE⟩ = ⟨PE⟩ = (E)/(2)
• average over displacement: ⟨PE⟩ = (E)/(3), ⟨KE⟩ = (2E)/(3)
• the PE–x graph is the parabola ½kx², with the total energy a horizontal line

How to apply. Decide which average is asked. Over time the two energies share E equally; an average over displacement gives every position equal weight, but the time average weights positions by the time spent there, and the particle lingers near the extremes where PE is large; so PE averages E/3 over displacement but E/2 over time. The energy graph against time is KE = E cos² (ωt) and PE = E sin² (ωt) for a start at the mean position.

Example. In a SHM of E = 0.12 J, the time average KE = 0.06 J, but the average of KE over the positions (distance average) is 0.08 J; KE becomes zero twice in every period T.

Sudden change of mass or amplitude

• a mass m dropped on a mass M at the mean position (sticks): momentum conserved, v′ = (Mv)/(M + m), A′ = A√(M)/(M + m), T′ = T√(M + m)/(M)
• dropped at the extreme: v = 0, so A′ = A and only T changes
• energy lost in the first case (inelastic)
• at the extreme the total energy ½kA² stays the same

How to apply. Decide the instant (for a block on a horizontal smooth surface; for a vertical spring the equilibrium position also shifts by mg/k): at the mean position the speed is maximum and momentum is conserved; at an extreme the speed is zero and nothing changes except the period. New A follows from E′ = ½(M + m)v′².

Example. M = 2 kg oscillating with A = 0.1 m, a 1 kg lump sticks at the mean position: A′ = 0.1 × √(2/3) = 0.0816 m and T increases by √(3/2) = 1.22 times. At the extreme the amplitude stays 0.1 m.

Spring systems

Spring–mass oscillator

• F = −kx
• T = 2π√(m)/(k), ω = √(k)/(m)
• T depends on m and k, not on A or on g
• valid for an ideal (massless) spring and a smooth surface
• T ∝ √m, T ∝ 1/√k
• the same T for a horizontal, an inclined or a vertical arrangement (only the position of the mean point changes)

How to apply. Identify k (of the spring or the combination) and the mass that moves. If the system is accelerated as a whole (lift, car) the period does not change; only the equilibrium position shifts.

Example. m = 0.5 kg, k = 200 N/m: T = 2π√(0.5 ÷ 200) = 0.314 s. Quadrupling the mass doubles T to 0.628 s.

Springs in parallel and series

• parallel (both springs stretched by the same x): k = k₁ + k₂
• series (same force): (1)/(k) = (1)/(k₁) + (1)/(k₂), so k = (k₁k₂)/(k₁ + k₂)
• a mass between two springs on opposite sides (on a smooth line): k = k₁ + k₂
• then T = 2π√(m)/(k)
• series gives a smaller k (larger T), parallel a larger k

How to apply. Decide how the springs are connected by asking whether they have the same extension (parallel) or carry the same force (series). A mass held between two springs on opposite sides is the parallel case, since both springs push the same way.

Example. k₁ = 100 N/m and k₂ = 300 N/m: parallel k = 400 N/m, series k = 100 × 300 ÷ 400 = 75 N/m. With m = 1 kg the periods are 0.314 s and 0.726 s.

Cutting a spring

• k × L = constant, so k ∝ (1)/(L) (natural length L, same wire)
• a spring cut into n equal parts: each part has k′ = nk
• cut in the ratio p : q : the pieces have k₁ = k(p + q)/(p), k₂ = k(p + q)/(q)
• the pieces joined back in parallel give a spring with k′ = k₁ + k₂

How to apply. A longer spring is softer. Give the cut length as a fraction of L, take k′ = k ÷ fraction, then use T = 2π√(m/k′). The period of a mass on a part is shorter than on the whole spring.

Example. k = 120 N/m cut in the ratio 1 : 2: the shorter piece (1/3 of L) has k = 360 N/m and the longer piece (2/3) has k = 180 N/m. Cut into 4 equal pieces: each has 480 N/m.

Vertical spring

• at equilibrium kΔ = mg, extension Δ = (mg)/(k)
• T = 2π√(m)/(k) = 2π√(Δ)/(g)
• the motion is SHM about the equilibrium point (not about the natural length)
• released from the natural length: A = Δ and the greatest extension = 2Δ = (2mg)/(k)
• gravity PE and spring PE combine to ½kx² measured from the equilibrium position

How to apply. Measure x from the equilibrium point, not from the natural length; then the equation is again F = −kx. The static extension Δ gives T directly without k or m.

Example. A mass hangs with a static extension of 0.1 m: T = 2π√(0.1 ÷ 10) = 0.628 s. If it is released from the natural length it oscillates with A = 0.1 m between extensions 0 and 0.2 m.

Two masses on a spring, and a block on a vibrating platform

• two masses m₁ and m₂ joined by a spring (no external force): T = 2π√(μ)/(k) with μ = (m₁m₂)/(m₁ + m₂)
• a block on a platform vibrating vertically stays in contact if ω²A ≤ g, i.e. A ≤ (g)/(ω²)
• on a horizontal platform without slipping: ω²A ≤ μ(s)g

How to apply. The reduced mass replaces m when both ends are free to move. For a contact problem, find the condition at the position of greatest acceleration and compare with g or μg.

Example. Two 1 kg blocks joined by a spring of k = 100 N/m: μ = 0.5 kg, T = 2π√(0.5 ÷ 100) = 0.444 s. A platform vibrating vertically at f = 2 Hz: the greatest amplitude without the block leaving is g ÷ ω² = 10 ÷ (4π)² = 0.063 m.

