Three harmonic waves having equal frequency v and same intensity I0, have phase angles 0,4π and −4π respectively. When they are superimposed the intensity of the resultant wave is close to:
(A)3I0
(B)I0
(C)0.2I0
(D)5.8I0✓ Correct
Answer: (D)
Solution
Treat each wave as a phasor of the same length A, with I0∝A2. The three phasors point at 0, +4π and −4π; the two tilted ones have equal and opposite vertical parts, so they add along the horizontal to 2Acos4π=A2. The resultant amplitude is therefore A(1+2), and the resultant intensity is (1+2)2I0=(3+22)I0≈5.8I0.
JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option