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MHT-CET Physics · Superposition of Waves

Superposition of Two Waves

Two waves of the same frequency add into one of the same frequency whose amplitude depends on their phase difference: R² = a₁² + a₂² + 2a₁a₂cos φ — largest when in step, smallest when opposite.

Why this matters

8 PYQs, one HARD. Two shapes: the resultant amplitude for a given phase difference (or the phase difference for a given amplitude), and the resultant intensity, with the extra care needed when one wave is written as a sine and the other as a cosine.

Concept 1 of 2: The Resultant Amplitude

Draw each wave's amplitude as an arrow at its phase angle and add the arrows. In phase they add straight up; opposite they subtract; at 90° they form a right triangle; at 120°, two equal arrows make a third of the same length.

Definition

  • R2=a12+a22+2a1a2cos⁡ϕR^2 = a_1^2 + a_2^2 + 2a_1a_2\cos\phi.
  • ϕ=0\phi = 0: a1+a2a_1 + a_2; ϕ=π\phi = \pi: ∣a1−a2∣|a_1 - a_2|; ϕ=π2\phi = \dfrac{\pi}{2}: a12+a22\sqrt{a_1^2 + a_2^2}.
  • Equal amplitudes aa: R=2acos⁡ϕ2R = 2a\cos\dfrac{\phi}{2}; R=aR = a when ϕ=120∘\phi = 120^\circ, i.e. cos⁡ϕ=−12\cos\phi = -\dfrac{1}{2}.
  • b1sin⁡ωt±b2cos⁡ωtb_1\sin\omega t \pm b_2\cos\omega t is a 90∘90^\circ pair: R=b12+b22R = \sqrt{b_1^2 + b_2^2} either sign.

Resultant amplitude

R2=a12+a22+2a1a2cos⁡ϕR^2 = a_1^2 + a_2^2 + 2a_1a_2\cos\phi

Worked example

Waves of amplitude 3 and 4 units superpose. Resultant amplitude in phase, at 90∘90^\circ, and in opposition?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q36Moderate

Example 1 · Superposition of Waves · Superposition, Phase/Path Difference, and Interference

Two sound waves each of wavelength 'λ\lambda' and having the same amplitude 'A' from two sources 'S1S_1' and 'S2S_2' interfere at a point P. If the path difference, S2P−S1P=λ/3S_2P - S_1P = \lambda/3 then the amplitude of resultant wave at point 'P' will be [cos⁡120°=−0.5\cos 120° = -0.5]

Adding amplitudes at 90°

Two waves a quarter-cycle apart do not give 2A; they give 2A\sqrt{2}A. Only waves in step add their amplitudes directly.

Concept 2 of 2: Resultant Intensity and Mixed Sine–Cosine Waves

Intensity goes as amplitude squared, so the intensity rule is the amplitude rule with the square roots of the intensities. For equal sources it becomes 4I₀cos²(φ/2). Before using any of it, write both waves as sines — a cosine is a sine a quarter-cycle ahead.

Definition

  • I=I1+I2+2I1I2cos⁡ϕI = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi; equal sources: I∝cos⁡2ϕ2I \propto \cos^2\dfrac{\phi}{2}.
  • Sources I and 4I at ϕ=π\phi = \pi: I+4I−4I=II + 4I - 4I = I.
  • acos⁡(θ+ϕ)=asin⁡(θ+ϕ+π2)a\cos(\theta + \phi) = a\sin\left(\theta + \phi + \dfrac{\pi}{2}\right): phase difference ϕ+π2\phi + \dfrac{\pi}{2}, path difference λ2π(ϕ+π2)\dfrac{\lambda}{2\pi}\left(\phi + \dfrac{\pi}{2}\right).

Resultant intensity

I=I1+I2+2I1I2cos⁡ϕI = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi

Worked example

Sources of intensity I and 9I meet with phase difference π3\dfrac{\pi}{3}. Resultant intensity?
Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q13Moderate

Example 2 · Superposition of Waves · Superposition, Phase/Path Difference, and Interference

Considering interference between two sources of intensities 'I' and '4I', the intensity at a point where the phase difference is π\pi is (cos⁡π=−1)(\cos\pi = -1)

Reading φ off a sine–cosine pair

a1sin⁡(ωt−kx)a_1\sin(\omega t - kx) and a2cos⁡(ωt−kx+ϕ)a_2\cos(\omega t - kx + \phi) differ by ϕ+π2\phi + \frac{\pi}{2}, not ϕ\phi. Convert the cosine first.

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)

Watch out for (2)

Test yourself on Superposition of Waves

20 past MHT-CET questions from this chapter, timed at 18 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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