Pendulums

Simple pendulum

• T = 2π√(L)/(g), ω = √(g)/(L)
• valid for a point mass on a light inextensible string, small angles (about 10° or less, so sin θ ≈ θ)
• independent of mass and (for small amplitude) of the amplitude
• T ∝ √L, T ∝ 1/√g
• L is the distance from the point of suspension to the centre of the bob

How to apply. Apply T = 2π√(L/g) with the length from the pivot to the centre of the bob. For ratios write T ∝ √(L/g) and cancel everything which does not change.

Example. L = 0.9 m, g = 10 m/s²: T = 2π√0.09 = 1.885 s. Doubling the length increases T by √2 = 1.41 times.

Effective g in an accelerating frame

• T = 2π√(L)/(g(eff)) with g(eff) = g + a (lift going up with acceleration a, or down while retarding)
• g(eff) = g − a (lift going down with acceleration a < g)
• free fall: g(eff) = 0, T → ∞
• horizontal acceleration a (a car): g(eff) = √(g² + a²), the mean position is tilted by tan⁻¹(a)/(g) from the vertical
• pendulum fixed in a trolley or box sliding freely down a smooth incline of angle α: g(eff) = g cos α (the equilibrium string is perpendicular to the incline)

How to apply. Write the pseudo force ma in the frame of the pendulum, add it to mg as vectors to get the effective gravity, and use its magnitude in T. The mean position of the string lies along the effective gravity.

Example. In a lift accelerating upwards at 2 m/s², g(eff) = 12 m/s²: T′ = T√(10 ÷ 12) = 0.913T. In a car accelerating at 10 m/s² horizontally g(eff) = 14.1 m/s².

Pendulum in a liquid or an electric field

• a bob of density ρ in a liquid of density σ (σ < ρ): g(eff) = g(1 − (σ)/(ρ)), T′ = T / √(1 − σ/ρ) (buoyancy as an upward force; neglect the drag)
• a bob of positive charge q in a downward field E: g(eff) = g + (qE)/(m); in an upward field: g − (qE)/(m) (oscillates only if qE < mg; for a negative charge swap the two cases)
• a horizontal field: g(eff) = √(g² + (qE/m)²)
• if g(eff) = 0 (the net extra force cancels the weight), the bob does not oscillate

How to apply. Write all the extra constant forces on the bob and divide the net extra force by the mass to get the change in g. Buoyancy per unit mass is gσ/ρ, so it lowers g(eff) and raises T.

Example. A bob of density 2000 kg/m³ swinging in water (1000 kg/m³): g(eff) = 5 m/s² and T′ = T√2 = 1.41T. A bob of 10 g with charge 10⁻⁴ C in a downward field of 1000 V/m: g(eff) = 10 + 10 = 20 m/s²; T′ = T/√2.

Length comparable to the Earth’s radius; a particle in a tunnel

• T = 2π√(1)/(g(1/L + 1/R))
• L ≪ R gives 2π√(L/g)
• L → ∞ gives T = 2π√(R)/(g) ≈ 84 min
• a particle in a straight tunnel through the Earth (or a fall through the centre): a = −(g)/(R)x, T = 2π√(R)/(g) ≈ 84 min
• the same for a satellite just above the surface (orbital period)
• valid for a uniform Earth

How to apply. Use the full formula only when the question says L is comparable with R. A tunnel through a uniform Earth gives SHM because the gravity inside is proportional to distance from the centre.

Example. R = 6.4 × 10⁶ m, g = 10 m/s²: T(∞) = 2π√(6.4 × 10⁵) = 5026 s = 83.8 min. For L = R: T = 2π√(1 ÷ (10 × 2 ÷ 6.4 × 10⁶)) = 3554 s = 59 min.

Seconds pendulum and a pendulum clock

• seconds pendulum: T = 2 s, so L = (g)/(π²) ≈ 1 m (0.99 m at g = 9.8)
• clock with length changed by ΔL: (ΔT)/(T) = (ΔL)/(2L) − (Δg)/(2g)
• for a rise of temperature Δθ: (ΔT)/(T) = (αΔθ)/(2), so the clock loses (1)/(2)αΔθ × 86400 s per day
• a clock runs slow when T increases (longer L, smaller g at height or depth, hotter)

How to apply. Compare the period of the clock with the correct period. Time lost per day = (ΔT ÷ T) × 86400 s when the period is too long. Use the small-change form only for changes of a few per cent.

Example. Brass pendulum, α = 1.2 × 10⁻⁵ K⁻¹, temperature rise of 20 °C: loses ½ × 1.2 × 10⁻⁵ × 20 × 86400 = 10.4 s per day. On a planet with g one sixth of Earth’s, the seconds pendulum is 0.99 ÷ 6 = 0.165 m long.

Other oscillators

Physical (compound) pendulum

• T = 2π√(I)/(mgd) where I is about the pivot and d = pivot to the centre of mass
• I = I(cm) + md²
• equivalent simple pendulum length L(eq) = (I)/(md)
• uniform rod about its end: T = 2π√(2L)/(3g)
• ring about the rim: T = 2π√(2R)/(g)
• disc about the rim: T = 2π√(3R)/(2g)
• valid for small angles

How to apply. Find I about the pivot with the parallel-axis theorem and the distance d to the centre of mass. Then the formula follows. A pivot and a point at distance L(eq) on the other side of the centre of mass are interchangeable.

Example. Uniform rod of length 1 m pivoted at one end: T = 2π√(2 ÷ 30) = 1.62 s and L(eq) = 2/3 m. A ring of radius 0.5 m on a nail at its rim: T = 2π√(1 ÷ 10) = 1.99 s.

