📝 Problem Types
JEENEETCBSE6 problem archetypes. Each solved with examiner intent, concept selection, step-by-step solution, and a shortcut insight. This is where rank separation happens.
Strategy Tip — How to Use This Page
For each problem: (1) Read the Given. (2) Cover the solution. (3) Try to solve yourself. (4) Reveal solution. (5) Compare your approach with the shortcut. This is the active recall method that improves retention by 60%.
📋 Given
A tuning fork of frequency 256 Hz produces 4 beats per second with another fork. When wax is applied to the second fork, the beats reduce to 2 per second. Find the original frequency of the second fork.
🎓 What Examiner Tests
Understanding that loading with wax DECREASES frequency. Using beat logic to determine whether unknown frequency is higher or lower.
Thinking Step
4 beats → unknown = 260 or 252 Hz. After wax: beats = 2. If unknown = 260 Hz, wax makes it 258 Hz → beats with 256 = 2 ✅. If unknown = 252, wax makes it ~250 → beats = 6 ❌. So unknown = 260 Hz.
Identify possibilities
Beat frequency = |f₁ − f₂| = 4. So f₂ = 256 ± 4 = 260 or 252 Hz
Apply wax logic
Wax ↓ mass → ↓ frequency. If f₂ = 260: after wax, f₂ decreases toward 256 → beats decrease from 4 → confirms f₂ = 260 Hz
Verify
After wax: beats = 2. So f₂_waxed = 256 + 2 = 258 Hz. Consistent with f₂ starting at 260 Hz. ✓
Answer
f₂ = 260 Hz
⚡ Shortcut Insight
Wax → f decreases. If beats DECREASE after wax → unknown was ABOVE reference. If beats INCREASE after wax → unknown was BELOW reference. Two-line logic, no algebra needed.
📋 Given
The displacement of a wave is y = 5 sin(4x − 8t + π/3) where x in m, t in s, y in cm. Find: (a) amplitude, (b) wavelength, (c) frequency, (d) wave speed, (e) direction of propagation.
🎓 What Examiner Tests
Ability to decode wave equation. This is a direct identity question that requires pattern recognition, not calculation.
Standard form
y = A sin(kx − ωt + φ). Compare: A = 5 cm, k = 4 rad/m, ω = 8 rad/s, φ = π/3
Extract each quantity
(a) A = 5 cm
(b) λ = 2π/k = 2π/4 = π/2 ≈ 1.57 m
(c) f = ω/2π = 8/2π = 4/π ≈ 1.27 Hz
(d) v = ω/k = 8/4 = 2 m/s
(e) Coefficient of x is +k, coefficient of t is −ω → wave moves in +x direction
⚡ Shortcut Insight
Wave speed = coeff(t)/coeff(x) = 8/4 = 2 m/s. Always positive. Direction: if (kx − ωt), wave moves in +x. If (kx + ωt), wave moves in −x. Takes 5 seconds to identify if you've internalized the pattern.
📋 Given
An open organ pipe of length 0.5 m and a closed organ pipe are in resonance with the 2nd harmonic of the open pipe. Speed of sound = 340 m/s. Find: (a) frequency of 2nd harmonic of open pipe, (b) length of the closed pipe in this resonance condition.
🎓 What Examiner Tests
Multi-concept application: harmonics of open pipe AND matching with closed pipe harmonics. Students must know which harmonic of the closed pipe matches.
Thinking Step
Open pipe 2nd harmonic → f₂ = 2v/2L. Closed pipe supports only ODD harmonics. The closed pipe must resonate at the same frequency. Find which odd harmonic n of closed pipe gives the same f. Then L_closed = (2n−1)v/4f.
Open pipe 2nd harmonic
f₂ = 2v/(2L) = v/L = 340/0.5 = 680 Hz
Closed pipe resonant frequency
Must equal 680 Hz: (2n−1)v/(4L_c) = 680
Simplest case: n=1 (fundamental of closed)
L_c = (2×1−1)×340/(4×680) = 340/2720 = 0.125 m = 12.5 cm
Verify: other harmonics of closed pipe
n=2: L_c = 3×340/(4×680) = 0.375 m. n=3: L_c = 5×340/(4×680) = 0.625 m. All valid — the question asks for the simplest (smallest), so L_c = 0.125 m.
⚡ Shortcut Insight
For the same frequency: open pipe L_open = nₐv/2f and closed pipe L_closed = (2n_c−1)v/4f. Ratio: L_open/L_closed = 2nₐ/(2n_c−1). For 2nd harmonic open (nₐ=2) and fundamental closed (n_c=1): L_open/L_closed = 4/1 → L_closed = 0.5/4 = 0.125 m ✓
📋 Given
A transverse wave travels in +x direction. At t=0, a y–x snapshot shows: at x=0, y=0 and slope ∂y/∂x is negative. Speed of wave = 2 m/s, ω = 4π rad/s. Find the particle velocity at x=0 at t=0.
🎓 What Examiner Tests
Relationship between particle velocity and slope of y–x graph. v_particle = −v_wave × (∂y/∂x). Most students apply this formula without understanding the sign.
Thinking Step
Particle velocity v_y = ∂y/∂t. From wave eqn: ∂y/∂t = −v_wave × ∂y/∂x. If wave moves in +x and slope is negative → particle velocity is positive (upward). This is the key physical insight.
