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Physics · Simple harmonic motion (S.H.M.) and its equation, phase

Three simple harmonic motions in the same direction having the same amplitude a

Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by \( 45^{\circ}, \) then : This question has multiple correct options

  • A. The resultant amplitude is \( (1+\sqrt{2}) a \)
  • B. The phase of the resultant motion relative to the first is \( 90^{\circ} \)
  • C. The energy associated with the resulting motion is \( (3+2 \sqrt{2}) \) times the energy associated with any single motion
  • D. The resulting motion is not simple harmonic

Step-by-step solution

The three SHMs have phases 0°, 45°, and 90° relative to the first. Using phasor addition, the components are a(1+1/√2) each, so resultant amplitude = a√2(1+1/√2) = a(1+√2). The phase is 45°, not 90°, so B is false. Energy is proportional to square of amplitude, so the ratio is (1+√2)² = 3+2√2, making C also correct. The superposition of same-frequency SHMs yields SHM, so D is false. Thus options A and C are correct, but the single best choice is A as it directly gives the amplitude.
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