Torsional pendulum and angular SHM

• restoring torque τ = −Cθ, so α = −(C)/(I)θ
• T = 2π√(I)/(C), ω = √(C)/(I)
• C = torsional constant of the wire = (πηr⁴)/(2l)
• θ(max) is the amplitude, angular speed ω(max)·θ₀
• energy: ½Cθ₀²
• small angles only for a wire in the elastic limit

How to apply. The same method as in a linear SHM with I in place of m and C in place of k. Both the angle and the angular velocity have SHM equations: θ = θ₀ sin(ωt + φ).

Example. A disc with I = 0.02 kg m² on a wire with C = 0.5 N m/rad: T = 2π√(0.02 ÷ 0.5) = 1.26 s. If θ₀ = 0.1 rad, the greatest angular speed is 0.1 × 5 = 0.5 rad/s.

Floating body and liquid in a U-tube

• a cylinder floating upright, pushed down by x: restoring force −Aρ(l)g·x, so T = 2π√(m)/(Aρ(l)g) = 2π√(h)/(g) (h = submerged depth)
• the body must stay in contact with the liquid, no drag
• liquid in a U-tube (total length L, uniform bore): restoring force 2Aρg·y, so T = 2π√(L)/(2g)
• valid for small displacement and ideal liquid; a piston of gas: k = (γPA²)/(V) (adiabatic)

How to apply. Find the extra restoring force for a displacement x and divide by the mass moved: the coefficient of x is ω². For the U-tube the column moves as a whole with a level difference 2y.

Example. A cylinder with 0.1 m submerged: T = 2π√(0.1 ÷ 10) = 0.628 s. Liquid of total length 0.5 m in a U-tube: T = 2π√(0.5 ÷ 20) = 0.993 s.

Composition, damped and forced

Two SHMs along the same line

• x₁ = A₁ sin ωt, x₂ = A₂ sin(ωt + δ), same ω: x = A sin(ωt + φ) with A = √(A₁² + A₂² + 2A₁A₂ cos δ) and tan φ = (A₂ sin δ)/(A₁ + A₂ cos δ)
• δ = 0: A = A₁ + A₂ (maximum)
• δ = π: A = |A₁ − A₂| (minimum)
• different frequencies: not simple SHM (beats if close)

How to apply. Treat each SHM as a rotating vector (phasor) and add them as vectors with the angle δ between. The result is again SHM of the same frequency.

Example. A₁ = 3 cm, A₂ = 4 cm: δ = 90° gives A = 5 cm, δ = 60° gives √37 = 6.08 cm, δ = 0 gives 7 cm and δ = 180° gives 1 cm.

Two perpendicular SHMs (briefly Lissajous)

• x = A sin ωt, y = B sin(ωt + δ), same frequency
• δ = 0 or π: a straight line y = ±(B)/(A)x
• δ = (π)/(2): ellipse (x²)/(A²) + (y²)/(B²) = 1 (circle if A = B)
• other δ: a tilted ellipse
• frequency ratio 1 : 2 or other simple ratio gives Lissajous figures (f(x) : f(y) = the number of points where the curve touches a vertical side : the number of points where it touches a horizontal side)

How to apply. Eliminate t between the two equations. For δ = 90° use sin² + cos² = 1. The direction of rotation on the ellipse follows from the sign of the phase difference.

Example. x = 3 sin ωt and y = 4 sin(ωt + π/2): path x²/9 + y²/16 = 1, an ellipse with semi-axes 3 and 4. With δ = 0 the path is the line y = (4/3)x.

Damped oscillation

• m (d²x)/(dt²) + b (dx)/(dt) + kx = 0 (damping force −bv)
• x = A₀ e^−bt/2m cos(ω′t + φ)
• ω′ = √(ω₀² − ((b)/(2m))²)
• amplitude A = A₀e^−bt/2m, energy E = E₀e^−bt/m
• time constant (energy) = (m)/(b)
• valid for light damping (b < 2√(km)); T is slightly larger than the undamped one

How to apply. Amplitude decays as an exponential and the energy decays twice as fast (E ∝ A²). For the time to reach a fraction of the amplitude, take logarithms. Heavy damping (b > 2√(km)) gives no oscillation.

Example. m = 0.2 kg, b = 0.05 kg/s: the energy falls to 1/e in m ÷ b = 4 s; the amplitude halves in (2m ÷ b) ln 2 = 8 × 0.693 = 5.55 s. With k = 20 N/m, ω₀ = 10 and ω′ = 9.9992, nearly the same.

Forced oscillation and resonance

• driving force F₀ cos ω(d)t
• after the transients, the system oscillates at the driving frequency ω(d), not at ω₀
• A = (F₀)/(√(m²(ω₀² − ω(d)²)² + b²ω(d)²))
• resonance: ω(d) ≈ ω₀, the amplitude is largest, A = (F₀)/(bω₀) at ω(d) = ω₀
• less damping gives a larger and sharper peak; at resonance the driving force is in phase with the velocity (energy absorbed at the greatest rate)

How to apply. Compare ω(d) with ω₀; the amplitude is largest when they are equal. Far below ω₀ the amplitude is F₀ ÷ k; far above it is small. If damping is zero the amplitude at resonance grows without limit.

Example. m = 1 kg, k = 100 N/m (ω₀ = 10 rad/s), F₀ = 1 N, b = 0.5 kg/s: at ω(d) = 10 rad/s, A = 1 ÷ (0.5 × 10) = 0.2 m. At ω(d) → 0 the amplitude is 1 ÷ 100 = 0.01 m.