Wave equation
y = A sin(kx − ωt). k = ω/v = 4π/2 = 2π rad/m. Amplitude A from graph (let's say readable).
Particle velocity relation
v_y = ∂y/∂t = −Aω cos(kx − ωt) = −v_wave × ∂y/∂x
Apply at x=0, t=0
∂y/∂x at (0,0) is negative (given). v_wave = +2 m/s.
v_y = −(+2) × (negative) = positive
Conclusion
Particle at x=0 is moving in +y direction (upward). Magnitude = v_wave × |slope|.
⚡ Shortcut Insight
Visual method: if wave moves RIGHT, a particle on the "rising slope" of the wave moves DOWN (it will be at trough next). A particle on the "falling slope" moves UP. Negative slope + rightward wave = particle moving UP. No formula needed if you visualize correctly.
📋 Given
Assertion (A): In a standing sound wave, the displacement nodes are pressure antinodes and vice versa.
Reason (R): This is because displacement and pressure variation are 90° out of phase in a sound wave.
🎓 What Examiner Tests
Understanding of the phase relationship between displacement and pressure in sound waves. This requires deeper conceptual clarity, not formula application.
Assertion Analysis
TRUE. At displacement nodes (particles don't move), the medium is compressed/expanded maximally → pressure change is maximum → pressure antinode.
Reason Analysis
TRUE. The pressure wave in sound lags displacement by π/2 (90°). So pressure antinodes appear where displacement nodes are, and vice versa.
Is R the correct explanation of A?
YES — the 90° phase relationship directly explains why nodes of displacement correspond to antinodes of pressure.
Answer
Both A and R are true, and R is the correct explanation of A. → Option (A)
⚡ Shortcut Insight
For sound standing waves: Displacement and pressure are complementary — where one is max, the other is zero. This is because pressure = −B(∂y/∂x): differentiating a sine gives cosine (90° shift). Memorize: DISPLACEMENT NODE = PRESSURE ANTINODE.
📋 Given
A train moving at 20 m/s approaches a stationary observer while blowing its whistle at 800 Hz. Speed of sound = 340 m/s.
(a) What frequency does the observer hear as the train approaches?
(b) What frequency does the observer hear after the train passes?
(c) What is the apparent change in frequency?
🎓 What Examiner Tests
Doppler formula application with changing sign convention for approach vs recession. Tests if student knows to change denominator signs only (source is moving, observer is stationary).
Setup Doppler formula
f' = f₀ × (v ± v_o)/(v ∓ v_s). Observer stationary → v_o = 0. Source (train) moving.
(a) Train approaching
Source moves toward → use (v − v_s) in denominator
f'_approach = 800 × 340/(340 − 20) = 800 × 340/320 = 850 Hz
(b) Train receding
Source moves away → use (v + v_s) in denominator
f'_recede = 800 × 340/(340 + 20) = 800 × 340/360 = 755.6 Hz ≈ 756 Hz
(c) Change in frequency
Δf = 850 − 756 = 94 Hz (the "pitch drop" you hear as train passes)
⚡ Shortcut Insight
For approaching source only: f' = f₀v/(v−v_s). For receding source only: f' = f₀v/(v+v_s). The ratio f_approach/f_recede = (v+v_s)/(v−v_s). This ratio appears directly in JEE integer-type questions. With v_s = 20, ratio = 360/320 = 1.125.
📋 Given
A wire of length 1 m, mass 10 g is stretched between two supports under tension 40 N. An organ pipe (open at both ends) of length 1 m is placed nearby. Speed of sound = 320 m/s. Do the string and pipe resonate at any common frequency? If yes, find it.
🎓 What Examiner Tests
Finding common frequencies in two different systems. Requires expressing both frequency series and finding intersection (LCM-style reasoning).
Thinking Step
String: f_n = n × f₁_string. Open pipe: f_m = m × f₁_pipe. Common resonance when n×f₁_string = m×f₁_pipe → n/m = f₁_pipe/f₁_string. Find the simplest integer ratio.
String fundamental frequency
μ = m/L = 0.01/1 = 0.01 kg/m. v_string = √(T/μ) = √(40/0.01) = √4000 = 63.25 m/s
f₁_string = v_string/(2L) = 63.25/(2×1) ≈ 31.6 Hz
Open pipe fundamental frequency
f₁_pipe = v_sound/(2L) = 320/(2×1) = 160 Hz
Find common frequencies
Ratio: f₁_pipe/f₁_string = 160/31.6 ≈ 5.06 ≈ 5. So 5th harmonic of string ≈ fundamental of pipe.
f₅_string = 5 × 31.6 = 158 Hz ≈ 160 Hz (small discrepancy due to approx μ)
Exact answer
For exact resonance: n_string × f₁_string = m_pipe × f₁_pipe → smallest common is at 160 Hz (1st harmonic of pipe = 5th harmonic of string if μ is adjusted). Both resonate at f = 160 Hz approximately.
⚡ Shortcut Insight
JEE Advanced loves this type. Find f₁ for both systems, then find LCM of the frequency series. If f₁_A = 32 Hz and f₁_B = 160 Hz, then 5th harmonic of A = fundamental of B. Draw a frequency ladder — first overlap point = resonance.
Common Mistake Alert
Students calculate v_string and confuse it with v_sound. These are DIFFERENT systems. The string vibration frequency must match the sound frequency in air. Don't mix v_string with v_sound in the same formula.