Wave equation and speed

Progressive wave equation

• y = A sin(kx − ωt + φ) travels towards +x; y = A sin(kx + ωt + φ) towards −x
• k = (2π)/(λ) (wave number), ω = 2πf = (2π)/(T)
• v = fλ = (ω)/(k)
• the argument (kx − ωt + φ) is the phase
• sign of the x and t terms opposite: moving along +x
• y = f(x − vt) in general
• wave equation (∂²y)/(∂t²) = v²(∂²y)/(∂x²)

How to apply. Compare the given y with the standard form to read A, k and ω; then λ, f and v follow. The speed and direction come from the ratio ω/k and the relative sign of the x and t terms.

Example. y = 0.05 sin(2πx − 100πt) m: A = 0.05 m, λ = 1 m, f = 50 Hz, T = 0.02 s, v = 100π ÷ 2π = 50 m/s along +x.

Particle velocity and wave velocity

• v(particle) = (∂y)/(∂t) = −Aω cos(kx − ωt + φ)
• v(max, particle) = Aω
• v(particle) = −v(∂y)/(∂x) (particle velocity = −wave speed × slope)
• the wave speed v = ω/k is fixed by the medium; the particle speed changes with position and time
• particle motion: transverse (waves on a string) or longitudinal (sound)

How to apply. Differentiate with respect to t at constant x for the particle, with respect to x at constant t for the slope. The particle speed Aω equals the wave speed when A = 1/k.

Example. For the wave above, Aω = 0.05 × 100π = 15.7 m/s is the greatest particle speed, while the wave speed is 50 m/s. A particle on the crest is momentarily at rest.

Phase difference, path difference and time difference

• Δφ = (2π)/(λ)Δx = kΔx
• Δφ = ωΔt = (2π)/(T)Δt
• two points Δx apart on a travelling wave: the later one lags in phase by kΔx
• Δx = λ gives 2π (in phase)
• Δx = (λ)/(2) gives π (opposite phase)
• a point in a standing wave: phase 0 or π only

How to apply. Use Δφ = (2π/λ) × path difference whenever two points on the same wave, or two sources, are compared. For time difference use Δφ = 2π Δt/T. The phase changes by π at a reflection from a denser (fixed) end.

Example. For f = 680 Hz and v = 340 m/s, λ = 0.5 m. Points 0.125 m apart differ in phase by (2π/0.5) × 0.125 = π/2; the time lag between them is 0.125 ÷ 340 = 0.37 ms.

Speed of a transverse wave on a string

• v = √(T)/(μ) where T = tension, μ = mass per unit length = (m)/(L)
• independent of the frequency, depends only on the string and its tension
• v ∝ √T, v ∝ 1/√μ, and for the same material v ∝ 1/r
• a heavy rope hanging vertically: v = √(gx), where x is the distance from the free end, time to run along it t = 2√(L)/(g)
• valid for small amplitude, a perfectly flexible string

How to apply. Find μ first (total mass divided by length) and T in newton. If the tension changes with position (a hanging rope), use v = √(gx) at that place and integrate for the time.

Example. T = 100 N, μ = 0.01 kg/m: v = √(100 ÷ 0.01) = 100 m/s. A hanging rope of length 5 m: a pulse takes 2√(5/10) = 1.41 s to travel along it.

Speed of waves in solids and liquids

• longitudinal wave in a solid rod: v = √(Y)/(ρ)
• in a liquid or a gas: v = √(B)/(ρ)
• transverse wave in a bulk solid: v = √(G)/(ρ)
• v(solid) > v(liquid) > v(gas) in general
• Y, B, G are the elastic moduli and ρ is the density

How to apply. Choose the modulus according to the kind of deformation: Young’s modulus for a thin rod, bulk modulus for a fluid and shear modulus for a transverse wave in a solid. Use SI units for ρ.

Example. Steel (Y = 2 × 10¹¹ Pa, ρ = 8000 kg/m³): v = √(2.5 × 10⁷) = 5000 m/s. Water (B = 2.2 × 10⁹ Pa): v = √(2.2 × 10⁶) = 1483 m/s.

Speed of sound in a gas (Newton–Laplace)

• v = √(γP)/(ρ) = √(γRT)/(M) (the change is adiabatic, not isothermal)
• Newton’s formula v = √(P)/(ρ) gave 280 m/s, too low
• v ∝ √T (T in kelvin)
• v ∝ (1)/(√M) at the same T
• independent of the pressure at a constant temperature (P/ρ is constant)
• v(T) ≈ v₀√(T)/(273)
• rises about 0.6 m/s per °C near room temperature

How to apply. Use kelvin for T. Compare speeds in different gases by v ∝ √(γ/M). A change in pressure alone does not change the speed if T is constant; the speed rises with humidity because the density falls.

Example. Air (γ = 1.4, P = 1.013 × 10⁵ Pa, ρ = 1.29 kg/m³): v = √(1.4 × 1.013 × 10⁵ ÷ 1.29) = 332 m/s. If 340 m/s at 300 K, then at 400 K: 340 × √(4/3) = 392.6 m/s.

Power carried by a wave on a string

• P(average) = ½μω²A²v = 2π²f²A²μv
• energy per unit length = ½μω²A²
• P ∝ A² f² and, at the same tension, ∝ √μ
• for a point source in a medium the intensity falls as 1/r², so A ∝ 1/r

How to apply. Find v = √(T/μ) first, then put A, f and μ. Doubling the amplitude or the frequency makes the power four times as large.

Example. μ = 0.01 kg/m, T = 100 N (v = 100 m/s), A = 2 mm, f = 50 Hz: P = ½ × 0.01 × (100π)² × (0.002)² × 100 = 0.197 W.

Intensity and superposition

Intensity and loudness (decibel)

• I = (P)/(A) (power per unit area), for a point source I = (P)/(4πr²) so I ∝ (1)/(r²) and amplitude ∝ (1)/(r)
• I = ½ρvω²A² ∝ A²f²
• loudness level β = 10 log₁₀(I)/(I₀) dB, I₀ = 10⁻¹² W/m²
• Δβ = 10 log₁₀(I₂)/(I₁)
• n identical incoherent sources: I = nI₁, β rises by 10 log₁₀ n (2 sources: +3 dB)
• valid in a medium with no absorption and a point source radiating equally

How to apply. Use the inverse-square law between two distances, then convert the ratio of intensities into decibels with the base-10 logarithm. Incoherent sources add in intensity; coherent sources add in amplitude.

Example. A 1 W source at 10 m: I = 1 ÷ (4π × 100) = 7.96 × 10⁻⁴ W/m² and β = 10 log(7.96 × 10⁸) = 89 dB. A 1000-fold rise of intensity is 30 dB; doubling is 3 dB.

Superposition and interference

• y = y₁ + y₂ for waves of the same frequency
• A = √(A₁² + A₂² + 2A₁A₂ cos φ) and I = I₁ + I₂ + 2√(I₁I₂) cos φ
• constructive: φ = 2nπ, Δx = nλ
• destructive: φ = (2n + 1)π, Δx = (n + ½)λ
• (I(max))/(I(min)) = ((A₁ + A₂)/(A₁ − A₂))²
• equal sources: I(max) = 4I₀, I(min) = 0
• sources must be coherent (constant phase difference) for a steady pattern

How to apply. Find the phase difference from the path difference (φ = 2πΔx/λ, plus π if one of the waves has been reflected). Then choose the formula. The average intensity is conserved: energy is only redistributed.

Example. Amplitudes in the ratio 3 : 1 give I(max) ÷ I(min) = (4/2)² = 4. Two sources f = 340 Hz (λ = 1 m) in phase: a point at 5 m and 5.5 m from them is in destructive interference (Δx = λ/2).

Beats

• two waves of nearly equal frequencies f₁ and f₂: y = 2A cos(π(f₁ − f₂)t) sin(π(f₁ + f₂)t)
• beat frequency f(b) = |f₁ − f₂|
• amplitude (and loudness) varies at f(b); beat period (1)/(f(b))
• I(max) = 4I₀ and I(min) = 0 for equal amplitudes
• loading a fork (wax) lowers its frequency; filing raises it
• a string: f ∝ √T, so a rise of tension by x% raises f by about x/2 %

How to apply. Use the change of the beats after loading or filing to decide which frequency is greater: if the number of beats per second falls after the unknown fork is loaded, the unknown fork had the higher frequency. The beats cannot be heard when |f₁ − f₂| exceeds about 10 Hz.

Example. A 512 Hz fork and an unknown fork give 4 beats per second; after loading the unknown fork the beats fall to 2: the unknown fork was 516 Hz (and 514 Hz after loading). A 200 Hz string with the tension raised by 2% rises by 2 Hz, giving 2 beats per second.

Standing waves

Standing waves and reflection

• y = 2A sin kx cos ωt (two identical waves opposite directions)
• nodes at x = n(λ)/(2), antinodes at (n + ½)(λ)/(2), distance between neighbouring nodes (λ)/(2)
• amplitude at x = 2A|sin kx|
• no energy is carried along; all points between two nodes are in phase and change phase by π across a node
• a fixed end is a node (reflected wave inverted, phase π), a free end an antinode (no inversion)

How to apply. To find a node, put sin kx = 0; for an antinode, |sin kx| = 1. A boundary condition (node or antinode at the ends) decides the allowed wavelengths. Energy sloshes between kinetic energy (concentrated near the antinodes) and potential energy (concentrated near the nodes, where the string is most stretched).

Example. Waves of A = 0.01 m, k = π m⁻¹ (λ = 2 m): nodes are 1 m apart and the antinode amplitude is 0.02 m. At x = 0.25 m the amplitude is 0.02 × sin(π/4) = 0.0141 m.

Vibrating string fixed at both ends

• λ = (2L)/(n), f(n) = (nv)/(2L) = (n)/(2L)√(T)/(μ) (n = 1, 2, 3, …)
• all harmonics present
• nth harmonic = (n − 1)th overtone
• f ∝ (1)/(L), f ∝ √T, f ∝ (1)/(√μ), same material f ∝ (1)/(Lr) (r radius)
• harmonics have f in ratio 1 : 2 : 3 : …
• n antinodes and n + 1 nodes (including the ends)

How to apply. Use the length of the vibrating part. A finger that stops the string at a point shortens L. The harmonics that can be excited depend on where the string is plucked (a pluck at the centre kills the even harmonics).

Example. L = 0.8 m, μ = 0.01 kg/m and T = 64 N: v = 80 m/s, f₁ = 80 ÷ 1.6 = 50 Hz, f₃ = 150 Hz. Halving the length doubles the frequency to 100 Hz.

Open organ pipe

• antinodes (displacement) at both open ends
• f(n) = (nv)/(2L), n = 1, 2, 3, … all harmonics present
• λ = (2L)/(n)
• the fundamental is v/2L
• overtones: 2f, 3f, …
• the pressure (and density) change is a node at an open end and an antinode at a closed end; in an open pipe the pressure antinode is at the middle

How to apply. Use L as the length of the air column, with the end correction added when the radius is given. Compare with a string of the same length: both have the same series of harmonics.

Example. L = 0.5 m, v = 340 m/s: f₁ = 340 Hz, f₂ = 680 Hz, f₃ = 1020 Hz. Number of harmonics below 2 kHz: 5.

Closed organ pipe

• node at the closed end, antinode at the open end
• f(n) = ((2n − 1)v)/(4L) = odd multiples only: f, 3f, 5f, …
• λ = (4L)/(2n − 1)
• the fundamental is v/4L, half that of an open pipe of the same length
• the harmonics are in the ratio 1 : 3 : 5 : …
• the 2nd overtone is the 5th harmonic

How to apply. For a closed pipe, the length holds an odd number of quarter wavelengths: L = (2n − 1)λ/4. Only odd harmonics exist; there is no 2f or 4f. A pipe closed at one end sounds an octave lower than an open pipe of equal length.

Example. L = 0.5 m, v = 340 m/s: f₁ = 170 Hz, then 510 Hz and 850 Hz. The same open pipe gives 340 Hz.

End correction and the resonance tube

• effective length L + e, e = 0.6r at each open end
• closed pipe: f = (v)/(4(L + 0.6r))
• open pipe: f = (v)/(2(L + 1.2r))
• resonance tube: first resonance L₁ + e = (λ)/(4), second L₂ + e = (3λ)/(4)
• λ = 2(L₂ − L₁), v = 2f(L₂ − L₁), e = (L₂ − 3L₁)/(2)
• v ∝ √T for the air in the tube

How to apply. Using the difference L₂ − L₁ removes the end correction e from the answer. The first resonance is shorter than λ/4 by e; the antinode lies slightly outside the open end.

Example. A closed pipe, L = 0.5 m and r = 0.025 m: f₁ = 340 ÷ (4 × 0.515) = 165 Hz. For a 512 Hz fork with L₁ = 0.16 m and L₂ = 0.50 m: λ = 0.68 m, v = 348 m/s and e = 0.01 m.

Melde’s experiment (string driven by a tuning fork)

• transverse arrangement (prong vibrates perpendicular to the string): f(string) = f(fork) = (p)/(2l)√(T)/(μ) (p loops, length l)
• longitudinal (prong along the string): f(string) = (f(fork))/(2), so f(fork) = (p)/(l)√(T)/(μ)
• for the same l, T ∝ (1)/(p²) so p²T = constant
• μ = mass per unit length

How to apply. Decide the arrangement: in the longitudinal case the string vibrates at half the fork frequency, so for the same string and tension the number of loops is half as large (or, for the same loops, the tension needed is four times less). Use p²T = constant to compare two sets of loops.

Example. Transverse, f = 100 Hz, μ = 0.001 kg/m and T = 0.4 N: v = 20 m/s, so 4 loops need l = 4 × 20 ÷ 200 = 0.4 m. If the tension is raised to 1.6 N, the number of loops falls from 4 to 2 (p²T constant).

Longitudinal vibrations of rods

• v = √(Y)/(ρ)
• rod clamped at its middle with free ends: antinodes at the ends, f = ((2n − 1)v)/(2L), f₁ = (v)/(2L)
• rod clamped at one end (free at the other): like a closed pipe, f = ((2n − 1)v)/(4L)
• rod fixed at both ends: f = (nv)/(2L)

How to apply. Find which ends are free (antinodes) and which are fixed (nodes), count quarter wavelengths, then use v = √(Y/ρ). In Kundt’s tube the rod and the air column vibrate at the same frequency, so v(air) ÷ v(rod) = (node spacing in the air) ÷ (node spacing in the rod).

Example. Aluminium rod, L = 1 m, Y = 7 × 10¹⁰ Pa and ρ = 2700 kg/m³, clamped at the centre: v = 5092 m/s, f₁ = 5092 ÷ 2 = 2546 Hz.

Doppler effect

Doppler effect (general formula)

• take the direction from source S to observer O as positive: f′ = f (v − v(o))/(v − v(s)) with v(o), v(s) the velocities of the observer and source along that direction
• source approaching: f′ = (fv)/(v − v(s)); receding: (fv)/(v + v(s))
• observer approaching: f′ = f(v + v(o))/(v); receding: f(v − v(o))/(v)
• the wavelength is changed only by the motion of the source: λ′ = (v − v(s))/(f)
• v = speed of sound in still air; speeds measured with respect to the medium, v(s) < v

How to apply. Draw the line from the source to the observer, mark the positive direction towards the observer, and give each speed its sign. Approach raises f, recession lowers it. Only components along the line joining them count.

Example. Source 340 Hz approaching a stationary observer at 34 m/s: f′ = 340 × 340 ÷ 306 = 377.8 Hz. Observer approaching a stationary source at 34 m/s: f′ = 340 × 374 ÷ 340 = 374 Hz. Source 20 m/s towards and observer 10 m/s away: 350.6 Hz.

Reflection from a moving surface (echo)

• the reflector first acts as an observer and then as a source
• reflector approaching at u, stationary source and observer: f″ = f(v + u)/(v − u)
• moving away: f″ = f(v − u)/(v + u)
• the beats between the source and the echo f(b) = f″ − f ≈ (2uf)/(v) for u ≪ v
• a source and observer moving together: no Doppler shift (equal speed in the same direction)

How to apply. Do the problem in two steps: the frequency received by the reflector, then the frequency re-emitted. The approach case gives a higher echo frequency.

Example. An ultrasonic wave of 100 kHz reflected from a car approaching at 20 m/s (v = 340 m/s): f″ = 100 × 360 ÷ 320 = 112.5 kHz, so the beat is 12.5 kHz.

Wind and motion at an angle

• wind of speed w along source → observer: replace v by v + w (the medium moves) in all terms: f′ = f (v + w − v(o))/(v + w − v(s))
• if the motion is not along the line of sight, use only the components along the line: v(s) cos θ and v(o) cos θ′
• a source on a circle: the frequency heard by a distant observer varies between f(v)/(v + v(s)) and f(v)/(v − v(s))
• wind does not change the frequency when source and observer are both at rest

How to apply. Wind blowing along the sound adds to the speed of sound in the medium for all the terms. For oblique motion drop the component perpendicular to the line of sight.

Example. A source moving at 40 m/s at 60° to the line to a stationary observer: component 20 m/s, so f′ = f × 340 ÷ 320 = 1.0625f. For a source on a circle at 20 m/s and f = 340 Hz: the observer hears 340 × 340 ÷ 360 = 321 Hz to 340 × 340 ÷ 320 = 361 Hz.

Most asked question types

How do I solve “Reading an SHM equation” questions in JEE Main?

An equation x = A sin(ωt + φ) or x = a sin ωt + b cos ωt is given. Find A, T, φ, or the position and velocity at an instant. 1. Compare with the standard form. 2. Put the time into x and v = Aω cos(ωt + φ). 3. For a sum of sine and cosine A = √(a² + b²). Example: x = 0.1 sin(20πt + π/6) m: A = 0.1 m, T = 0.1 s, x(0) = 0.05 m.

How do I solve “Velocity, acceleration and time between positions” questions in JEE Main?

The speed at a position, the position at a given speed, or the time to go between two points is asked. 1. v = ω√(A² − x²), a = −ω²x. 2. Convert positions to angles on the reference circle. 3. Use the standard times: T/12, T/8, T/6, T/4. Example: T = 2.4 s: from x = 0 to A/2 takes 0.2 s.

How do I solve “Composition of two SHMs” questions in JEE Main?

Two SHMs of the same frequency along one line or at right angles are added. Find the resultant amplitude or the path. 1. Along a line: A² = A₁² + A₂² + 2A₁A₂ cos δ. 2. At right angles: eliminate t to get a line, ellipse or circle. 3. Check δ = 0, π/2 and π. Example: Amplitudes 3 cm and 4 cm at 90°: A = 5 cm.

How do I solve “Energy and KE = PE” questions in JEE Main?

The kinetic or potential energy at a position, the position where they are equal, or an energy fraction is asked. 1. E = ½kA². 2. KE = ½k(A² − x²), PE = ½kx². 3. KE = PE at x = A/√2. Example: A = 0.1 m, k = 20 N/m: E = 0.1 J, and KE = PE = 0.05 J at x = 0.0707 m.

How do I solve “Combinations and cuts of springs” questions in JEE Main?

Springs in series, in parallel or cut into pieces carry a mass. Find k or T. 1. Parallel adds k; series adds 1/k. 2. A part of length fraction f has k/f. 3. T = 2π√(m/k). Example: 100 and 300 N/m: parallel 400, series 75 N/m.

How do I solve “Vertical spring and two masses” questions in JEE Main?

A mass hangs from a spring, or two masses are joined by a spring. Find T, the amplitude or the greatest extension. 1. Static extension Δ = mg/k and T = 2π√(Δ/g). 2. Release from the natural length: A = Δ. 3. Two masses: use the reduced mass. Example: Δ = 0.1 m: T = 0.628 s; extension reaches 0.2 m.

How do I solve “Simple pendulum and effective gravity” questions in JEE Main?

A pendulum is in a lift, a car, an incline, or a field. Find the new period or the time lost. 1. Add the pseudo force to mg to get g(eff). 2. T = 2π√(L/g(eff)). 3. For a clock: time lost = (ΔT/T) × 86400 s. Example: Lift up at 2 m/s²: T′ = 0.913T.

How do I solve “Physical, torsional, floating and U-tube oscillators” questions in JEE Main?

A rod or ring swings about a point, a disc twists on a wire, a cylinder floats, or liquid oscillates in a tube. 1. Find the restoring force or torque. 2. Read ω² as the coefficient of x (or of θ). 3. Use T = 2π√(I/mgd), 2π√(I/C), 2π√(h/g), 2π√(L/2g). Example: Rod of 1 m pivoted at an end: T = 1.62 s.

How do I solve “Damped and forced oscillations” questions in JEE Main?

The amplitude or energy after some time, the time constant, or the resonance condition is asked. 1. A = A₀e^−bt/2m, E = E₀e^−bt/m. 2. After the transient the system oscillates at the driving frequency, not at ω₀. 3. Resonance at ω(d) ≈ ω₀, A = F₀/(bω₀). Example: m = 0.2 kg, b = 0.05 kg/s: the amplitude halves in 5.55 s.

How do I solve “Wave equation and phase difference” questions in JEE Main?

y(x, t) is given. Find A, λ, f, v, the direction, the particle velocity or the phase difference between two points. 1. Compare with y = A sin(kx − ωt). 2. v = ω/k and v(particle) = ∂y/∂t. 3. Δφ = 2πΔx/λ. Example: y = 0.05 sin(2πx − 100πt): v = 50 m/s along +x.

How do I solve “Intensity and the decibel scale” questions in JEE Main?

A source of given power or a change of distance is considered. Find I, the loudness in dB or the new amplitude. 1. I = P/4πr². 2. β = 10 log₁₀(I/I₀). 3. Δβ = 10 log₁₀(I₂/I₁). Example: 1 W at 10 m is 89 dB; doubling the sources adds 3 dB.

How do I solve “Speed of waves in strings, solids and gases” questions in JEE Main?

The speed of a wave on a string, in a rod, a liquid or a gas at a temperature, or a ratio of speeds is asked. 1. String √(T/μ); solid √(Y/ρ); fluid √(B/ρ). 2. Gas √(γRT/M), with T in kelvin. 3. For ratios cancel the constants. Example: T = 100 N, μ = 0.01 kg/m gives v = 100 m/s.

How do I solve “Standing waves on strings and in pipes” questions in JEE Main?

A string with fixed ends, or an open or closed pipe, vibrates. Find the frequencies, harmonics, overtones or the lengths. 1. String and open pipe: f = nv/2L. 2. Closed pipe: f = (2n − 1)v/4L. 3. Count harmonics below a given frequency. Example: L = 0.5 m, v = 340 m/s: open 340 Hz, closed 170 Hz.

How do I solve “End correction, resonance tube and Melde” questions in JEE Main?

A resonance tube gives two lengths, a pipe has a radius, or a string vibrates in loops under a fork. 1. v = 2f(L₂ − L₁). 2. e = 0.6r, e = (L₂ − 3L₁)/2. 3. Melde: p²T = constant. Example: 512 Hz, L₁ = 0.16 m, L₂ = 0.50 m: v = 348 m/s.

How do I solve “Superposition, interference and beats” questions in JEE Main?

Two waves meet. Find the resultant amplitude or intensity, the points of maximum or minimum, or the beat frequency. 1. Phase difference from path difference. 2. A² = A₁² + A₂² + 2A₁A₂ cos φ. 3. Beats f₁ − f₂; loading lowers the frequency. Example: 512 Hz fork with 4 beats; loading reduces it to 2: unknown is 516 Hz.

How do I solve “Doppler effect” questions in JEE Main?

A source and an observer move. Find the observed frequency, the beat with an echo or the speed of a moving body. 1. Take S to O positive. 2. f′ = f(v − v(o))/(v − v(s)). 3. Echo: f″ = f(v + u)/(v − u). Example: 340 Hz source approaching at 34 m/s: 377.8 Hz.

Common myths and the truth

• Myth: The period of a pendulum depends on the mass of the bob. Truth: It depends on L and g(eff) only; the mass cancels (for small angles).
• Myth: Every periodic motion is SHM. Truth: SHM needs a restoring force proportional to the displacement; a bouncing ball is periodic but not SHM.
• Myth: In SHM the acceleration is greatest at the mean position. Truth: It is zero at the mean position and maximum at the extremes; the speed is the opposite.
• Myth: The period of a spring changes on the Moon. Truth: T = 2π√(m/k) does not contain g, so it is the same.
• Myth: KE and PE of an oscillator oscillate with the same frequency as x. Truth: They have frequency 2f, and their sum stays constant.
• Myth: Damping changes the period a lot. Truth: For light damping ω′ is almost ω₀; the amplitude, not the period, decays.
• Myth: A forced oscillator vibrates at its natural frequency. Truth: After the transient it vibrates at the driving frequency; the amplitude is largest when this equals ω₀.
• Myth: A wave carries the medium along with it. Truth: Particles only oscillate about their mean positions; the energy and the pattern move.
• Myth: The speed of sound in air increases with pressure. Truth: At constant temperature it does not, since P/ρ is constant; it rises with √T.
• Myth: Louder sound has a higher frequency. Truth: Loudness depends on intensity (amplitude), pitch on frequency.
• Myth: A standing wave transports energy along the string. Truth: The two travelling waves carry equal energy in opposite directions, so there is no net flow.
• Myth: The Doppler effect depends only on the relative speed of source and observer. Truth: It depends on which of them moves with respect to the medium, so the two cases give different formulas.

Mistakes to avoid

• The period of a spring–mass system does not depend on g, but a pendulum’s does.
• The pendulum formula is valid only for small angles.
• The vertical spring oscillates about the equilibrium, not the natural length.
• Series springs make k smaller; parallel makes it larger.
• Wave speed and particle speed are different quantities.
• Use kelvin in v ∝ √T and the decibel needs log₁₀, not ln.
• A closed pipe has only odd harmonics; the 2nd overtone is the 5th harmonic.
• End correction is added to each open end; the difference L₂ − L₁ removes it.
• Moving source and moving observer are not symmetric in the Doppler formula.

Simulations for Oscillations and Waves

• Coupled pendulums — Join two pendulums with a weak spring. Energy slowly swaps between them in beats.
• Damped harmonic motion — Add velocity-dependent resistance to a spring–mass system.
• Driven oscillations and resonance — Drive a damped oscillator near its natural frequency and watch its steady-state response build in amplitude.
• Simple pendulum — Set the length and release angle to watch a pendulum swing.
• Energy exchange in SHM — Watch kinetic and potential energy trade places as a mass oscillates.
• Displacement, velocity, acceleration — Follow the three SHM curves and see how they line up in phase.
• Mass–spring oscillation — Watch a mass oscillate around a spring’s equilibrium position.
• Oscillating liquid in a U-tube — Displace the liquid in a U-tube and release it. The column oscillates in simple harmonic motion.
• Mass on a vertical spring — Hang a mass on a spring: it settles at a new equilibrium and oscillates about it.
• Beats: two notes, one pulsing sound — Superpose two nearby frequencies and see the amplitude rise and fall in beats.
• Doppler effect — Move a sound source and watch its wavefronts crowd together ahead.
• Open and closed organ pipes — Choose a pipe type and harmonic. See the standing wave of air displacement inside the pipe.
• Pulse reflection at a boundary — Send a pulse between two strings and compare its reflected amplitude and sign.
• Resonance tube: speed of sound — Hold a vibrating tuning fork over a tube whose air column is set by a water level. Find the first and second…
• Ripple tank: two-source interference — Two dippers make circular waves on water. Watch lines of calm water appear where the waves cancel.
• Sonometer: frequency and length — Press a vibrating tuning fork on the sonometer box and move the bridge until the paper rider on the wire…
• Standing waves on a string — Form nodes and antinodes between two fixed ends.
• Wave speed on a stretched rope — Flick a rope and time the pulse. The speed depends on tension and mass per unit length.
• Superposition of two waves — Add two travelling waves and watch the resultant form.
• Travelling wave — Change amplitude, wavelength and frequency to see a wave move through a medium.

More: Class 11 Physics simulations · all JEE Main chapters.